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Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 5 Gravitation Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 5 Gravitation

Question 1.
Mention some main characteristics of gravitational force.
Answer:
Characteristics of gravitational force:

1. Every massive object in the universe experiences gravitational force.
2. It is the force of mutual attraction between any two objects by virtue of their masses.
3. It is always an attractive force with infinite range.
4. It does not depend upon the intervening medium.
5. It is much weaker than other fundamental forces. Gravitational force is 10 times weaker than strong nuclear force.

Question 2.
State and explain Kepler’s law of orbits.
Answer:
Statement:
All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.
Explanation:

1. The figure M shows the orbit of a around the planet P Sun.
   
2. Here, S and S’ are the foci of the ellipse and the Sun is situated at S.
3. P is the closest point along the orbit from S and is called perihelion.
4. A is the farthest point from S and is called aphelion.
5. PA is the major axis whose length is 2a. PO and AO are the semimajor axes with lengths ‘a’ each.
   MN is the minor axis whose length is 2b. MO and ON are the semiminor axes with lengths ‘b’ each.

Question 3.
State and prove Kepler’s law of equal areas.
Answer:
Statement:

The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
Explanation:
i)Kepler observed that planets move faster when they are nearer to the Sun while they move slower when they are farther from the Sun.

ii) Suppose the Sun is at the origin. The position of planet is denoted by \(\vec{r}\) and its momentum is denoted by \(\vec{p}\) (component ⊥ \(\vec{r}\)).

vi) For central force the angular momentum is conserved. Hence, from equations (4) and (5),
\(\frac{\overrightarrow{\Delta \mathrm{A}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{L}}}{2 \mathrm{~m}}\) = constant
This proves the law of areas.

Question 4.
What is a central force?
Answer:
A central force on an object is a force which is always directed along the line joining the position of object and a fired point usually taken to the origin of the coordinate system.

Question 5.
State and explain Kepler’s law of periods.
Answer:
Statement:
The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.
Explanation:
If r is length of semi major axis then, this law states that.
T² × r³ or \(\frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}\) = constant

Solved Examples

Question 6.
What would be the average duration of year if the distance between the Sun and the Earth becomes
i) thrice the present distance.
ii) twice the present distance.
Solution:
i) Consider r₁ be the present distance between the Earth and Sun
We know, T₁ = 365 days.
When the distance is made thrice, r₂ = 3r₁
According to Kepler’s law of period,

i) The duration of the year would be 1897 days when distance is made thrice.
ii) The duration of the year would be 1032 days when distance is made twice.

Question 7.
What would have been the duration of the year if the distance between the Earth and the Sun were half the present distance?
Solution:
Given: r₂ = \(\frac{\mathrm{r}_{1}}{2}\)
∴ \(\frac{r_{2}}{r_{1}}=\frac{1}{2}\)
To find. Time period (T₂)
Formula: \(\left(\frac{T_{2}}{T_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}\)
Calculation: We know. T₁ = 365 days
From formula,

= 365 × 0.3536
= 129.1
T₂ = 129.1 days
The duration of the year would be 129.1 days.

Question 8.
Calculate the period of revolution of Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of the Earth’s orbit is 5.
(Period of revolution of the Earth is 1 year.)
Solution:
Given: \(\frac{\mathrm{r}_{\mathrm{J}}}{\mathrm{r}_{\mathrm{E}}}=\frac{5}{\mathrm{l}}\), T_E = 1 year
To find: Period of revolution of Jupiter (T_J)
Formula: T² ∝ r³
Calculation: From formula,
\(\left(\frac{T_{J}}{T_{E}}\right)^{2}=\left(\frac{r_{J}}{r_{E}}\right)^{3}\)
∴ T_J = 5^3/2
= 5 x \(\sqrt {5}\)
= 11.18 years.
Period of revolution of Jupiter around the Sun is 11.18 years.

Question 9.
A Saturn sear is 29.5 times the Earth’s year. How far is the Saturn from the Sun if the Earth is 1.50 × 10⁸ km away from the Sun?
Solution:
Given: T_S = 29.5 T_E,
r_E = 1.50 × 10⁸ km
To find: Distance of Saturn form the Sun (r_S)
Formula: T² ∝ r³
Calculation: From formula,

∴ r_S = 14.32 × 10⁸ km
Saturn is 14.32 × 10⁸ km away from the Sun.

Question 10.
The distances of two planets from the Sun are 10¹³ m and 10¹² m respectively. Find the ratio of time periods of the two planets.
Solution:
Given: r₁ = 10¹³ m, r₂ = 10¹² m

Question 11.
Heavy and light objects are released from same height near the Earth’s surface. What can we conclude about their acceleration?
Answer:
Heavy and light objects, when released from the same height, fall towards the Earth at the same speed., i.e., they have the same acceleration.

Question 12.
Explain how Newton concluded that gravitational force F ∝ = \(\frac{\mathrm{Mm}}{\mathrm{r}^{2}}\)
Answer:
Before generalising and stating universal law of gravitation, Newton first studied the motion of moon around the Earth.

1. The known facts about the moon were,
   • the time period of revolution of moon around the Earth (T) = 27.3 days.
   • distance between the Earth and the moon (r) = 3.85 × 10⁵ km.
   • the moon revolves around the Earth in almost circular orbit with constant angular velocity ω.
2. Thus, the centripetal force experienced by moon (directed towards the centre of the Earth) is given by,
   F = mrω² …………… (1)
   Where, m = mass of the moon
3. From Newton’s laws of motion,
   F = ma
   ∴ a = rω² ……………… (2)
4. 
5. This is the acceleration of the moon directed towards the centre of the Earth.
6. This acceleration is much smaller than the acceleration felt by bodies near the surface of the Earth (while falling on Earth).
7. The value of acceleration due to Earth’s gravity at the surface is 9.8 m/s².
   
8. Newton therefore concluded that the acceleration of an object towards the Earth is inversely proportional to the square of distance of object from the centre of the Earth.
   ∴ a ∝ \(\frac{1}{r^{2}}\)
   x. As, F = ma
   Therefore, the force exerted by the Earth on an object of mass m at a distance r from it is
   F ∝ \(\frac{\mathrm{m}}{\mathrm{r}^{2}}\)
   Similarly, an object also exerts a force on the Earth which is
   F_E ∝ \(\frac{\mathrm{M}}{\mathrm{r}^{2}}\)
   Where M is the mass of the Earth. .
9. According to Newton’s third law of motion, F = F_E. Thus, F is also proportional to the mass of the Earth. From these observations, Newton concluded that the gravitational force between the Earth and an object of mass m is F ∝ \(\frac{\mathrm{Mm}}{\mathrm{r}^{2}}\)

Question 13.
Discuss the vector form of gravitational force between two masses with the help of diagram.
Answer:

1. Consider two point masses m₁ and m₂ having position vectors \(\overrightarrow{\mathrm{r}_{1}}\) and \(\overrightarrow{\mathrm{r}_{2}}\) from origin O respectively as shown in figure.
   
2. The position vector of mass m₂ with respect to m₁ is given by, \(\vec{r}_{2}-\vec{r}_{1}=\vec{r}_{21}\)
3. Similarly, position vector of mass m₁ with respect to m₂ is, \(\overrightarrow{\mathrm{r}}_{1}-\overrightarrow{\mathrm{r}_{2}}=\overrightarrow{\mathrm{r}_{12}}\)
4. Let \(\left|\overrightarrow{\mathrm{r}_{12}}\right|=\left|\overrightarrow{\mathrm{r}_{21}}\right|\) Then, the force acting on mass m₂ due to mass m₁ will be given as,
   \(\overrightarrow{\mathrm{F}_{21}}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\left(-\hat{\mathrm{r}}_{21}\right)\)
   where, \(\hat{\mathbf{r}}_{21}\) is the unit vector from m₁ to m₂.
   The force \(\overrightarrow{\mathrm{F}_{21}}\) is directed from m₂ to m₁.
5. Similarly, force experienced by m₁ due to m₂ is given as, \(\overrightarrow{\mathrm{F}}_{12}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\left(-\hat{\mathrm{r}}_{12}\right)\)
   ∴ \(\overrightarrow{\mathrm{F}}_{12}=-\overrightarrow{\mathrm{F}}_{21}\)
   [Note: As \(\hat{\mathbf{r}}_{21}\) is defined as unit vector from m₁ to m₂, conceptually force \(\overrightarrow{\mathrm{F}}_{21}\) is directed from m₂ lo m₁.]

Question 14.
Why Is the law of gravitation known as universal law of gravitation?
Answer:
The law of gravitation is applicable to all material objects in the universe. Hence it is known as the universal law of gravitation.

Question 15.
Give formula for the gravitational force due to a collection of masses and represent it diagrammatically.
Answer:

For a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by all the other point masses.
For n particles, force on i^th mass \(\overrightarrow{\mathrm{F}}_{\mathrm{i}}=\sum_{\mathrm{j}=1 \atop \mathrm{j} \neq \mathrm{i}}^{n} \overrightarrow{\mathrm{F}}_{\mathrm{ij}}\)
where, \(\vec{F}_{\mathrm{ij}}\) is the force on i^th particle due to j^th particle.

Question 16.
Discuss qualitative idea for the gravitational force of attraction due to a hollow, thin spherical shell of uniform density on a point mass situated inside it.
Answer:

1. Let us consider the case when the point mass A, is at the centre of the hollow thin shell.
2. In this case as every point on the shell is equidistant from A, all points exert force of equal magnitude on A but the directions of these forces are different.
3. Consider the forces on A due to two diametrically opposite points on the shell.
4. The forces on A due to them will be of equal magnitude but will be in opposite directions and will cancel each other.
5. Thus, forces due to all pairs of points diametrically opposite to each other will cancel and there will be no net force on A due to the shell.
6. When the point object is situated elsewhere inside the shell, the situation is not symmetric. However, gravitational force varies directly with mass and inversely with square of the distance.
7. When the point object is situated elsewhere inside the shell, the situation is not symmetric. However, gravitational force varies directly with mass and inversely with square of the distance.
8. Thus, some part of the shell may be closer to point A, but its mass is less. Remaining part will then have larger mass but its centre of mass is away from A.
9. In this way, mathematically it can be shown that the net gravitational force on A is still zero, so long as it is inside the shell.
10. Hence, the gravitational force at any point inside any hollow closed object of any shape is zero.

Question 17.
Discuss qualitative idea for the gravitational force of attraction between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside the shell / sphere.
Answer:

• Gravitational force caused by different regions of shell can be resolved into components along the line joining the point mass to the centre and along a direction perpendicular to this line.
• The components perpendicular to this line cancel each other and the resultant force remains along the line joining the point to the centre.
• Mathematical calculations show that, this resultant force on this point equals the force that would get exerted by the shell whose entire mass is situated at its centre.

Solved Problems

Question 18.
The gravitational force between two bodies is 1 N. If distance between them is doubled, what will be the gravitational force between them?
Solution:
Let m₁ and m₂ be masses of the given two bodies. If they are r distance apart initially, then the force between them will be,

The force between two bodies reduces to 0.25 N.

Question 19.
Calculate the force of attraction between two metal spheres each of mass 90 kg, if the distance between their centres is 40 cm. (G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: m₁ = 90 kg, m₂ = 90 kg,
r = 40 cm = 40 × 10^-2 m.
G = 6.67 × 10^-11 N m²/kg²
To find: Force of attraction (F)
Formula: F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
Calculation.
From formula,
F = \(\frac{6.67 \times 10^{-11} \times 90 \times 90}{\left(40 \times 10^{-2}\right)^{2}}\)
= \(\frac{6.67 \times 8100}{1600} \times 10^{-7}\)
∴ F = 3.377 × 10^-6 N
The force of attraction between the two metal spheres is 3.377 × 10^-6 N.

Question 20.
Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = 1. A fourth particle D is kept on the perpendicular bisector of AC at a distance ¡ from B. Determine the gravitational force on D.
Solution:

From figure,
distance of particle D, from particles A and C is.
ADCD = \(\sqrt{\mathrm{AB}^{2}+\mathrm{BD}^{2}}\)
= \(\sqrt{(l)^{2}+(l)^{2}}=l \sqrt{2}\)
Gravitational force on particle D is the vector sum of forces due to particles A, B, and C. Gravitational force due to A,
F_A = \(\frac{\mathrm{Gmm}}{(l \sqrt{2})^{2}}=\frac{\mathrm{Gm}^{2}}{2 l^{2}}\) = along \(\overrightarrow{\mathrm{DA}}\)
This force can he resolved into horizontal and vertical components using rectangular unit vectors as shown in figure.
From figure,


Negative sign indicates net force is acting along DB.
The net force acting on particle D is \(\frac{\mathbf{G m}^{2}}{l^{2}}\left(\frac{1}{\sqrt{2}}+1\right)\) directed along \(\overrightarrow{\mathbf{D B}}\).

[Note: When force, in given case is resolved into its components, its horizontal component contains cos argument while vertical component contains sine argument.]

Question 21.
Three balls of masses 5 kg each are kept at points P(1, 2, 0) Q(2, 3, 0) and R(2, 2, 0). Find the gravitational force exerted on the ball at point R.
Solution:
The net force acting on ball placed at R will be vector sum of forces acting due to balls at P and Q.

= 2.358 × 10^-9 N
The net force acting on the ball at point R will be 2.358 × 10^-9 N.

Question 22.
For what purpose Cavendish balance is used?
Answer:
Cavendish balance is used to find the magnitude of the gravitational constant G.

Question 23.
Describe the construction of Cavendish balance with the help of neat labelled diagram.
Answer:

1. The Cavendish balance consists of a light rigid rod. It is supported at the centre by a fine vertical metallic fibre about 100 cm long.
2. Two small spheres, s₁ and s₂ of lead having equal mass m and diameter about 5 cm are mounted at the ends of the rod and a small mirror M is fastened to the metallic fibre as shown in figure.
   
3. The mirror can be used to reflect a beam of light onto a scale and thereby measure the angel through which the wire will be twisted.
4. Two large lead spheres L₁ and L₂ of equal mass M and diameter of about 20 cm are kept close to the small spheres on opposite side as shown in figure.

Question 24.
Describe the working of the experiment performed to measure the value of gravitational constant.
Answer:

1. In the experiment performed to find the magnitude of gravitational constant (G), Cavendish balance is used.
2. The large spheres in the balance attract the nearby smaller spheres by equal and opposite force \(\overrightarrow{\mathrm{F}}\). Hence, a torque is generated without exerting any net force on the bar.
3. Due to the torque the bar turns and the suspension wire gets twisted till the restoring torque due to the elastic property of the wire becomes equal to the gravitational torque.
4. If r is the initial distance of separation between the centres of the large and the neighbouring small sphere, then the magnitude of the force between them is, F = G\(\frac{\mathrm{mM}}{\mathrm{r}^{2}}\)
5. If length of the rod is L, then the magnitude of the torque arising out of these forces is
   τ = FL = G\(\frac{\mathrm{mM}}{\mathrm{r}^{2}}\)L
6. At equilibrium, it is equal and opposite to the restoring torque.
   ∴ G\(\frac{\mathrm{mM}}{\mathrm{r}^{2}}\)L = Kθ
   Where, K is the restoring torque per unit angle and θ is the angle of twist.
7. By knowing the values of torque τ₁ it and corresponding angle of twist a, the restoring torque per unit twist can be determined as K = τ₁/α.
8. Thus, in actual experiment measuring θ and knowing values of τ, m, M and r, the value of G can be calculated from equation (2).

Question 25.
Derive the expression for the acceleration due to gravity on the surface of the Earth.
Answer:

1. The Earth is an extended object and can be assumed to be a uniform sphere.
2. If the mass of the Earth is M and that of any point object is m, the distance of the point object from the centre of the Earth is r then the force of attraction between them is given by,
   F = \(\text { G } \frac{M m}{r^{2}}\) …. (1)
3. If the point object is not acted upon by any other force, it will be accelerated towards the centre of the Earth under the action of this force. Its acceleration can be calculated by using Newton’s second law,
   F = ma … (2)
4. From equations (1) and (2),
   ma = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
   ∴ Acceleration due to the gravity of the Earth
   \(=\frac{\mathrm{GMm}}{\mathrm{r}^{2}} \times \frac{1}{\mathrm{~m}}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}\)
   This is denoted by g.
5. If the object is close to the surface of the Earth, r ≈ R, then,
   g_{Earth’s surface} = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 26.
Explain why the Earth doesn’t appear to move even though the object of mass m (m << M) kept on the Earth exerts equal and opposite gravitational force on it.
Answer:

1. An object of mass m (much smaller than the mass of the Earth) is attracted towards the Earth and falls on it.
2. At the same time, the Earth is also attracted by the equal and opposite force towards the mass m.
3. However, its acceleration towards m will be,
   
4. As m << M, a_Earth << g and is nearly zero. As, a result, practically only the mass m moves towards the Earth and the Earth doesn’t appear to move.

Solved Examples

Question 27.
Calculate mass of the Earth from given data, Acceleration due to gravity g = 9.81 m/s², Radius of the Earth R_E = 6.37 × 10⁶ m, G = 6.67 × 10^-11 N m²/kg²
Solution:
Given: g = 9.81 m/s², R_E = 6.37 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²
To find: Mass of the Earth (M_E)
Formula: g = \(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^{2}}\)
Calculation: From formula,

= antilog {0.9912 + 0.8041 + 0.8041 – 0.8241) × 10²³
= antilog {1.7753} × 10²³
= 59.61 × 10²³
= 5.961 × 10²⁴ kg
Mass of the Earth is 5.961 × 10²⁴ kg.

Question 28.
Calculate the acceleration due to gravity at the surface of the Earth from the given data. (Mass of the Earth = 6 × 10²⁴ kg, Radius of the Earth = 6.4 × 10⁶ m, G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given. M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10^-11 N m²/kg²
Tofind. Acceleration due to gravity (g)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,

The acceleration due to gravity at the surface of the Earth is 9.77 m/s².

Question 29.
Calculate the acceleration due to gravity on the surface of moon if mass of the moon is 1/80 times that of the Earth and diameter of the moon is 1/4 times that of the Earth (g = 9.8 m/s²)
Solution:
Given: M_m = \(\frac{\mathrm{M}_{\mathrm{E}}}{80}\), R_m = \(\frac{\mathrm{R}_{\mathrm{E}}}{4}\), g = 9.8 m/s²
To find: Acceleration due to gravity on the surface of moon (g_m)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: For moon, from formula,

Acceleration due to gravity on surface of the planet is 0.245 m/s².

Question 30.
Find the acceleration due to gravity on a planet that is 10 times as massive as the Earth and with radius 20 times of the radius of the Earth (g = 9.8 m/s²).
Solution:
Given: M_P = 10 × Mass of the Earth = 10 M_E,
R_P = 20 × radius of the Earth = 20 R_E, g = 9.8 m/s²
To find: Acceleration due to gravity on surface of the planet (g_P)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,

Acceleration due to gravity on surface of the planet is 0.245 m/s².

Question 31.
Acceleration due to gravity on the Earth is g. A planet has mass and radius half that of the Earth. How much will be percentage change in the acceleration due to gravity on the planet?
Solution:

The percentage change in acceleration due to gravity between the planet and the Earth will be 100%.

Question 32.
Explain the graph showing variation of acceleration due to gravity with altitude and depth.
Answer:
The value of acceleration due to gravity is calculated to be maximum at the surface of the Earth. The value goes on decreasing with
i) increase in depth below the Earth’s surface. [varies linearly with (R – d) = r]

ii) increase in height above the Earth’s surface. [varies inversely with (R + h)² = r²].

Graph of g, as a function of r, the distance from the centre of the Earth, is plotted as shown in figure.
For r< R,
g_d = g\(\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
if r = R – d,
g(r) = g\(\left(\frac{r}{R}\right) \Rightarrow g(r) \propto r\)
Hence, the graph shows a straight line passing through origin and having slope \(\frac{\mathrm{g}}{\mathrm{R}}\).

which is represented in the graph.

Question 33.
Why does the weight of a body of a finite mass m is zero at the centre of the Earth?
Answer:
Acceleration at depth d due to gravity is given
by,
g_d = g\(\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
As, at centre of the Earth d = R ⇒ g_d = 0.
Hence, the weight of a body of a finite mass m is zero at the centre of the Earth.

Question 34.
Discuss the variation of acceleration due to gravity at poles and equator due to latitude of the Earth.
Answer:

1. Effective acceleration due to gravity at P is given as,
   g’ = g – Rω²cos²θ.
2. As the value of θ increases, cos θ decreases. Therefore g’ will increase as we move away from equator towards any pole due to the rotation of the Earth.
3. At equator θ = 0°
   ∴ cos θ = 1
   ∴ g’_e = g – Rω²
   The effective acceleration due to gravity (g’_e) is minimum at equator, as here it is reduced by Rω²
4. At poles θ = 90° cos θ = 0
   ∴ g’_p = g – Rω² cos θ
   = g – 0
   = g
   There is no reduction in acceleration due to gravity at poles, due to the rotation of the Earth as the poles are lying on the axis of rotation and do not revolve.

Question 35.
If g = 9.8 m/s² on the surface of the Earth, find its value at h = \(\frac{\mathbf{R}}{\mathbf{2}}\) from the surface of the Earth.
Solution:
Given: g = 9.8 m/s², h = \(\frac{\mathrm{R}}{2}\)
To find: Acceleration due to gravity (g_h)
Formula: \(\frac{\mathrm{g}_{\mathrm{h}}}{\mathrm{g}}\) = \(\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}\)
Calculation:
From formula,

At h = \(\frac{\mathrm{R}}{2}\) from the surface of the Earth, the value of g is 4.35 m/s².

Question 36.
At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface? (Radius of Earth = 6400 km)
Solution:
Given: g_h = 90% of g i.e., \(\frac{g_{h}}{g}\) = 0.9,
R = 6400 km = 6.4 × 10⁶ m
To find: Distance above the surface of the Earth (h)

Question 37.
Find the altitude at which the acceleration due to gravity is 25% of that at the surface of the Earth.
(Radius of the Earth = 6400 km)
Solution:
Given: g_h = 25% of g = \(\frac{25}{100} \times \mathrm{g}=\frac{\mathrm{g}}{4}\), R = 6400 km = 6.4 × 10⁶ m
To find: Height (h)
Formula: g_h = g\(\left(\frac{R}{R+h}\right)^{2}\)
Calculation: From formula,
\(\frac{\mathrm{g}}{4}\) = g\(\left(\frac{R}{R+h}\right)^{2}\)
(R + h)² = 4R²
R + h = 2R
∴ h = 2R – R
∴ h = R
∴ h = 6400 km

Question 38.
A hole is drilled half way to the centre of the Earth. A body is dropped into the hole. How much will it weigh at the bottom of the hole if the weight of the body on the Earth’s surface is 350 N?
Solution:
Given: W = mg = 350 N, d = \(\frac{\mathrm{R}}{2}\)
To find: Weight at certain depth (W_d)
Formula: g_d = \(\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]\)
Calculation: Since W_d = mg_d,
from formula,

Question 39.
Assuming the Earth to be a homogeneous sphere, determine the density of the Earth from following data. (g = 9.8 m/s², G = 6.673 × 10^-11 N m²/kg², R = 6400 km)
Solution:
Given g = 9.8 m/s²,
G = 6.673 × 10^-11 N m²/kg², R = 6400 km = 6.4 × 10⁶ m
To find: Density (ρ)
Formula: g = \(\frac{4}{3} \pi \mathrm{R} \rho \mathrm{G}\)
Calculation: From formula,

Question 40.
If the Earth were a perfect sphere of radius 6.4 × 10⁶ m rotating about its axis with the period of one day (8.64 × 10⁴ s), what is the difference in acceleration due to gravity from poles to equator?
Solution:
Given: R = 6.4 × 10⁶ m, T = 8.64 × 10⁴ s
To find: Difference in acceleration due to gravity (g_P – g_E)
Formula: g’ = g – Rω² cos²θ
Calculation: Since ω = \(\frac{2 \pi}{\mathrm{T}}\)
∴ ω = \(\frac{2 \times 3.14}{8.64 \times 10^{4}}\) = \(\frac{6.28}{8.64 \times 10^{4}}\)
= 0.7268 × 10^-4
ω = 7.268 × 10^-5 rad/s
At poles, θ = 90°
From formula,
g_P = g – Rω²cos² (90°)
= g – 0 ….(∵ cos 90° = 0)
∴ g_P = g …. (i)
At equator, θ = 0°,
∴ g_E = g – Rω²cos² 0°
g_E = g – Rω² …. (ii)
Subtracting equation (ii) from equation (i), we have,
g_P – g_E = g – (g – Rω²)
∴ g_P – g_E = Rω²
= 6.4 × 10⁶ × (7.268 × 10^-5)²
= 6.4 × 10⁶ × 52.82 × 10^-10
= 338 × 10^-4
∴ g_P – g_E = 3.38 × 10^-2 m/s²

Question 41.
The Earth is rotating with angular velocity ω about its own axis. R is the radius of the Earth. If Rω² = 0.03386 m/s², calculate the weight of a body of mass 100 gram at latitude 25°. (g = 9.8 m/s²)
Solution:
Given: Rω² = 0.03386 m/s², θ = 25°,
m = 0.1 kg, g = 9.8 m/s²
To find: Weight (W)

Formulae:
i) g’ = g – Rω² cos² θ
ii) W = mg

Calculation:
From formula (i),
g’ = 9.8 – [0.03386 – cos² (25°)]
∴ g’ = 9.8 – [0.03386 × (0.9063)²]
∴ g’ = 9.8 – 0.02781
∴ g’ = 9.772 m/s²
From formula (ii),
W = 0.1 × 9.772
∴ W = 0.9772 N

Question 42.
If the angular speed of the Earth is 7.26 × 10^-5 rad/s and radius of the Earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.
Solution:
Given: R = 6.4 × 10⁶ m, ω = 7.26 × 10^-5 rad/s
To find: Change in weight (∆W)
Formulae:
i) ∆g = g_p – g_eq = Rω²
ii) ∆W = m∆g

Calculation: From formula (i) and (ii),
∆W = m(Rω²)
= 1 × 6.4 × 10⁶ × (7.26 × 10^-5)²
= 3373 × 10^-5 N

Question 43.
Define potential energy.
Answer:
Potential energy is the work done against conservative force (or forces) in achieving a certain position or configuration of a given system.

Question 44.
Explain with examples the universal law which states that “Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy”.
Answer:
Example 1:

1. A spring in its natural state, possesses minimum potential energy. Whenever we stretch it or compress it, we perform work against the conservative force.
2. Due to this work, the relative distances between the particles of the system change i.e., configuration changes and potential energy of the spring increases.
3. The spring finally regains its original configuration of minimum potential energy on removal of the applied force.
   This explains that the spring always try to rearrange itself in order to attain minimum potential energy.

Example 2:

1. When an object is lying on the surface of the Earth, the system of that object and the Earth has minimum potential energy.
2. This is the gravitational potential energy of the system as these two are bound by the gravitational force. While lifting the object to some height, we do work against the conservative gravitational force.
3. In its new position, the object is at rest due to balanced forces. However, now, the object has a capacity to acquire kinetic energy, when allowed to fall.
4. This increase in the capacity is the potential energy gained by the system. The object falls on the Earth to achieve the configuration of minimum potential energy on dropping it from the new position.

Question 45.
Obtain an expression for change in gravitational potential energy of any object displaced from one point to another.
Answer:
i) Work done against gravitational force in displacing an object through a small displacement, stored in the system in the form of increased potential energy of the system.
∴ dU = –\(\overrightarrow{\mathrm{F}}_{\mathrm{g}} \cdot \overrightarrow{\mathrm{dr}}\)
Negative sign appears because dU is the work done against the gravitational force \(\overrightarrow{\mathrm{F}_{\mathrm{g}}}\).

ii) For displacement of the object from an initial position \(\overrightarrow{\mathrm{r}_{\mathrm{i}}}\) to the final position \(\overrightarrow{\mathrm{r}_{\mathrm{f}}}\), the change in potential energy ∆U, can be obtained by integrating dU.

 

iii) Gravitational force of the Earth, \(\overrightarrow{\mathrm{F}}_{\mathrm{g}}\) = –\(\frac{\mathrm{GMm}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\) where \(\hat{r}\) is the unit vector in the direction of \(\overrightarrow{\mathrm{r}}\).
Negative sign appears here because \(\overrightarrow{\mathrm{F}_{\mathrm{g}}}\) is directed towards centre of the Earth and opposite to \(\overrightarrow{\mathrm{r}}\).

iv) For Earth and mass system,

Hence, change in potential energy corresponds to the work done against conservative forces.

Question 46.
Using expression for change in potential energy, show that gravitational potential energy of the system of object of mass m and the Earth with separation of r is, –\(\frac{\text { GMm }}{\text { r }}\)
Answer:

1. Change in P.E. for a system of Earth and mass is given by,
   ∆U = GMm\(\left(\frac{1}{r_{i}}-\frac{1}{r_{f}}\right)\)
2. For gravitational force, point of zero potential energy is taken to be at r = ∞.
3. Hence, U(r_i) = 0 at r_i = ∞
   Final point r_f is the point where the potential energy of the system is to be determined.
4. At r_f = r
   
   This is gravitational potential energy of the system of object of mass m and Earth of mass M having separation r (between their centres of mass).

Question 47.
Derive the formula for increase in gravitational potential energy of a Earth – mass system when the mass is lifted to a height h provided h << R.
Answer:

1. If the object is on the surface of Earth, r = R
   U₁ = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
   If the object is lifted to height h above the surface of Earth, the potential energy becomes _ GMm 12 ~~ R+h
   U₂ = –\(\frac{G M m}{R+h}\)
2. Increase in the potential energy is given by
   
3. If g is acceleration due to gravity on the surface of Earth. GM = gR²
   ∴ ∆U = mgh\(\left(\frac{R}{R+h}\right)\) … (1)
4. Equation (1) gives the work to be done to raise an object of mass rn to a height h, above the surface of the Earth.
5. If h << R, we can use R + h ≈ R.
   ∴ ∆U = mgh
   Thus, mgh is increase in the gravitational potential energy of the Earth – mass system if an object of mass m is lifted to a height h, provided h << R.

Question 48.
Write a short note on gravitational potential.
Answer:
The gravitational potential energy of the system of Earth and any mass m at a distance r from the centre of the Earth is given by,

The factor –\(\frac{\mathrm{GM}}{\mathrm{r}}\) = (V_E)_r is defined as the
gravitational potential of Earth at distance r from its centre.

As the potential depends only upon mass of the Earth and location of the object, it is same for any mass m bound to the Earth.

Question 49.
Explain the relation between the gravitational potential energy and the gravitational potential.
Answer:

1. In terms of potential, we can write the potential energy of the Earth-mass system as, Gravitational potential energy (U) = Gravitational potential (V_r) × mass (m)
2. Thus, gravitational potential is gravitational potential energy per unit mass.
   ∴ V_r = \(\frac{\mathrm{U}}{\mathrm{m}}\)
3. For any conservative force field, the concept of potential can be defined on similar lines.
4. Gravitational potential difference between any two points in gravitational field can be written as,
   V₂ – V₁ = \(\frac{U_{2}-U_{1}}{m}\)
   = \(\frac{\mathrm{dW}}{\mathrm{m}}\)
   This is the work done (or change in potential energy) per unit mass.
5. Therefore, in general, for a system of any two masses m₁ and m₂, separated by distance r, we can write,
   U = –\(\frac{\mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}}\)
   = (V₁)m₂
   = (V₂)m₁
   Here V₁ and V₂ are gravitational potentials at r due to m₁ and m₂ respectively.

Solved Exmaples

Question 50.
What will be the change in potential energy of a body of mass m when it is raised from height R_E above the Earth’s surface to 5/2 R_E above the Earth’s surface? R_E and M_E are the radius and mass of the Earth respectively.
Solution:
Change in potential energy (P.E.) of a body of mass m is given by,
∆U = GM_Em\(\left(\frac{1}{r_{i}}-\frac{1}{r_{f}}\right)\)
Here, r_i = R_E + R_E = 2 R_E

[Note: Answer calculated above is in accordance with textual methods of calculation.]

Question 51.
What will be the change in potential energy of a body of mass m when it is placed on the surface of the Earth from height R above the Earth’s surface?
Solution:
Change in potential energy (P.E.) of a body of mass m is given by,
∆U = GM_Em\(\frac{1}{r_{i}}-\frac{1}{r_{r}}\)
Here, r_i = R + R = 2R
Similarly, r_f = R
∴ ∆U = GM_Em \(\left[\frac{1}{2 R}-\frac{1}{R}\right]\) = GM_Em\(\left(-\frac{1}{2 R}\right)\)
∴ ∆U = –\(\frac{\mathbf{G} \mathbf{M} \mathbf{m}}{\mathbf{2 R}}\)
Negative sign indicates that potential energy is decreasing.

Question 52.
Determine the gravitational potential of a body of mass 80 kg whose gravitational potential energy is 5 × 10⁹ J on the surface of the Earth.
Solution:
Given: m = 80 kg, U = 5 × 10⁹ J
To find: Gravitational potential (V)
Formula: V = \(\frac{\mathrm{U}}{\mathrm{m}}\)
Calculation: From formula,
V = \(\frac{5 \times 10^{9}}{80}=\frac{25}{4}\) × 10⁷
= 6.25 × 10⁷ J kg^-1
Potential of the body at the surface of the Earth is 6.25 × 10⁷ J kg^-1.

Question 53.
Calculate the escape velocity of a body from the surface of the Earth.
(Average density of Earth = 5.5 × 10³ kg/m³, G = 6.67 × 10^-11 N m²/kg², radius of Earth R = 6.4 × 10⁶ m)
Solution:
Given: ρ = 5.5 × 10³ kg/m³, R = 6.4 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²
To find: Escape velocity (v_e)

= 2 × 6.4 × 10⁶ × 8.759 × 10^-4
∴ v_e = 11.21 × 10³ m/s = 11.21 km/s
The escape velocity of a body is 11.21 km/s.

Question 54.
What is a satellite?
Answer:
An object which revolves in an orbit around a planet is called as satellite.
Example:

• Moon is a natural satellite of the Earth.
• INSAT is an artificial satellite of the Earth.

Question 55.
Write a short note on Polar satellites.
Answer:

1. Polar Satellites are placed in lower polar orbits.
2. They are at low altitude 500 km to 800 km.
3. Period of revolution of polar satellite is nearly 85 minutes, so it can orbit the Earth 16 time per day.
4. They go around the poles of the Earth in a north-south direction while the Earth rotates in an east-west direction about its own axis.
5. The polar satellites have cameras fixed on them. The camera can view small stipes of the Earth in one orbit. In entire day the whole Earth can be viewed strip by strip.
6. Polar region and equatorial regions close to it can be viewed by these satellites.
7. Polar satellites are used for weather forecasting and meteorological purpose. They are also used for astronomical observations and study of Solar radiations.

Question 56.
Derive the expression for the critical velocity of a satellite revolving close to the surface of the Earth in terms of acceleration due to gravity.
Answer:

1. When the satellite is revolving close to the surface of the Earth, the height is very small as compared to the radius of the Earth.
2. Hence the height can be neglected and radius of the orbit is nearly equal to R, i.e., R + h ≈ R
3. The critical speed of the satellite then becomes,
   v_c = \(\sqrt{\frac{G M}{R}}\)
4. G is related to acceleration due to gravity by the relation,
   g = \(\frac{G M}{R^{2}}\)
   ∴ GM = gR²
5. Thus, critical speed in terms of acceleration due to gravity, neglecting the air resistance, can be obtained as,
   v_c = \(\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}\)

Question 57.
From an inertial frame of reference, explain the apparent weight for a person standing in a lift having zero acceleration.
Answer:

1. A passenger inside a lift experiences only two forces:
   • Gravitational force mg directed vertically downwards and
   • normal reaction force N directed vertically upwards, exerted by the floor of the lift.
2. As these forces are oppositely directed, the net force in the downward direction will be F = ma – N.
3. Though the weight of a passenger is the gravitational force acting upon it, the person experiences his weight only due to the normal reaction force N exerted by the floor.
4. A lift has zero acceleration when the lift is at rest or is moving upwards or downwards with constant velocity.
5. Thus, a net force acting on the passenger inside the lift will be,
   F = 0 = mg – N
   ∴ mg = N
   Hence, in this case the passenger feels his normal weight mg.

Question 58.
What happens to the apparent weight of the person inside the lift moving with net upward acceleration?
Answer:

1. The lift is said to be moving with net upward acceleration in two possible conditions:
   • when the lift just starts moving upwards or
   • is about to stop at a lower floor during its downward motion.
2. As the net acceleration is upwards, the upward force must be greater.
   ∴ F = ma = N – mg
   ∴ N = mg + ma
   ∴ N > mg
3. Thus, for a passenger inside this lift, his apparent weight is more than his actual weight when the lift was not accelerated.

Question 59.
Why does a passenger feel lighter when the lift is about to stop at a higher floor during its upward motion?
Answer:

1. When the lift is about to stop at a higher floor during its upward motion it has a net downward acceleration.
2. As the net acceleration is downwards, the downward force must be greater.
   ∴ F = ma = mg – N
   ∴ N = mg – ma
   i.e., N < mg
   Hence, a passenger feels lighter when the lift is about to stop at a higher floor during its upward motion.

Question 60.
When does a weighing machine will record zero for a passenger in a lift?
Answer:
If the cables of the lift are cut, the downward acceleration of the lift, a_d = g. In this case, we get,
N = mg – ma_d = 0
Thus, there will not be any feeling of weight and the weighing machine will record zero.

Question 61.
Define time period of a satellite.
Obtain an expression for the period of a satellite in a circular orbit round the Earth. Show that the square of the period of revolution of a satellite is directly proportional to the cube of the orbital radius.
Answer:
Definition:
The time taken by a satellite to complete one revolution around the Earth is called its time period.
Expression for time period:
i) Consider, m = mass of satellite, h = altitude of satellite. Thus, r = R + h = radius of orbit of the satellite.
ii) In one revolution, distance traced by satellite is equal to circumference of its circular orbit.
iii) If T is the time period of satellite, then

Since π², G and M are constant,
∴ T² ∝ r³
Hence, square of the period of revolution of a satellite is directly proportional to the cube of the radius of its orbit.

vi) Taking square roots on both the sides of equation (4), we get,
T = 2π\(\sqrt{\frac{\mathrm{r}^{3}}{\mathrm{GM}}}\)
T = 2π\(\sqrt{\frac{(R+h)^{3}}{G M}}\)
This is the required expression for period of satellite orbiting around the Earth in circular path.

Question 62.
For an orbiting satellite very close to surface of the Earth, show that T = 2π \(\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\).
Answer:

1. Time period of an orbiting satellite at certain height is given by, T = 2π \(\sqrt{\frac{(R+h)^{3}}{G M}}\)
2. If satellite is orbiting very close to the Earth’s surface, then h ≈ 0
   

Solved Examples

Question 63.
Show that the critical velocity of a body revolving in a circular orbit very close to the surface of a planet of radius R and
mean density ρ is 2R\(\sqrt{\frac{G \pi \rho}{3}}\).
Solution:
Since the body is revolving very close to the surface of a planet,
∴ h << R
R = radius of planet
ρ = mean density of planet
Critical velocity of a body very close to Earth is given by,

Question 64.
Find the orbital speed of the satellite w hen satellite is revolving round the Earth in circular orbit at a distance 9 × 10⁶ m from its centre. (Given: Mass of Earth = 6 × 10²⁴ kg, G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: r = 9 × 10⁶ m, M = 6 × 10²⁴kg,
G = 6.67 × 10^-11 N m²/kg²
To find: Orbital speed (v_c)
Formula: v_c = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\)
Calculation: From formula,

∴ v_c = 6.668 × 10³ m/s
The orbital speed of the satellite is 6.668 × 103 m/s.

Question 65.
Taking radius of the Earth as 6400 km and g at the Earth’s surface as 9.8 m/s², calculate the speed of revolution of a satellite orbiting close to the Earth’s surface.
Solution:
Given: R = 6400 km = 6.4 × 10⁶ m, g = 9.8 m/s²
To find: Critical velocity (v_c)
Formula: v_c = \(\sqrt{\mathrm{gR}}\)
Calculation: From formula,
v_c = \(\sqrt{9.8 \times 6.4 \times 10^{6}}\)
= \(\sqrt{98 \times 64 \times 10^{4}}\)
= 7\(\sqrt{2}\) × 8 ×10²
= 7.92 × 10³ m/s
The speed of revolution of the satellite orbiting close to the Earth’s surface is 7.92 × 10³ m/s.

Question 66.
The critical velocity of a satellite revolving around the Earth is 10 km/s at a height where g_h = 8 m/s ². Calculate the height of the satellite from the surface of the Earth. (R = 6.4 × 10⁶ m)
Solution:
Given: v_c = 10 km/s = 10 × 10³ m/s,
g_h = 8 m/s³, R = 6.4 × 10⁶ m
To find: Height of the satellite (h)
Formula: v_c = \(\sqrt{g_{\mathrm{h}}(R+h)}\)
Calculation: From formula,
10 × 10³ = \(\sqrt{8 \times(\mathrm{R}+\mathrm{h})}\)
Squaring both the sides, we get,
100 × 10⁶ = 8(R + h)
∴ 8(R + h) = 100 × 10⁶
∴ R + h = \(\frac{100}{8}\) × 10⁶
∴ h = 12.5 × 10⁶ – R
= 12.5 × 10⁶ – 6.4 × 10⁶
= 6.1 × 10⁶m
∴ h = 6100 km
The height of the satellite from the surface of the Earth is 6100 km.

Question 67.
An artificial satellite revolves around a planet in circular orbit close to its surface. Obtain the formula for period of the satellite in terms of density p and radius R of planet.
Solution:
Time period of a satellite revolving around the planet at certain height is given by,

Question 68.
Calculate the period of revolution of a polar satellite orbiting close to the surface of the Earth. Given R = 6400 km, g = 9.8 m/s².
Solution:
Given: For satellite close to Earth surface,
R + h ≈ R
R = 6400 km = 6.4 × 10⁶ m, g = 9.8 m/s²
To find: Time period of satellite (T)
Formula: T = 2π \(\sqrt{\frac{R}{g}}\)
Calculation: From Formula,
T = 2 × 3.14 × \(\sqrt{\frac{6.4 \times 10^{6}}{9.8}}\)
= 6.28 × 8.081 × 10²
= 5.075 × 10³ sec
≈ 85 min
The time period of satellite very close to the Earth’s surface is nearly 85 minute.

Question 69.
A satellite orbits around the Earth at a height equal to R of the Earth. Find its period. (R = 6.4 × 10⁶ m, g = 9.8 m/s²)
Solution:
Given: h = R = 6.4 × 10⁶m, g = 9.8 m/s²
To find: Time period (T)

The time period of the satellite is 1.435 × 10⁴ s.

Question 70.
Calculate the height of the communication satellite. (Given: G = 6.67 × 10^-11 N m²/kg², M = 6 × 10²⁴ kg, R = 6400 km)
Solution:
For communication satellite, T = 24 × 60 × 60 s,
Given: M = 6 × 10²⁴ kg,
G = 6.67 × 10^-11 N m²/kg²,
R = 6400 km = 6.4 × 10⁶m
To find: Height (h)

The height of the communication satellite is 35910 km.

Question 71.
How will you ‘weigh the Sun’, that is estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 × 10⁸ km.
Solution:
Given: r = 1.5 × 10⁸ × 10³m,
T = 365 days = 365 × 24 × 60 × 60 s
To find: Mass (M)
Formula: T = 2π \(\sqrt{\frac{r^{3}}{G M}}\)
Calculation:
From formula,
M = \(\frac{4 \pi^{2} r^{3}}{G T^{2}}\)
= \(\frac{4 \times(3.14)^{2}\left(1.5 \times 10^{11}\right)^{3}}{\left(6.67 \times 10^{-11}\right)(365 \times 24 \times 60 \times 60)^{2}}\)
∴ M = 2.01 × 10³⁰kg
The mass of the Sun is 2.01 × 10³⁰ kg.
[Trick: To ‘weigh the Sun’, i.e., estimate its mass, one needs to know the period of one of its planets and the radius of the planetary orbit.]

Question 72.
Calculate the B.E. of a satellite of mass 2000 kg moving in an orbit very close to the surface of the Earth. (G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: m = 2 × 10³ kg, R = 6.4 × 10⁶ m,
R = 6.4 × 10⁶ m, M = 6 × 10²⁴kg,
G = 6.67 × 10^-11 N m²/kg²,
M = 6 × 10²⁴ kg
To find: Binding Energy (B.E.)
Formula: For satellite very close to Earth,
B.E. = \(\frac{1}{2} \times \frac{\mathrm{GMm}}{\mathrm{R}}\)
Calculation: From formula,
B.E. = \(\frac{1}{2} \times \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{3}}{6.4 \times 10^{6}}\)
∴ B.E. = 6.25 × 10¹⁰ joule
The binding energy of the satellite is 6.25 × 10¹⁰ joule.

Question 73.
Find the binding energy of a body of mass 50 kg at rest on the surface of the Earth. (Given: G = 6.67 × 10^-11 N m²/kg², R = 6400 km, M = 6 × 10²⁴ kg)
Solution:
Given: G = 6.67 × 10^-11 N m²/kg²,
R = 6400 km = 6.4 × 10⁶m,
M = 6 × 10²⁴ kg, m = 50 kg
To find: Binding energy (B.E.)
Formula: B.E. = \(\frac{\text { GMm }}{\mathrm{R}}\)
Calculation: From formula,
B.E. = \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 50}{6.4 \times 10^{6}}\)
= \(\frac{2001}{6.4}\) × 10⁷
∴ B.E. = 3.127 × 10⁹ J
The binding energy of the body 3.127 × 10⁹ J.

Question 74.
Find the total energy and binding energy of an artificial satellite of mass 1000 kg orbiting at height of 1600 km above the Earth’s surface.
(Given: G = 6.67 × 10^-11 N m²/kg², R = 6400 km, M = 6 × 10²⁴ kg)
Solution:
Given: h = 1600 km = 1.6 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²,
R = 6400 km = 6.4 × 10⁶m,
m = 1000 kg, M = 6 × 10²⁴kg
To find:
i) Total Energy (T.E.)
ii) Binding Energy (B.E.)
Formulae: i. T.E. = –\(\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
ii. B.E. = -T.E.

Calculation: From formula (i),
T.E = – \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2(6.4+1.6) \times 10^{6}}\)
= –\(\frac{40020 \times 10^{7}}{2 \times 8}\)
∴ T.E. = -2.501 × 10¹⁰J
From formula (ii),
B.E. = 2.501 × 10¹⁰ J
i) The total energy of the artificial satellite is -2.501 × 10¹⁰ J.
ii) The binding energy of the artificial satellite is 2.501 × 10¹⁰ J.

Question 75.
Determine the binding energy of satellite of mass 1000 kg revolving in a circular orbit around the Earth when it is close to the surface of Earth. Hence find kinetic energy and potential energy of the satellite. (Mass of Earth = 6 × 10²⁴ kg, radius of Earth = 6400 km; gravitational constant G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: m = 1000 kg, M = 6 × 10²⁴ kg,
R = 6400 km, G = 6.67 × 10^-11 N m²/kg²
To find:
i) Binding Energy (B.E.)
ii)Kinetic Energy (K.E.)
iii) Potential Energy (P.E.)

Formulae: For satellite very close to Earth,
i) B.E. = \(\frac{1}{2} \times \frac{\mathrm{GMm}}{\mathrm{R}}\)
ii) K.E.= B.E.
iii) P.E. = -2 K.E.

Calculation: From formula (i),
B.E. = \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 6.4 \times 10^{6}}\)
∴ B.E. = 3.1265 × 10¹⁰ J
From formula (ii),
K.E. = 3.1265 × 10¹⁰
∴ K.E. = 3.1265 × 10¹⁰ J
From formula (iii),
P.E. = -2(3.1265 × 10¹⁰)
∴ P.E. = -6.2530 × l0¹⁰J
i) The binding energy of the satellite is 3.1265 × 10¹⁰ J.
ii) The kinetic energy of the satellite is 3.1265 × 10¹⁰ J.
iii) The potential energy of the satellite is -6.2530 × 10¹⁰J.

Apply Your Knowledge

Question 76.
How are Kepler’s law of periods and Newton’s law of gravitation related?
Answer:
Consider a planet of mass m revolving around the Sun of mass M in a circular orbit.
Let,
r = radius of the circular orbit of the planet.
T = Time period of revolution of planet around the Sun.
ω = angular velocity of planet.
F = Centripetal force exerted by the Sun on the planet.
Centripetal force is given by,
F = mrω²;
But ω = \(\frac{2 \pi}{T}\)
∴ F = mr (\(\frac{2 \pi}{\mathrm{T}}\))²
∴ F = \(\frac{4 \pi^{2} \mathrm{mr}}{\mathrm{T}^{2}}\) …(i)
According to Kepler’s third law,
T² ∝ r³
T² = Kr³ ……….. (where, K = constant) (ii)
Substituting equation (ii) in equation (i),
F = \(\frac{4 \pi^{2} \mathrm{mr}}{\mathrm{Kr}^{3}}\)
∴ F = \(\frac{4 \pi^{2}}{\mathrm{~K}} \frac{\mathrm{m}}{\mathrm{r}^{2}}\)
∴ F ∝ \(\frac{\mathrm{m}}{\mathrm{r}^{2}}\) ….(∵ \(\frac{4 \pi^{2}}{\mathrm{~K}}\) is a constant quantity)
Since, the gravitational attraction between the Sun and the planet is mutual, force exerted by the planet on the Sun will be proportional to the mass M of the Sun.
∴ F ∝ \(\frac{\mathrm{Mm}}{\mathrm{r}^{2}}\)
∴ F = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
The above equation represents Newton’s law of gravitation. In this manner, Newton’s law of gravitation is derived from Kepler’s law of periods.

Question 77.
Represent graphically the variation of total energy, kinetic energy and potential energy of a satellite with its distance from the centre of the Earth.
Answer:
For a satellite,
Potential energy (U) = \(\)
Kinetic energy (K) = \(\) and
Total energy (E) = \(\), where, r = R + h
Also, U and E remain negative whereas K remains positive.
Hence, the graph will be:

Multiple Choice Questions

Question 1.
Kepler’s law of equal areas is an outcome of
(A) conservation of energy
(B) conservation of linear momentum
(C) conservation of angular momentum
(D) conservation of mass
Answer:
(C) conservation of angular momentum

Question 2.
Amongst given statements, choose the correct statement.
(I) Kepler derived the laws of planetary motion.
(II) Newton provided the reason behind the laws of planetary motion.
(A) (I) is correct.
(B) (II) is correct.
(C) Both (I) and (II) are correct.
(D) Neither (I) nor (II) is correct.
Answer:
(B) (II) is correct.

Question 3.
The figure shows the motion of a planet satellite in terms of mean density of Earth. around the Sun in an elliptical orbit with Sun at the focus. The shaded areas A and B are also shown in the figure which can be assumed to be equal. If t₁ and t₂ represent the time for the planet to move from a to b and d to c respectively, then

(A) t₁ < t₂
(B) t₁ >t₂
(C) t₁ = t₂
(D) t₁ ≤ t₂
Answer:
(C) t₁ = t₂

Question 4.
A planet is revolving around a star in a circular orbit of radius R with a period T. If the gravitational force between the planet and the star is proportional to R^-3/2, then
(A) T^{2 ∝} R^5/2
(B) T² ∝ R^-7/2
(C) T² ∝ R^3/2
(D) T² ∝ R⁴
Answer:
(A) T² ∝ R^5/2

Question 5.
Time period of revolution of a satellite around a planet of radius R is T. Period of revolution around another planet whose radius is 3R is
(A) T
(B) 9T
(C) 3T
(D) 3\(\sqrt{3}\) T
Answer:
(D) 3\(\sqrt{3}\) T

Question 6.
Newton’s law of gravitation is called universal law because
(A) force is always attractive.
(B) it is applicable to lighter and heavier bodies.
(C) it is applicable at all times,
(D) it is applicable at all places of universe for all distances between all particles.
Answer:
(D) it is applicable at all places of universe for all distances between all particles.

Question 7.
If the mass of a body is M on the surface of the Earth, the mass of the same body on the surface of the moon is
M
(A) 6M
(B) \(\frac{M}{6}\)
(C) M
(D) Zero
Answer:
(C) M

Question 8.
Which of the following statements about the gravitational constant is true?
(A) It has no units.
(B) It has same value in all systems of units.
(C) It is a force.
(D) It does not depend upon the nature of medium in which the bodies lie.
Answer:
(D) It does not depend upon the nature of medium in which the bodies lie.

Question 9.
The gravitational force between two bodies is ______
(A) attractive at large distance only
(B) attractive at small distance only
(C) repulsive at small distance only
(D) attractive at all distances large or small
Answer:
(D) attractive at all distances large or small

Question 10.
Mass of a particle at the centre of the Earth is
(A) infinite.
(B) zero.
(C) same as at other places.
(D) greater than at the poles.
Answer:
(C) same as at other places.

Question 11.
Which of the following is not a property of gravitational force?
(A) It is an attractive force.
(B) It acts along the line joining the two bodies.
(C) The forces exerted by two bodies on each other form an action-reaction pair.
(D) It has a very finite range of action.
Answer:
(D) It has a very finite range of action.

Question 12.
If the distance between Sun and Earth is made two third times of the present value, then gravitational force between them will become
(A) \(\frac{4}{9}\)times
(B) \(\frac{2}{3}\)times
(C) \(\frac{1}{3}\)times
(D) \(\frac{9}{4}\) times
Answer:
(D) \(\frac{9}{4}\) times

Question 13.
The gravitational constant G is equal to 6.67 × 10^-11 N m²/kg² in vacuum. Its value in a dense matter of density 10¹⁰ g/cm³ will be
(A) 6.67 × 10^-1 N m²/kg²
(B) 6.67 × 10^-11 N m²/kg²
(C) 6.67 × 10^-10 N m²/kg²
(D) 6.67 × 10^-21 N m²/kg²
Answer:
(B) 6.67 × 10^-11 N m²/kg²

Question 14.
Acceleration due to gravity above the Earth’s surface at a height equal to the radius of the Earth is ______
(A) 2.5 m/s²
(B) 5 m/s²
(C) 9.8 m/s²
(D) 10 m/s²
Answer:
(A) 2.5 m/s²

Question 15.
If R is the radius of the Earth and g is the acceleration due to gravity on the Earth’s surface, the mean density of the Earth is

Answer:
(D) \(\frac{3 \mathrm{~g}}{4 \pi \mathrm{RG}}\)

Question 16.
Variation of acceleration due to gravity (g) with distance x from the centre of the Earth is best represented by (R → Radius of the Earth)

Answer:
(D)

Question 17.
Which of the following statements is not correct for the decrease in the value of acceleration due to gravity?
(A) As we go down from the surtce of the Earth towards its centre.
(B) As we go up from the surface of the Earth.
(C) As we go from equator to the poles on the surface on the Earth.
(D) As the rotational velocity of the Earth is increased.
Answer:
(C) As we go from equator to the poles on the surface on the Earth.

Question 18.
Calculate angular velocity of Earth so that acceleration due to gravity at 60° latitude becomes zero. (Radius of Earth = 6400 km, gravitational acceleration at poles = 10 m/s², cos60° = 0.5)
(A) 7.8 × 10^-2 rad/s
(B) 0.5 × 10^-3 radis
(C) 1 × 10^-3 radis
(D) 2.5 × 10^-3 rad/s
Answer:
(D) 2.5 × 10^-3 rad/s

Question 19.
The gravitational potential energy per unit mass at a point gives ________ at that point.
(A) gravitational field
(B) gravitational potential
(C) gravitational potential energy
(D) gravitational force
Answer:
(B) gravitational potential

Question 20.
A satellite is orbiting around a planet at a constant height in a circular orbit. If the mass of the planet is reduced to half, the satellite would
(A) fall on the planet.
(B) go to an orbit of smaller radius.
(C) go to an orbit of higher radius,
(D) escape from the planet.
Answer:
(D) escape from the planet.

Question 21.
How does the escape velocity of a particle depend on its mass?
(A) m²
(B) m
(C) m⁰
(D) m^-1
Answer:
(C) m⁰

Question 22.
Escape velocity on a planet is ve. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes
(A) 4v_e
(B) 2v_e
(C) v_e
(D) 0.5 v_e
Answer:
(B) 2v_e

Question 23.
If the escape velocity of a body on Earth is 11.2 km/s, the escape velocity of the body thrown at an angle 45° with the horizontal will be
(A) 11.2 km/s
(B) 22.4 km/s
(C) \(\frac{11.2}{\sqrt{2}}\)km/s
(D) 11.2 \(\sqrt{2}\) km/s
Answer:
(A) 11.2 km/s

Question 24.
Potential energy of a body in the gravitational field of planet is zero. The body must be
(A) at centre of planet.
(B) on the surface of planet.
(C) at infinity.
(D) at distance equal to radius of Earth.
Answer:
(C) at infinity.

Question 25.
If gravitational force of Earth disappears, what will happen to the satellite revolving round the Earth?
(A) Satellite will come back to Earth.
(B) Satellite will continue to revolve.
(C) Satellite will escape in tangential path.
(D) Satellite will start falling towards centre.
Answer:
(C) Satellite will escape in tangential path.

Question 26.
If v_e and v_o represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R respectively, then
(A) v_e = v_o
(B) \(\sqrt{2}\)v_o = v_e
(C) v_e = \(\frac{1}{\sqrt{2}}\)v_o
(D) v_e and v_o are not related
Answer:
(B) \(\sqrt{2}\)v_o = v_e

Question 27.
If the kinetic energy of a satellite is 2 × 10⁴ J, then its potential energy will be
(A) – 2 × 10⁴ J
(B) 4 × 10⁴ J
(C) -4 × 10⁴ J
(D) -10⁴J
Answer:
(C) -4 × 10⁴ J

Competitive Corner

Question 1.
A body weighs 200 N on the surface of the Earth. How much will it weigh half way down to the centre of the Earth?
(A) 250 N
(B) 100 N
(C) 150 N
(D) 200 N
Answer:
(B) 100 N
Hint:
Acceleration due to gravity at depth d,
g_d = g (1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
= g(1 – \(\frac{1}{2}\)) …(∵ d = \(\frac{1}{2}\))
∴ g_d = \(\frac{\mathrm{g}}{2}\)
Weight of the body at depth d = R/2,
W_d = mg_d = m × g/2 = \(\frac{1}{2}\) × 200
∴ W_{d = 100 N}

Question 2.
The work done to raise a mass m from the surface of the Earth to a height h, which is equal to the radius of the Earth, is:
(A) \(\frac{1}{2}\) mgR
(B) \(\frac{3}{2}\) mgR
(C) mgR
(D) 2mgR
Answer:
(A) \(\frac{1}{2}\) mgR
Hint:
Initial potential energy on Earth’s surface,
U_i = \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
Final potential energy at height h = R
U_f = \(\frac{-\mathrm{GMm}}{2 \mathrm{R}}\)
Work done, W = U_f – U_i

∴ W = \(\frac{1}{2}\) mgR

Question 3.
The time period of a geostationary satellite is 24 h, at a height 6R_E (R_E is radius of Earth) from surface of Earth. The time period of another satellite whose height is 2.5 R_E from surface will be,
(A) \(\frac{12}{2.5}\)h
(B) 6\(\sqrt{2}\) h
(C) 12\(\sqrt{2}\) h
(D) \(\frac{24}{2.5}\)h
Answer:
(B) 6\(\sqrt{2}\) h
Hint:
By Kepler’s third law,

Question 4.
Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass m, when taken to a height h from the surface of Earth (of radius R), is given by,
(A) \(\frac{\text { GMm }}{R+h}\)
(B) – \(\frac{\text { GMm }}{R+h}\)
(C) \(\frac{\text { GMmh }}{R(R+h)}\)
(D) mgh
Answer:
(C) \(\frac{\text { GMmh }}{R(R+h)}\)
Hint:
Potential energy of object of mass m on the surface of Earth,
P.E = \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
Potential energy of object of mass m at a height h from the surface of the Earth,
P.E.’ = \(\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\)
∴ Change in potential energy
= P.E.’ – P.E.

Question 5.
A body mass ‘m’ is dropped from height \(\), from Earth’s surface, where ‘R’ is the radius of Earth. Its speed when it will hit the Earth’s surface is (v_e = escape velocity from Earth’s surface)

Answer:
(B) \(\frac{\mathbf{v}_{\mathrm{e}}}{\sqrt{3}}\)
Hint:

Question 6.
According to Kepler’s Law, the areal velocity of the radius vector drawn from the Sun to any planet always
(A) decreases.
(B) first increases and then decreases.
(C) remains constant.
(D) increases.
Answer:
(C) remains constant.

Question 7.
A body is thrown from the surface of the Earth with velocity ‘u’ m/s. The maximum height in m above the surface of the Earth upto which it will reach is (R = radius of Earth, g = acceleration due to gravity)

Answer:
(A) \(\frac{u^{2} R}{2 g R-u^{2}}\)
Hint:
(T.E.) on surface = (T.E.) at height ‘h’
∴ (K.E.)₁ + (P.E.)₁ = (K.E.)₂ + (P.E.)₂

Question 8.
A satellite is revolving in a circular orbit at a height ‘h’ above the surface of the Earth of radius ‘R’. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the Earth. The relation between ‘h’ and ‘R’ is
(A) h = 2R
(B) h = 3R
(C) h = 5R
(D) h = 7R
Answer:
(D) h = 7R
Hint:

Question 9.
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
(A) keep floating at the same distance between them.
(B) move towards each other.
(C) move away from each other.
(D) will become stationary.
Answer:
(B) move towards each other.

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 4 Laws of Motion Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 4 Laws of Motion

Question 1.
‘Rest and motion are relative concepts.’ Explain the statement with an example.
Answer:

1. A body can be described to be at rest or in motion with respect to a system of co¬ordinate axes known as the frame of reference.
2. A body is in motion if it changes its position with respect to a fixed reference point in a frame of reference. On the other hand, a body is at rest if it does not change its position with respect to a fixed reference point in a frame of reference.
3. An object can be said to be at rest with respect to a frame of reference while the same object can be said to be in motion with respect to a different frame of reference.
   Example: In a running train, all the travellers in the train are in a state of rest if the train is taken as the frame of reference. On the other hand, all the travellers in the train are in a state of motion if ground (or platform) is taken as the frame of reference.
4. Thus, motion and rest always need a frame of reference to be described. Hence, rest and motion are relative concepts.

Question 2.
Explain how acceleration and initial velocity decides trajectory of a motion.
Answer:

1. The resultant motion is linear if:
   • initial velocity \(\overrightarrow{\mathrm{u}}\) = 0 (starting from rest) and acceleration \(\overrightarrow{\mathrm{a}}\) is in any direction.
   • initial velocity \(\overrightarrow{\mathrm{u}}\) ≠ 0 and acceleration a is in line with the initial velocity (same or opposite direction).
2. The resultant motion is circular if initial velocity \(\overrightarrow{\mathrm{u}}\) ≠ 0 and acceleration \(\overrightarrow{\mathrm{a}}\) is perpendicular to the velocity throughout.
3. The resultant motion is parabolic if the initial velocity \(\overrightarrow{\mathrm{u}}\) is not in line with the acceleration \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{a}}\) = constant.
   e.g., trajectory of a projectile motion.
4. Similarly, various other combinations of initial velocity and acceleration will result into more complicated motions.

Question 3.
State Newton’s first law of motion.
Answer:
Statement: Every inanimate object continues to be in a state of rest or of uniform unaccelerated motion along a straight line, unless it is acted upon by an external, unbalanced force.

Question 4.
An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 m s^-2. What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.) (NCERT)
Answer:

1. Assuming absence of stars in the vicinity, the only gravitational force exerted on astronaut is by the spaceship.
2. But this force is negligible.
3. Hence, once astronaut is out of the spaceship net external force acting on him can be taken as zero.
4. From the first law of motion, the acceleration of astronaut is zero.

Question 5.
Give the magnitude and direction of the net force acting on:

1. a drop of rain falling down with a constant speed.
2. a cork of mass 10 g floating on water.
3. a kite skilfully held stationary in the sky.
4. a car moving with a constant velocity of 30 kmh^-1 on a rough road.
5. a high speed electron in space far from all gravitating objects, and free of electric and magnetic fields. (NCERT)

Answer:

1. The drop of rain falls down with a constant speed, hence according to the first law of motion, the net force on the drop of rain is zero.
2. Since the 10 g cork is floating on water, its weight is balanced by the up thrust due to water. Therefore, net force on the cork is zero.
3. As the kite is skilfully held stationary in the sky, in accordance with first law of motion, the net force on the kite is zero.
4. As the car is moving with a constant velocity of 30 km/h on a road, the net force on the car is zero.
5. As the high-speed electron in space is far from all material objects, and free of electric and magnetic fields, it doesn’t accelerate and moves with constant velocity. Hence, net force acting on the electron is zero.

Question 6.
State Newton’s second law of motion and its importance.
Answer:
Statement: The rate of change of linear momentum of a rigid body is directly proportional to the applied (external unbalanced) force and takes place in the direction of force.
\(\overrightarrow{\mathrm{F}}=\frac{\mathrm{d} \overrightarrow{\mathrm{p}}}{\mathrm{dt}}\)
Where, \(\overrightarrow{\mathrm{F}}\) = Force applied
p = m\(\overrightarrow{\mathrm{v}}\) = linear momentum

Importance of Newton’s second law:

1. It gives mathematical formulation for quantitative measure of force as rate of change of linear momentum.
2. It defines momentum instead of velocity as the fundamental quantity related to motion.
3. It takes into consideration the resultant unbalanced force on a body which is used to overcome Aristotle’s fallacy.

Question 7.
Explain why a cricketer moves his hands backwards while holding a catch. (NCERT)
Answer:

1. In the act of catching the ball, by drawing hands backward, cricketer allows longer time for his hands to stop the ball.
2. By Newton’s second law of motion, force applied depends on the rate of change of momentum.
3. Taking longer time to stop the ball ensures smaller rate of change of momentum.
4. Due to this the cricketer can stop the ball by applying smaller amount of force and thereby not hurting his hands.

Question 8.
Large force always produces large change in momentum on a body than a small force. Is this correct?
Answer:
No. From Newton’s second law, we have.
\(\frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{F}\) …. (i)
dP = Fdt …. (ii)
From equation (ii), we can infer that a small force acting for a longer time can produce same change in momentum of a body as a large force acting in the same direction for a short time. Hence, the given statement is incorrect.

Question 9.
Newton’s first law is contained in the second law. Prove it.
Answer:

1. From Newton’s second law of motion, we have, \(\overrightarrow{\mathrm{F}}=\frac{\mathrm{d} \overrightarrow{\mathrm{p}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{m} \overrightarrow{\mathrm{v}})\)
2. For a given body, mass m is constant.
   ∴ \(\overrightarrow{\mathrm{F}}=\mathrm{m} \frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=\mathrm{m} \overrightarrow{\mathrm{a}}\)
3. If \(\overrightarrow{\mathrm{F}}\) = zero, \(\overrightarrow{\mathrm{v}}\) is constant. Hence if there is no force, velocity will not change. This is nothing but Newton’s first law of motion.

Question 10.
State Newton’s third law of motion
Answer:
Statement: To every action (force) there is always an equal and opposite reaction force).

Question 11.
State the importance of Newton’s third law of motion.
Answer:

1. Newton’s third law of motion defines action and reaction as a pair of equal and opposite forces acting along the same line.
2. Action and reaction forces always act on different objects.

Question 12.
State the consequences of Newton’s third law of motion.
Answer:

1. Two interacting bodies exert forces which are always equal in magnitude, have the same line of action and are opposite in direction, upon each other. Thus, forces always occur in pairs.
2. If a body A exerts an action force \(\overrightarrow{\mathrm{F}}_{\mathrm{BA}}\) on body B, then body B also exerts an equal and opposite reaction force \(\overrightarrow{\mathrm{F}}_{\mathrm{AB}}\) on body A, simultaneously.
3. Body A experiences the force \(\overrightarrow{\mathrm{F}}_{\mathrm{AB}}\) only and
   body B experiences the force \overrightarrow{\mathrm{F}}_{\mathrm{BA}} only.
4. Both the forces, action and reaction act at the same instant.
5. Both the forces always act on different bodies. Hence, they never cancel each other.
6. Both the forces do not necessarily arise due to contact i.e., they can be non-contact forces. Example: Repulsive forces between two magnets.

Question 13.
If a constant force of 800 N produces an acceleration of 5 m/s² in a body, what is its mass? If the body starts from rest, how much distance will it travel in 10 s?
Solution:
Given: F = 800 N, a = 5 m/s², u = 0, t = 10 s
To find: mass (m), distance travelled (s)
Formulae:

i. F = ma
ii. s = ut + \(\frac{1}{2} \mathrm{at}^{2}\)

Calculation:
From formula (i),
∴ m = \(\frac{\mathrm{F}}{\mathrm{a}}=\frac{800}{5}\) = 160 kg
From formula (ii),
s = \(\frac{1}{2}\) × 5 × (10)² [∵ u = 0]
∴ s = 250 m
Answer:
Mass of the body is 160 kg and the distance travelled by the body is 250 m.

Question 14.
A constant force acting on a body of mass 3 kg changes its speed from 2 m s^-1 to 3.5 m/s in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force? (NCERT)
Solution:
Given: u = 2 ms^-1, m = 3 kg,
v = 3.5 m s^-1, t = 25s
To find: Force (F)
Formula: F = ma
Calculation: Since, v = u + at
∴ 3.5 = 2 + a × 25
a = \(\frac{3.5-2}{25}\) = 0.06 m s^-2
From formula,
F = 3 × 0.06 = 0.18 N
Since, the applied force increases the speed of the body, it acts in the direction of the motion.
Answer:
The applied force is 0.18 N along the direction of motion.

Question 15.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop? (NCERT)
Solution:
Given: m = 20 kg, u = 15ms^-1, v = 0,
F = – 50 N (retarding force)
To find: Time (t)
Formula: v = u + at
Calculation: Since, F = ma
∴ a = \(\frac{\mathrm{F}}{\mathrm{m}}=\frac{-50}{20}\) = -2.5 m s^-2
From formula,
0 = 15 + (-2.5) × t
∴ t = 6s
Answer:
Time taken to stop the body is 6 s.

Question 16.
A hose pipe used for gardening is ejecting water horizontally at the rate of 0.5 m/s. Area of the bore of the pipe is 10 cm2. Calculate the force to be applied by the gardener to hold the pipe horizontally stationary.
Solution:
Let ejecting water horizontally be considered as the action force on the water, then the water exerts a backward force (called recoil force) on the pipe as the reaction force.

Where, V = volume of water ejected
A = area of cross section of bore = 10 cm²
ρ = density of water = 1 g/cc
l = length of the water ejected in time t
\(\frac{\mathrm{d} l}{\mathrm{dt}}\) = v = velocity of water ejected
= 0.5 m/s = 50 cm/s
F = \(\frac{\mathrm{dm}}{\mathrm{dt}} \mathrm{v}\)
= (Aρv) v
= Aρv²
= 10 × 1 × 50²
∴ F = 25000 dyne = 0.25 N
Answer:
The gardener must apply an equal and opposite force of 0.25 N.

Question 17.
What does the term frame of reference mean?
Answer:
A system of co-ordinate axes with reference to which the position or motion of an object is described is called a frame of reference.

Question 18.
Explain the terms inertial and non-inertial frame of reference.
Answer:

1. Inertial frame of reference:
   • A frame of reference in which Newton ‘s first law of motion is applicable is called inertia/frame of reference.
   • A body moves with a constant velocity (which can be zero) in the absence of a net force. The body does not accelerate.
   • Example: A rocket in inter-galactic space (gravity free space between galaxies) with all its engine shut.
2. Non-inertial frame of reference:
   • A frame of reference in which an object suffers acceleration in absence of net force is called non-inertial frame of reference.
   • The body undergoes acceleration.
   • Example: If a car just start its motion from rest, then during the time of acceleration the car will be in a non-inertial frame of reference.

Question 19.
State the limitations of Newton’s laws of motion.
Answer:

1. Newton’s laws of motion are not applicable in a non-inertial (accelerated) frame of reference.
2. Newton’s laws are only applicable to point objects.
3. Newton’s laws are only applicable to rigid bodies.
4. Results obtained by applying Newton’s laws of motion for objects moving with speeds comparable to that of light do not match with the experimental results and Einstein special theory of relativity has to be used.
5. Newton’s laws of motion fail to explain the behaviour and interaction of objects having atomic or molecular sizes, and quantum mechanics has to be used.

Question 20.
Name the different types of fundamental forces in nature.
Answer:
Fundamental forces in nature are classified into four types:

1. Gravitational force
2. Electromagnetic force
3. Strong nuclear force
4. Weak nuclear force

Question 21.
Define gravitational force. Give its examples.
Answer:
Force of attraction between two (point) masses separated by a distance is called as gravitational force.
F = \(\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
where ‘G’ is constant called the universal gravitational constant = 6.67 × 10^-11 Nm²/kg²

Examples:

1. The motion of moon, artificial satellites around the earth and motion of planets around the sun is due to gravitational force of attraction.
2. The concept of weight of a body in the earth’s gravitational field is due to gravitational force exerted by the earth on a body.

Question 22.
Write down the main characteristics of gravitational force.
Answer:
Characteristics of gravitational force:

1. It is always attractive.
2. It is the weakest of the four basic forces in nature.
3. Its range is infinite.
4. Structure of the universe is governed by this force.

Question 23.
Write a note on electromagnetic (EM) force.
Answer:
Electromagnetic force:

1. The attractive and repulsive force between electrically charged particles is called electromagnetic force.
2. It can be attractive or repulsive.
3. It is stronger than the gravitational force.
4. Example: force of friction, normal reaction, tension in strings, collision forces, elastic forces, fluid friction etc. are electromagnetic in nature.
5. Reaction forces are a result of the action of electromagnetic forces.
6. Since majority of forces are electromagnetic in nature, our life is practically governed by these forces.

Question 24.
Write a note on strong and weak nuclear force.
Answer:

1. Strong nuclear force: The strong force which binds protons and neutrons (nucleons) together in the nucleus of an atom is called strong nuclear force.
   Characteristics of strong nuclear force:
   • It is a very strong attractive force.
   • It is a short range force of the order of 10^-14 m.
   • it is charge independent.
2. Weak nuclear force: The force of interaction between subatomic particles which results in the radioactive decay of atoms is called weak nuclear force.

Characteristics of weak nuclear force:

• It acts between any two elementary particles (pair of subatomic particles).
• It is a stronger force than gravitational force.
• It is much weaker than electromagnetic force or strong nuclear force.
• It is a short range force of the order of 10^-16m.

Question 25.
Three identical point masses are fixed symmetrically on the periphery of a circle. Obtain the resultant gravitational force on any point mass M at the centre of the circle. Extend this idea to more than three identical masses symmetrically located on the periphery. How far can you extend this concept?
Solution:

i. Consider three identical points A, B and C of mass m on the periphery of a circle of radius r. Mass M is at the centre of the circle.
Gravitational forces on M due to these masses are attractive and are given as,
In magnitude, \(\mathrm{F}_{\mathrm{MA}}=\mathrm{F}_{\mathrm{MB}}=\mathrm{F}_{\mathrm{MC}}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)

ii. Forces \(\overrightarrow{\mathrm{F}}_{\mathrm{MB}}\) and \(\overrightarrow{\mathrm{F}}_{\mathrm{MC}}\) are resolved along \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\) and perpendicular to \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\). Components perpendicular to \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\) cancel each other. Components along \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\) are,
F_MB cos 60° = F_MC cos 60° = \(\frac{1}{2} F_{M A}\) each.

Magnitude of their resultant is F_MA and its direction is opposite to that of F_MA. Thus, the
resultant force on mass M is zero. For any even number of equal masses, the force due to any mass m is balanced (cancelled) by diametrically opposite mass. For any odd number of masses, the components perpendicular to one of them cancel each other while the components parallel to one of these add up in such a way that the resultant is zero for any number of identical masses m located symmetrically on the periphery.

As the number of masses tends to infinity, their collective shape approaches circumference of the circle, which is nothing but a ring. Thus, the gravitational force exerted by a ring mass on any other mass at its centre is zero.

iii. This concept can be further extended to three-dimensions by imagining a uniform hollow sphere to be made up of infinite number of such rings with a common diameter. Thus, the gravitational force for any mass kept at the centre of a hollow sphere is zero.

Question 26.
A car of mass 1.5 ton is running at 72 kmph on a straight horizontal road. On turning the engine off, it stops in 20 seconds. While running at the same speed, on the same road, the driver observes an accident 50 m in front of him. He immediately applies the brakes and just manages to stop the car at the accident spot. Calculate the braking force.
Solution:
Given: m = 1.5 ton = 1500 kg,
u = 72 kmph = 72 × \(\frac{5}{18} \mathrm{~m} / \mathrm{s}\)m/s = 20 m
s^-1 (on turning engine off),
v = 0, t = 20 s, s = 50 m
To find: Braking force (F)

Formula:

i. v = u + at
ii. v² – u² = 2as
iii. F = ma

Calculation:
On turning the engine off,
From formula (i),
a = \(\frac{0-20}{20}\) = -1 m s^-2
This is frictional retardation (negative acceleration).
After seeing the accident,
From formula (ii),
a₁ = \(\frac{0^{2}-20^{2}}{2(50)}\) = -4 m s^-2
This retardation is the combined effect of braking and friction
∴ braking retardation =4 – 1 = 3 m s^-2
From formula (iii), the braking force, F = 1500 × 3 = 4500 N
Answer:
The braking force is 4500 N.

Question 27.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \(\overrightarrow{\mathbf{F}}=(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) N where \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), \(\hat{\mathbf{k}}\) are unit vectors along the x, y and z axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis? (NCERT)
Solution:
Given: \(\overrightarrow{\mathbf{F}}=(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) N, \(\overrightarrow{\mathrm{s}}=4 \hat{\mathrm{k}}\)
To find: work done (W)

Formula: W = \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}\)
Calculation: From formula,
W = \((-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{k}})\)
= \(12 \hat{\mathrm{k}} \cdot \hat{\mathrm{k}}\) = 12 J
Answer:
The work done by the force is in moving the body 12 J.

Question 28.
Over a given region, a force (in newton) varies as F = 3x² – 2x + 1. In this region, an object is displaced from x₁ = 20 cm to x₂ = 40 cm by the given force. Calculate the amount of work done.
Solution:
Given: F = 3x² – 2x + 1, x₁ = 20 cm = 0.2 m,
x₂ = 40 cm = 0.4 m.
To find: Work done (W)
Formula: W = \(\int_{A}^{B} \vec{F} \cdot \overrightarrow{d s}\)
Calculation:
From formula,
W = \(\int_{x_{1}}^{x_{2}} F \cdot d x=\int_{0.2}^{0.4}\left(3 x^{2}-2 x+1\right) d x\)
= [x³ – x² + x]^0.4
= [0.4³ – 0.4² + 0.4] – [0.2³ – 0.2² + 0.2]
= 0.304 – 0.168 = 0.136 J
The work done is 0.136 J.

Question 29.
A position dependent force f = 7 – 2x + 3x² newton acts on a small body of mass 2 kg and displaces from x = 0 m to x = 5 m, calculate the work done.
Solution:
Given: F = 7 – 2x + 3x², x = 0 at A and x = 5 at B.
To find: Work done (W)
Formula: W = \(\int_{A}^{B} \vec{F} \cdot \overrightarrow{d s}\)
Calculation: From formula,

∴ W = 135 J
Answer:
The work done is 135 J.

Question 30.
State the principle of work-energy theorem in case of a conservative force and explain.
OR
Show that work done on a body by a conservative force is equal to the change in its kinetic energy
Answer:
Principle: Decrease in the potential energy due to work done by a conservative force is entirely converted into kinetic energy. Vice versa, for an object moving against a conservative force, its kinetic energy decreases by an amount equal to the work done against the force.

Work-energy theorem in case of a conservative force:

1. Consider an object of mass m moving with velocity u experiencing a constant opposing force F which slows it down to v during displacement s.
2. The equation of motion can be written as, v² – u² = -2as (negative acceleration for
   opposing force.)
   Multiplying throughout by \(\frac{\mathrm{m}}{2}\), we get,
   \(\frac{\mathrm{1}}{2}\)mu² – \(\frac{\mathrm{1}}{2}\)mv² = (ma)s …. (1)
3. According to Newton’s second law of motion,
   F = ma … (2)
4. From equations (1) and (2), we get,
   \(\frac{\mathrm{1}}{2}\)mu² – \(\frac{\mathrm{1}}{2}\)mv² = F.s
5. But, \(\frac{\mathrm{1}}{2}\)mv²= K_f = final K.E of the body,
   \(\frac{\mathrm{1}}{2}\)mu² = K_i = initial K.E of the body. and, work done by the force = F.s
   ∴ work done by the force = k_f – k_i
   = decrease in K.E of the body.
6. Thus, work done on a body by a conservative force is equal to the change in its kinetic energy.

Question 31.
Explain the work-energy theorem in case of an accelerating conservative force along with a retarding non-conservative force.
Answer:

1. Consider an object dropped from some point at height h.
2. While coming down its potential energy decreases.
   ∴ Work done = decrease in P.E of the body.
3. But, in this case, some part of the energy is used in overcoming the air resistance. This part of energy appears in some other forms such as heat, sound, etc. Thus, the work is not entirely converted into kinetic energy. In this case, the work-energy theorem can mathematically be written as,
   ∴ ∆ PE = ∆ K.E. + W_{air resistance}
   ∴ Decrease in the gravitational P.E. = Increase in the kinetic energy + work done against non-conservative forces.

Question 32.
A liquid drop of 1.00 g falls from height of cliff 1.00 km. It hits the ground with a speed of 50 m s^-1. What is the work done by the unknown force? (Take g = 9.8 m/s²)
Solution:
Given: m = 1.0 g = 1.0 × 10^-3 kg,
h = 1 km = 10³ m, v = 50 ms^-1
To find: Work done (W_f)
Formula: W_f = ∆ K.E – W_g

Calculation:

i. The change in kinetic energy of the drop
∆ K.E = (K.E.)_final (K.E.)_initial
∴ ∆ K.E. = \(\frac{1}{2} \mathrm{mv}^{2}-0\)
= \(\frac{1}{2} \times 1.0 \times 10^{-3} \times(50)^{2}\)
∴ ∆ K.E.= 1.25 J

ii. Work done by the gravitational force is,
W_g = mgh = 1.0 × 10^-3 × 9.8 × 10³ = 9.8 J
∴ W_g = 9.8J
From formula,
W_f = ∆K.E. – W_g = 1.25 – 9.8
W_f = -8.55 J
Answer:
Work done by the unknown force is – 8.55 J.

Question 33.
A body of mass 0.5 kg travels in a straight line with velocity y = ax^3/2, where a = 5 m^1/2s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2m? (NCERT)
Solution:
Given: M = 0.5 kg, y = ax^3/2,
where a = 5 m^-1/2s^-1
Let v₁ and v₂ be the velocities of the body, when x = 0 and x = 2 m respectively. Then,
v₁ = 5 × 0^3/2 = 0, v₂ = 5 × 2^3/2 = \(10 \sqrt{2}\) m
To find: Work done (W)
Formula: Work done = Increase in kinetic energy
W = \(\frac{1}{2} \mathrm{M}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right)\)
Calculation: From formula,
W = \(\frac{1}{2}\) × 0.5 × [latex](10 \sqrt{2})^{2}-0^{2}[/latex]
∴ W = 50J
Answer:
Work done by the net force on the body is 50 J.

Question 34.
A particle of mass 12 kg is acted upon by a force f = (100 – 2x²) where f is in newton and ‘x’ is in metre. Calculate the work done by this force in moving the particle x = 0 to x = -10 m. What will be the speed at x = 10 m if it starts from rest?
Solution:
Given: F = 100 – 2x²
at A, x = 0 and at B, x = -10 m
To find: Work done (W), speed (v)

Formulae:

i. W = \(\int_{A}^{B} \vec{F} \cdot d s\)
ii. W = K.E. = \(\frac{1}{2} \mathrm{mv}^{2}\)

Calculation:
From formula (i),
W = \(\int_{A}^{B} \vec{F} \cdot \overline{d s}=\int_{x=0}^{x=-10} F d x\)

Answer:

1. Work done by the force on the particle is 333.3 J.
2. The speed of the particle at x = 10 will be 7.45 m/s.

Question 35.
Define free body diagram. In the figure given below, draw the free body diagrams for mass of 2 kg, 4 kg and 5 kg and hence state their force equations.

Answer:
i. The diagram showing the forces acting on only one body at a time along-with its acceleration is called a free body diagram.

ii. The free body diagram for the mass of 2 kg is

Free body diagram for 2 kg mass
The force equation is given as,
2a = T₃ – 2g

iii. The free body diagram for the mass of 4 kg is,

The force equation is given as, 4a = T₁ + 4g – T₂

iv. The free body diagram for the mass of 5kg is,

The force equation for the mass of 5 kg is given as,
N + F sin 60° = 5g, along the vertical direction.
T₁ + 10 = F cos 60°, along the horizontal direction (Considering the mass is in equilibrium).

Question 36.
Figure shows a fixed pulley. A massless inextensible string with masses m₁ and m₂ > m₁ attached to its two ends is passing over the pulley. Such an arrangement is called an Atwood machine. Calculate accelerations of the masses and force due to the tension along the string assuming axle of the pulley to be frictionless.

Solution:
Method I: As m₂ > m₁, mass m₂ is moving downwards and mass m₁ is moving upwards.
Net downward force = F = (m₂) g – (m₁) g
= (m₂ – m₁)g
the string being inextensible, both the masses travel the same distance in the same time. Thus, their accelerations are equal in magnitude (one upward, other downward). Let it be a.
Total mass in motion, M = m₂ + m₁
∴ a = \(\frac{F}{M}=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right) g\) …. (i)

For mass m₁, the upward force is the force due to tension T and downward force is mg. It has upward acceleration a. Thus, T – m₁g = m₁a
∴ T = m₁(g + a)
Using equation (i), we get,

From the free body equation for the first body,
T – m₁g = m₁a .. (i)

From the free body equation for the second body,
m₂g – T = m₂a … (ii)
Adding (i) and (ii), we get,
a = \(\left(\frac{\mathrm{m}_{2}-\mathbf{m}_{1}}{\mathrm{~m}_{2}+\mathrm{m}_{1}}\right) \mathbf{g}\) ….(iii)
Solving equations. (ii) and (iii) for T, we get,
T = m₂(g – a) = \(\left(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\right) g\)

Question 37.
Write a note on elastic collision.
Answer:

1. Collision between two bodies in which kinetic energy of the entire system is conserved along with the linear momentum is called as elastic collision.
2. In an elastic collision,
   \(\mathrm{m}_{1} \overrightarrow{\mathrm{u}_{1}}+\mathrm{m}_{2} \overrightarrow{\mathrm{u}_{2}}=\mathrm{m}_{1} \overrightarrow{\mathrm{v}_{1}}+\mathrm{m}_{2} \overrightarrow{\mathrm{v}}_{2}\)
3. In an elastic collision,
   \(\sum \mathrm{K} \cdot \mathrm{E}_{\text {‘initial }}=\sum \mathrm{K} \cdot \mathrm{E}_{\text {. final }}\)
4. An elastic collision is impossible in daily life.
5. However, in many situations, the interatomic and intermolecular collisions are considered to be elastic.

Question 38.
Write a note on inelastic collision.
Answer:

1. A collision is said to be inelastic if there is a loss in the kinetic energy during collision, but linear momentum is conserved.
2. In an inelastic collision, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
3. In an inelastic collision,
   \(\sum \mathrm{K} \cdot \mathrm{E}_{\text {.initial }} \neq \sum \mathrm{K} \cdot \mathrm{E}_{\text {.final }}\)
4. The loss in kinetic energy is either due to internal friction or vibrational motion of atoms causing heating effect.

Question 39.
Define perfectly inelastic collision. Give an example of it.
Answer:

1. Collision in which the colliding bodies stick together after collision and move with a common velocity is called perfectly inelastic collision.
2. The loss in kinetic energy is maximum in perfectly elastic collision.
3. Example: Lump of mud thrown on a wall sticks to the wall due to the loss of kinetic energy.

Question 40.
In case of an elastic head on collision between two bodies, derive an expression for the final velocities of the bodies in terms of their masses and velocities before collision.
Answer:
Head on elastic collision of two spheres:

i. Consider two rotating smooth bodies A and B of masses m1 and m2 respectively moving
along the same straight line.

ii. Let \(\overrightarrow{\mathrm{u}}_{1}\) = initial velocity of the sphere A before collision.
\(\overrightarrow{\mathrm{u}}_{2}\) = initial velocity of the sphere B before collision.
\(\overrightarrow{\mathrm{v}}_{1}\) = velocity of the sphere A after collision.
\(\overrightarrow{\mathrm{v}}_{2}\) = velocity of the sphere B after collision.

iii. After the elastic collision, the spheres separate and move along the same straight line without rotation.

iv. According to the law of conservation of momentum,
m₁\(\overrightarrow{\mathrm{u}}_{1}\) + m₂\(\overrightarrow{\mathrm{u}}_{2}\) = m₁\(\overrightarrow{\mathrm{v}}_{1}\) + m₂\(\overrightarrow{\mathrm{v}}_{2}\) ….(i)
According to the law of conservation of energy (as kinetic energy is conserved during elastic collision),

v. Since kinetic energy is a scalar quantity, the terms involved in the above equations are scalars.

vi. The equation (1) can be written in scalar form as,

vii. Also the equation (2) can be written as,

viii. Now dividing equation (4) by (3) we get,
(u₁ + v₁) = (u₂ + v₂)
∴ u₁ + v₁ = u₂ + v₂
∴ v₂ = u₁ – u₂ + v₁ … (5)

ix. Comparing equation (3) and (5),

Equations, (6) and (7), represent the final velocities of two spheres after collision.

Question 41.
Are you aware of elasticity of materials? Is there any connection between elasticity of materials and elastic collisions?
Answer:
(Students should answer the question as per their understanding).

Question 42.
Two bodies undergo one-dimensional, inelastic, head-on collision. State an expression for their final velocities in terms
of their masses, initial velocities and coefficient of restitution.
Answer:
If e is the coefficient of restitution, the final velocities after an inelastic, head on collision are given as,

Question 43.
Two bodies undergo one-dimensional, inelastic, head-on collision. State an expression for the loss in the kinetic energy.
Answer:

ii. As e < 1, (1 – e²) is always positive. Thus, there is always a loss of kinetic energy in an inelastic collision.

Question 44.
Two bodies undergo one-dimensional, inelastic, head-on collision. Obtain an expression for the magnitude of impulse.
Answer:
i. When two bodies undergo collision, the linear momentum delivered by the first body to the second body must be equal to the change in momentum or impulse of the second body and vice versa.
∴ Impulse,
|J| = |∆p₁| = |∆p₂|
= |m₁v₁ – m₁u₁| = |m₂v₂ – m₂u₂| ….(1)

In equation (1) and solving, we get,

Question 45.
Two bodies undergo one-dimensional, perfectly inelastic, head-on collision. Derive an expression for the loss in the kinetic energy.
Answer:
i. Let two bodies A and B of masses m₁ and m₂ move with initial velocity \(\overrightarrow{\mathrm{u}}_{1}\) and \(\overrightarrow{\mathrm{u}}_{2}\), respectively such that particle A collides head on with particle B i.e., u₁ > u₂.

ii. If the collision is perfectly inelastic, the particles stick together and move with a common velocity \(\overrightarrow{\mathrm{v}}\) after the collision along the same straight line.
loss in kinetic energy = total initial kinetic energy – total final kinetic energy.

iii. By the law of conservation of momentum,
m₁u₁ + m₂u₂ = (m₁ + m₂)v
∴ v = \(\frac{\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)

iv. Loss of Kinetic energy,

iv. Both the masses and the term (u₁ – u₂)² are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as e = 0, the loss is maximum.

Question 46.
Distinguish between elastic and inelastic collision.
Answer:

No.	Elastic Collision	Inelastic Collision
i.	In an elastic collision, both momentum and kinetic energy are conserved.	In an inelastic collision, momentum is conserved but kinetic energy is not conserved.
ii.	The total kinetic energy after collision is equal to the total kinetic energy before collision.	The total kinetic energy after the collision is not equal to the total kinetic energy before collision.
iii.	Coefficient of restitution (e) is equal to one.	Coefficient of restitution (e) is less than one. For a perfectly inelastic collision coefficient of restitution is equal to zero.
iv.	Bodies do not stick together in elastic collision.	Bodies stick together in a perfectly inelastic collision.
v.	Sound, heat and light are not produced.	Sound or light or heat or all of these may be produced.

Question 47.
Explain elastic collision in two dimensions.
Answer:
i. Suppose a particle of mass mi moving with initial velocity \(\overrightarrow{\mathrm{u}_{1}}\), undergoes a non head-on collide with another particle of mass m₂ and initial velocity \(\overrightarrow{\mathrm{u}_{2}}\).

ii. Let us consider two mutually perpendicular directions; Common tangent at the point of impact, along which there is no force (or no change of momentum).
Line of impact which is perpendicular to the common tangent through the point of impact, in the two-dimensional plane of initial and final velocities.

iii. Applying the law of conservation of linear momentum along the line of impact, we have, m₁u₁ cos α₁ + m₂u₂ cos α₂ = m₁v₁ cos β₁ + m₂v₂ cos β₂
As there is no force along the common tangent,
m₁u₁ sin α₁ = m₁u₁ sin β₁ and m₂u₂ sin α₂ = m₂v₂ sin β₂
iv. Coefficient of restitution (e) along the line of impact is given as

Question 48.
Two bodies undergo a two-dimensional collision. State an expression for the magnitude of impulse along the line of impact and the loss in kinetic energy.
Answer:
i. For two bodies undergoing a two-dimensional collision, the magnitude of impulse along the line of impact is given as, Magnitude of the impulse, along the line of impact,

ii. The loss in the kinetic energy is given as Loss in the kinetic energy = ∆ (K.E.)

Question 49.
0ne marble collides head-on with another identical marble at rest. If the collision is partially inelastic, determine the ratio of their final velocities in terms of coefficient of restitution e.
Solution:
According to conservation of momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
As m₁ = m₂, we get, u₁ + u₂ = v₁ + v₂
∴ If u₂ = 0, we get, v₁ + v₂ = u₁ ….. (i)
Coefficient of restitution,

Question 50.
A 10 kg mass moving at 5 m/s collides head- on with a 4 kg mass moving at 2 m/s in the same direction. If e = \(\frac{1}{2}\), find their velocity after impact.
Solution:
Given: m₁ = 10 kg, m₂ = 4 kg
u₁ = 5 m/s, u₂ = 2 m/s, e = \(\frac{1}{2}\)
To find: Velocity after impact (v₁ and v₂)

Formulae:

i. m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
ii. e = \(\left(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\right)\)

Calculation:

From formula (i),
10 × 5 + 4 × 2 = 10v₁ + 4v₂
∴ 5v₁ + 2v₂ = 29 … (1)

From formula (ii),
v₂ – v₁ = e(u₁ – u₂) = \(\frac{1}{2}\) (5 – 2) = \(\frac{3}{2}\)
∴ 2v₂ – 2v₁ = 3 … (2)
Solving (1) and (2), we have
∴ v₁ = \(\frac{26}{7}\) m/s and v₂ = \(\frac{73}{14}\) m/s
Answer:
The respective velocities of the two masses are \(\frac{26}{7}\) m/s and \(\frac{73}{14}\) m/s.

Question 51.
A metal ball falls from a height 1 m on a steel plate and jumps upto a height of 0.81 m. Find the coefficient of restitution.
Solution:
As the ball falls to the steel plate P.E changes to kinetic energy.

As ground is stationary, both its initial and final velocities are zero.

Question 52.
Two bodies of masses 5 kg and 3 kg moving in the same direction along the same straight line with velocities 5 m s^-1 and 3 m s^-1 respectively suffer one-dimensional elastic collision. Find their velocities after the collision.
Solution:
Given: m₁ = 5kg, u₁ = 5ms^-1, m₂ = 3kg, u₂ = 3 m s^-1
To find: velocities after collision (v₁ and v₂)

Answer:
The velocities of the two bodies after collision are 3.5 m/s and 5.5 m/s.

Question 53.
A 20 g bullet leaves a machine gun with a velocity of 200 m/s. If the mass of the gun is 20 kg, find its recoil velocity. If the gun fires 20 bullets per second, what force is to be applied to the gun to prevent recoil?
Solution:
Given: m₁ = 20g = 0.02 kg, m₂ = 20 kg, v₁ = 200 m/s, t = \(\frac{1}{20}\) s,
To find: Recoil velocity (v₂), applied force (F)

Formulae:

i. v₂ = \(-\frac{\mathrm{m}_{1} \mathrm{v}_{1}}{\mathrm{~m}_{2}}\)
ii. F = ma

Calculation: From formula (i),
∴ v₂ = \(-\frac{0.02}{20} \times 200\)
= -0.2 m/s

Negative sign shows that the machine gun moves in a direction opposite to that of the bullet.

From formula (ii),
∴ F = m₂ × a = 20 × 4 = 80N
Answer:
The recoil velocity of gun is 0.2 m/s and the required force to prevent recoil is 80 N.

Question 54.
A shell of mass 3 kg is dropped from some height. After falling freely for 2 seconds, it explodes into two fragments of masses 2 kg and 1 kg. Kinetic energy provided by the explosion is 300 J. Using g = 10 m/s², calculate velocities of the fragments. Justify your answer if you have more than one options.
Solution:
Total mass = m₁ + m₂ = 3 kg
Initially, when the shell falls freely for 2 seconds, v = u+ at = 0 + 10(2) = 20 ms^-1 = u₁ = u₂
According to conservation of linear momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

There are two possible answers since the positions of two fragments can be different as explained below.
Case 1: v₁ = 30 m s^-1 and v₂ = 0 with the lighter fragment 2 above.
Case 2: v₁ = 10 m s^-1 and v₂ = 40 m s^-1 with the lighter fragment 2 below, both moving downwards.

Question 55.
Bullets of mass 40 g each, are fired from a machine gun at a rate of 5 per second towards a firmly fixed hard surface of area 10 cm². Each bullet hits normal to the surface at 400 m/s and rebounds in such a way that the coefficient of restitution for the collision between bullet and the surface is 0.75. Calculate average force and average pressure experienced by the surface due to this firing.
Solution:
For the collision,
u₁ = 400 m s^-1, e = 0.75
For the firmly fixed hard surface, u₂ = v₂ = 0
e = 0.75 = \(\frac{v_{1}-v_{2}}{u_{2}-u_{1}}=\frac{v_{1}-0}{0-400}\)
∴ v₁ = -300 m/s.
Negative sign indicates that the bullet rebounds in exactly opposite direction.
Change in momentum of each bullet = m(v₁ – u₁)
The same momentum is transferred to the surface per collision in opposite direction.
∴ Momentum transferred to the surface, per collision,
p = m (u₁ – v₁) = 0.04(400 – [-300]) = 28 Ns
The rate of collision is same as rate of firing.
∴ Momentum received by the surface per second, \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = average force experienced by the surface = 28 × 5 = 140 N

This is the average force experienced by the surface of area A = 10 cm² = 10^-3 m²
∴ Average pressure experienced,
P = \(\frac{\mathrm{F}}{\mathrm{A}}=\frac{140}{10^{-3}}\) = 1.4 × 10⁵ N m^-2
∴ P ≈ 1.4 times the atmospheric pressure.
Answer:
The average force and average pressure experienced by the surface due to the firing is 140 N and 1.4 × 10⁵ N m^-2 respectively.

Question 56.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s^-1, what is the recoil speed of the gun? (NCERT)
Solution:
Given: m₁ = 0.02 kg, m₂ = 100 kg, v₁ = 80 m s^-1
To find: Recoil speed (v₂)
Formula: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Calculation: Initially gun and shell are at rest.
∴ m₁u₁ + m₂u₂ = 0
Final momentum = m₁v₁ – m₂v₂
Using formula,
0 = 0.02 (80) – 100(v₂)
∴ v₂ = \(\frac{0.02 \times 80}{100}\) = 0.016 ms^-1
Answer:
The recoil speed of the gun is 0.016 m s^-1.

Question 57.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s^-1 collide and rebound with the same speed. What is the impulse imparted to each bail due to the other? (NCERT)
Solution:
Given: m = 0.05 kg. u = 6 m/s, v = -6 m/s
To find: Impulse (J)
Formula: J = m (v – u)
calculation: From formula,
J = 0.05 (-6 – 6) = -0.6 kg m s^-1
Answer:
Impulse received by each ball is -0.6 kg m s^-1.

Question 58.
A bullet of mass 0.1 kg moving horizontally with a velocity of 20 m/s strikes a target and brought to rest in 0.1 s. Find the impulse and average force of impact.
Solution:
Given: m = 0.1 kg, u = 20 m/s, t = 0.1 s
To find: Impulse (J), Average force (F)

Formulae:

i. J = mv – mu
ii. F = \(m \frac{(v-u)}{t}\)

Calculation:

From formula (i).
J = m(v – u) = 0.1 (0 – 20) = -2 Ns
From formula (ii),
F = \(\frac{m(v-u)}{t}=\frac{2}{0.1}=20 N\)
Answer:
Magnitude of impulse is 2 Ns, average force of impact is 20 N.

Question 59.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg) (NCERT)
Answer:
Let the point B represents the position of bat. The ball strikes the bat with velocity v along the path AB and gets deflected with same velocity along BC. such that ∠ABC = 45°


Thus, impulse imparted to the ball is 4.157 kg ms^-1

Question 60.
A cricket ball of mass 150 g moving with a velocity of 12 m/s is turned back with a velocity of 20 m/s on hitting the bat. The force of the ball lasts for 0.01 s. Find the average force exerted on the ball by the bat.
Solution:
m = 0.150 kg, v = 20 m/s,
u = -12 m/s and t = 0.01 s
To find: Average force (F)
Formula: F = \(\frac{m(v-u)}{t}\)
Calculation: From formula,
F = \(\frac{0.150[20-(-12)]}{0.01}\) = 480 N
Answer:
The average force exerted on the ball by the bat is 480 N.

Question 61.
Mass of an Oxygen molecule is 5.35 × 10^-26 kg and that of a Nitrogen molecule is 4.65 × 10^-26 kg. During their Brownian motion (random motion) in air, an Oxygen molecule travelling with a velocity of 400 m/s collides elastically with a nitrogen molecule travelling with a velocity of 500 m/s in the exactly opposite direction. Calculate the impulse received by each of them during collision. Assuming that the collision lasts for
1 ms, how much is the average force experienced by each molecule?
Solution:
Let, m₁ = m_O = 5.35 × 10^-26 kg,
m₂ = m_N = 4.65 × 10^-26 kg,
∴ u₁ = 400 ms^-1 and u₂ = -500 ms_-1 taking direction of motion of oxygen molecule as the positive direction.
For an elastic collision,


Hence, the net impulse or net change in momentum is zero.

Answer:
The average force experienced by the nitrogen molecule and the oxygen molecule are
-4.478 × 10^-20 N and 4.478 × 10^-20 N.

Question 62.
Explain rotational analogue of the force. On what factors does it depend? Represent it in vector form.
Answer:

1. Rotational analogue of the force is called as moment of force or torque.
2. It depends on the mass of the object, the point of application of the force and the angle between direction of force and the line joining the axis of rotation with the point of application.
3. In its mathematical form, torque or moment of a force is given by
   \(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)
   where \(\overrightarrow{\mathrm{F}}\) is the applied force and \(\overrightarrow{\mathrm{r}}\) is the position vector of the point of application of the force from the axis of rotation.

Question 63.
Illustrate with an example how direction of the torque acting on any object is determined.
Answer:
i. Consider a laminar object with axis of rotation perpendicular to it and passing through it as shown in figure (a).

ii. Figure (b) indicates the top view of the object when the rotation is in anticlockwise direction

iii. Figure (c) shows the view from the top, if rotation is in clockwise direction.

iv. The applied force \(\vec{F}\) and position vector \(\vec{r}\) of the point of application of the force are in the plane of these figures.

v. Direction of the torque is always perpendicular to the plane containing the vectors \(\vec{r}\) and \(\vec{F}\) and can be obtained from the rule of cross product or by using the right-hand thumb rule.

vi. In Figure (b), it is perpendicular to the plane of the figure and outwards while in the figure (c), it is inwards.

Question 64.
State the equation for magnitude of torque and explain various cases of angle between the direction of \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{F}}\).
Answer:
Magnitude of torque, \(\tau\) = r F sin θ
where θ is the smaller angle between the directions of \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{F}}\).

Special cases:

1. If θ = 90°, \(\tau\) = \(\tau\)_max = rF. Thus, the force should be applied along normal direction for easy rotation.
2. If θ = 0° or 180°, \(\tau\) = \(\tau\)_min = 0. Thus, if the force is applied parallel or anti-parallel to \(\overrightarrow{\mathrm{r}}\), there is no rotation.
3. Moment of a force depends not only on the magnitude and direction of the force, but also on the point where the force acts with respect to the axis of rotation. Same force can have different torque as per its point of application.

Question 65.
A force \(\overrightarrow{\mathbf{F}}=\mathbf{3} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{4} \hat{\mathbf{k}}\) is applied at a point (3, 4, -2). Find its torque about the point (-1, 2, 4).
Solution:

Question 66.
Define couple. Show that moment of couple is independent of the points of application of forces.
Answer:
A pair of forces consisting of two forces of equal magnitude acting in opposite directions along different lines of action is called a couple.

1. Figure shows a couple consisting of two forces \(\overrightarrow{\mathrm{F}}_{1}\) and \(\overrightarrow{\mathrm{F}}_{2}\) of equal magnitudes and opposite directions acting along different lines of action separated by a distance r.
2. Position vector of any point on the line of action of force \(\overrightarrow{\mathrm{F}}_{1}\) from the line of action of force \(\overrightarrow{\mathrm{F}}_{2}\) is \(\overrightarrow{\mathrm{r}}_{12}\). Similarly, the position vector of any point on the line of action of force \(\overrightarrow{\mathrm{F}}_{2}\) from the line of action of force \(\overrightarrow{\mathrm{F}}_{1}\) is \(\overrightarrow{\mathrm{r}}_{21}\).
3. Torque or moment of the couple is then given mathematically as
   
4. From the figure, it is clear that r₁₂ sinα = r₂₁ sin β = r.
5. If \(\left|\overrightarrow{\mathrm{F}}_{1}\right|=\left|\overrightarrow{\mathrm{F}}_{2}\right|\) = F, the magnitude of torque is given by
   \(\tau=\mathrm{r}_{12} \mathrm{~F}_{1} \sin \alpha\) = \(r_{21} F_{2} \sin \beta=r F\)
6. It clearly shows that the torque corresponding to a given couple, i.e., the moment of a given couple is constant, i.e., it is independent of the points of application of forces.

Question 67.
The figure below shows three situations of a ball at rest under the action of balanced forces. Is the bail in mechanical equilibrium? Explain how the three situations differ.

Answer:

1. In all these cases, as the ball is at rest under the action of balanced forces i.e, there is no net force acting on it. Hence, it is in mechanical equilibrium.
   However, potential energy-wise, the three situations show the different states of mechanical equilibrium.
2. Stable equilibrium: In situation (a), potential energy of the system is at its local minimum. If it is disturbed slightly from its equilibrium position and released, it tends to recover its position. In this situation, the ball is most stable and is said to be in stable equilibrium.
3. Unstable equilibrium: In situation (b), potential energy of the system is at its local maximum. If it is slightly disturbed from its equilibrium position, it moves farther from that position. This happens because initially, if disturbed, it tries to achieve the configuration of minimum potential energy. In this situation, the ball is said to be in unstable equilibrium.
4. Neutral equilibrium: in situation (e), potential energy of the system is constant over a plane and remains same at any position. Thus, even if the ball is disturbed, it still remains in equilibrium, practically at any position. In this situation, the ball is in neutral equilibrium.

Question 68.
Two weights 5 kg and 8 kg are suspended from a uniform rod of length 10 m and weighing 3 kg. The distances of the weights from one end are 2 m and 7 m. Find the point at which the rod balances.
Solution:

Let the rod balance at a point P, x m from the end A of the rod AB. The suspended weight and the centre of gravity of the rod G is shown in the figure.
P = 5 + 3 + 8 = 16kg
Taking moments about A,
16 × x = 5 × 2 + 3 × 5 + 8 × 7
16x = 10 + 15 + 56 = 81
∴ x = \(\frac{81}{16}\) = 5.1 m
Answer:
The rod balances at 5.1 m from end A.

Question69.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? (NCERT)

Consider the metre scale AB. Let its mass be concentrated at C at 50 cm. mark. Upon placing coins, balancing point of the scale and the coins system is shifted to C’ at 45 cm mark.
For equilibrium about C’
10(45 – 12) = m (50 – 45)
m = \(\frac{10 \times 33}{5}\) = 66g
Answer:
The mass of the metre scale is 66 g.

Question 70.
The diagram shows a uniform beam of length 10 m, used as a balance. The beam is pivoted at its centre. A 5.0 N weight is attached to one end of the beam and an empty pan weighing 0.25 N Is attached to the other end of the beam.

i. What is the moment of couple at pivot?
ii. If pivot is shifted 2 ni towards left, then what will be moment of couple at new position?
Answer:

Answer:

1. The moment of couple at center is 23.75 N-m.
2. The moment of couple at new position is 13.25 N-m.
   ∴ For equilibrium,
   40 × x = 20 × 0.5
   ∴ x = \(\frac{1}{4}\) = 0.25 m

Hence, the total distance walked by the person is 1.25 m.

Ans
The person can walk 1.25 m before the plank topples.

Question 71.
Define centre of mass.
Answer:
Centre of mass of a body is a point about which the summation of moments of masses in the system is zero.

Question 72.
Derive an expression for the position of centre of mass of a system of n particles and for continuous mass distribution.
Answer:
System of n particles;
i. Consider a system of n particles of masses m₁, m₂, …, m_n having position vectors \(\overrightarrow{\mathrm{r}_{1}}\), \(\overrightarrow{\mathrm{r}_{2}}\),….., \(\overrightarrow{\mathrm{r}_{n}}\) from the origin O.
The total mass of the system is,

Centre of mass for n particles

ii. Position vector \(\vec{r}\) of their centre of mass from the same origin is then given by

iii. If the origin is at the centre of mass, \(\overrightarrow{\mathbf{r}}\) = 0
∴ \(\sum_{1}^{\mathrm{n}} \mathrm{m}_{\mathrm{i}} \overrightarrow{\mathrm{r}}_{\mathrm{i}}\) = 0,

iv. In this case, \(\sum_{1}^{n} m_{i} \vec{r}_{i}\) gives the moment of masses (similar to moment of force) about the centre of mass.

v. If (x₁, x₂, …… x_n), (y₁, y₂, …..y_n), (z₁, z₂, …. z_n) are the respective x, y and z – coordinates of (r₁, r₂,…….. r_n) then x,y and z – coordinates of the centre of mass are given by,

vi. Continuous mass distribution: For a continuous mass distribution with uniform density, the position vector of the centre of mass is given by,
r = \(\frac{\int \vec{r} d m}{\int d m}=\frac{\int \vec{r} d m}{M}\)
Where \(\int \mathrm{dm}=\mathrm{M}\) is the total mass of the object.

vii. The Cartesian coordinates of centre of mass are

Question 73.
State the expression for velocity of the centre of mass of a system of n particles and for continuous mass distribution.
Answer:
Let v₁, v₂,…..v_n be the velocities of a system of point masses m₁, m₂, … m_n. Velocity of the centre of mass of the system is given by

x, y and z components of \(\overrightarrow{\mathbf{v}}\) can be obtained similarly.
For continuous distribution, \(\overrightarrow{\mathrm{v}}_{\mathrm{cm}}\) = \(\frac{\int \vec{v} \mathrm{dm}}{\mathrm{M}}\)

Question 74.
State the expression for acceleration of the centre of mass of a system of n particles and for continuous mass distribution.
Answer:
Let a₁, a₂,…. a_n be the accelerations of a system of point masses m₁, m₂ … m_n.
Acceleration of the centre of mass of the system is given by

x, y and z components of can be obtained similarly.
For continuous distribution, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}\) = \(\frac{\int \vec{a} d m}{\mathrm{M}}\)

Question 75.
State the characteristics of centre of mass.
Answer:

1. Centre of mass is a hypothetical point at which entire mass of the body can be assumed to he concentrated.
2. Centre of mass is a location, not a physical quantity.
3. Centre of mass is particle equivalent of a given object for applying laws of motion.
4. Centre of mass is the point at which applied force causes only linear acceleration and not angular acceleration.
5. Centre of mass is located at the centroid, for a rigid body of uniform density.
6. Centre of mass is located at the geometrical centre, for a symmetric rigid body of uniform density.
7. Location of centre of mass can be changed only by an external unbalanced force.
8. Internal forces (like during collision or explosion) never change the location of centre of mass.
9. Position of the centre of mass depends only upon the distribution of mass, however, to describe its location we may use a coordinate system with a suitable origin.
10. For a system of particles, the centre of mass need not coincide with any of the particles.
11. While balancing an object on a pivot, the line of action of weight must pass through the centre of mass and the pivot. Quite often, this is an unstable equilibrium.
12. Centre of mass of a system of only two particles divides the distance between the particles in an inverse ratio of their masses, i.e., it is closer to the heavier mass.
13. Centre of mass is a point about which the summation of moments of masses in the system is zero.
14. If there is an axial symmetry for a given object, the centre of mass lies on the axis of symmetry.
15. If there are multiple axes of symmetry for a given object, the centre of mass is at their point of intersection.
16. Centre of mass need not be within the body.
    Example: jumper doing fosbury flop.

Question 76.
The mass of moon is 0.0 123 times the mass of the earth and separation between them is 3.84 × 10⁸ m. Determine the location of C.M as measured from the centre of the earth.

Solution:

Answer:
The location of centre of mass as measured from the center of the earth is 4.67 × 10⁶ m.

Question 77.
Three particles of masses 3 g, 5 g and 8 g are situated at point (2, 2, 2), (-3, 1, 4) and (-1, 3, -2) respectively. Find the position vector of their centre of mass.
Solution:
m₁ = 3 g, m₂ = 5 g, m₃ = 8 g, x₁ = 2, y₁ = 2, z₁ = 2
x₂ = -3, y₂ = 1, z₂ = 4,
x₃ = -1, y₃ = 3, z₃ = -2
Let (X, Y, Z) be co-ordinates of C.M., then

The co-ordinates of the C.M. are \(\left(-\frac{17}{16}, \frac{35}{16}, \frac{5}{8}\right)\)
Answer:
The position vector of the C.M. is \(-\frac{17}{16} \hat{\mathbf{i}}+\frac{35}{16} \hat{\mathbf{j}}+\frac{5}{8} \hat{\mathbf{k}}\)

Question 78.
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. (NCERT)
Solution:

Claculation: From formula

Answer:
The location of centre of mass from the nucleus of hydrogen atom is 1.235 A.

Question 79.
Three thin walled uniform hollow spheres of radii 1 cm, 2 cm and 3 cm are so located that their centres are on the three vertices of an equilateral triangle ABC having each side 10 cm. Determine centre of mass of the system.

Solution:
Mass of a thin walled uniform hollow sphere is proportional to its surface area. (as density is constant) hence proportional to r².

Thus, if mass of the sphere at A is m_A = m, then m_B = 4m and m_C = 9m. By symmetry of the spherical surface, their centres of mass are at their respective centres, i.e., at A, B and C. Let us choose the origin to be at C, where the largest mass 9m is located and the point B with mass 4m on the positive x-axis. With this, the co-ordinates of C are (0, 0) and that of B are (10, 0). If A of mass m is taken in the first quadrant, its co-ordinates will be \([5,5 \sqrt{3}]\)

Question 80.
Locate the centre of mass of three particles of mass m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg at the corner of an equilateral triangle of each side of 1 m.
Solution:



Answer:
The centre of mass of the system of three particles lies at \(\left(\frac{2}{3} m, \frac{\sqrt{3}}{6} m\right)\) with respect to the particle of mass 1 kg as the origin.

Question 81.
A letter ‘E’ is prepared from a uniform cardboard with shape and dimensions as shown in the figure. Locate its centre of mass.

Solution:
As the sheet is uniform, each square can be taken to be equivalent to mass m concentrated at its respective centre. These masses will then be at the points labelled with numbers 1 to 10, as shown in figure. Let us select the origin to be at the left central mass m₅, as shown and all the co-ordinates to be in cm.

By symmetry, the centre of mass of m₁, m₂ and m₃ will be at m₂ (1, 2) having effective mass 3m. Similarly, effective mass 3m due to m₈, m₉ and m₁₀ will be at m₉ (1, -2). Again, by symmetry, the centre of mass of these two (3m each) will have co-ordinates (1, 0). Mass m₆ is also having co-ordinates (1, 0). Thus, the
effective mass at (1, 0) is 7m.

Using symmetry for m4, m5 and m7, there will be effective mass 3m at the origin (0, 0). Thus, effectively, 3m and 7m are separated by 1 cm along X-direction. Y-coordinate is not required.

Question 82.
A hole of radius r is cut from a uniform disc of radius 2r. Centre of the hole is at a distance r from centre of the disc. Locate centre of mass of the remaining part of the disc.
Solution:
Before cutting the hole, c.în. of the full disc was at its centre. Let this be our origin O. Centre of mass of the cut portion is at its centre D. Thus, it is at a distance x₁ = r form the origin. Let C be the centre of mass of the
remaining disc, which will he on the extension of the line DO at a distance x₂ = x from the origin. As the disc is uniform, mass of any of its part is proportional to the area of that part.

Thus, if m is the mass of the cut disc, mass of the entire disc must be 4m and mass of the remaining disc will be 3m.

Alternate method: (Using negative mass):

Let \(\overrightarrow{\mathrm{R}}\) be the position vector of the centre of mass of the uniform disc of mass M. Mass m is with centre of mass at position vector \(\overrightarrow{\mathrm{r}}\) from the centre of the disc be cut out from the complete disc. Position vector of the centre of mass of the remaining disc is then given by

Question 83.
Define centre of gravity of a body. Under what conditions the centre of gravity and centre of mass coincide?
Answer:

1. centre of gravity of a body is the point around which the resultant torque due to force of gravity on the body is zero.
2. The centre of mass coincides with centre of gravity when the body is in a uniform gravitational field.

Question 84.
Explain how to find the location of centre of mass or centre of gravity of a laminar object.
Answer:

1. A laminar object is suspended from a rigid support at two orientations.
2. Lines are drawn on the object parallel to the plumb line as shown in the figure.
3. Plumb line is always vertical, i.e., parallel to the line of action of gravitational force.
4. Intersection of the lines drawn is then the point through which line of action of the gravitational force passes for any orientation. Thus, it gives the location of the c.g. or c.m.
   

Question 85.
Why do cricketers wear helmet and pads while playing? Is it related with physics?
Answer:
Helmet and pads used by cricketers protects the head, using principles of physics.

1. The framing between the interior of the helmet and pads increases the time over which the impulse acts on the head resulting into reduction of force.
   (Impulse = force × time = constant)
2. The pads spread the force applied by the ball over a wider area reducing pressure at a point.

Question 86.
The diagram shows four objects placed on a flat surface.

The centre of mass of each object is marked M. Which object is likely to fall over?
Answer:
Object C will fall because its centre of mass is not exactly at centre, in turn applying more force on one side of the object resulting in to unbalance force. Whereas, in other objects center of mass is exactly at center resulting into zero rotational or translational motion maintaining their equilibrium.

Question 87.
A boy is about to close a large door by applying force at A and B as shown. State with a reason, which of the two positions, A or B, will enable him to close the door with least force.

Answer:
Boy has to apply more force at point B as compared to point A, because point B is at least distance from the hinge of the door while point A is at maximum distance from hinge of the door. Opening of door applies a moment, which is. given by,
M = F × perpendicular distance
F ∝ \(\frac{1}{\text { distance }}\)
More is the distance from the axis of rotation less will be the force.

Question 88.
How is a seat belt useful for safety?
Answer:
When car hits another car or an object with high speed it applies a high impulse on the driver and due to inertia driver tends to move in forward direction towards the steering. Seat belts spreads the force over large area of the body and holds the driver and protects him from crashing at the steering.

Question 89.
According to Newton’s third law of motion for every action there is equal and opposite reaction, why two equal and opposite forces don’t cancel each other?
Answer:
Forces of action and reaction always act on different bodies, hence they never cancel each other.

Question 90.
Linear momentum depends on frame of reference, but principle of conservation of linear momentum is independent of frame of reference. Why?
Answer:
Observers in different frame find different values of linear momentum of a system, hence linear momentum depends upon frame of reference, but each would observe that the value of linear momentum does not change with time (provided the system is isolated), hence principle of conservation of linear momentum is independent of frame of reference.

Question 91.
Multiple choice Questions

Question 1.
A body of mass 2 kg moving on a horizontal surface with initial velocity of 4 m s^-1 comes to rest after two seconds. If one wants to keep this body moving on the same surface with a velocity of 4 m s^-1, the force required is
(A) 2 N
(B) 4 N
(C) 0
(D) 8 N
Answer:
(B) 4 N

Question 2.
What force will change the velocity of a body of mass 1 kg from 20 m s^-1 to 30 m s^-1 in two seconds?
(A) 1N
(B) 5 N
(C) 10 N
(D) 25 N
Answer:
(B) 5 N

Question 3.
A force of 5 newton acts on a body of weight 9.80 newton. What is the acceleration produced in m/s²?
(A) 0.51
(B) 1.96
(C) 5.00
(D) 49.00
Answer:
(C) 5.00

Question 4.
A body of mass m strikes a wall with velocity v and rebounds with the same speed. Its change in momentum is
(A) 2 mv
(B) mv/2
(C) – mv
(D) Zero
Answer:
(A) 2 mv

Question 5.
A force of 6 N acts on a body of mass 1 kg initially at rest and during this time, the body attains a velocity of 30 m/s. The time for which the force acts on a body is
(A) 10 second
(B) 8 second
(C) 7 second
(D) 5 second
Answer:
(D) 5 second

Question 6.
A bullet of mass 10 g is fired from a gun of mass 1 kg with recoil velocity of gun = 5 m/s. The muzzle velocity will be
(A) 30 km/min
(B) 60 km/min
(C) 30 m/s
(D) 500 m/s
Answer:
(D) 500 m/s

Question 7.
The velocity of rocket with respect to ground is v₁ and velocity of gases ejecting from rocket with respect to ground is v₂ Then velocity of gases with respect to rocket is given by
(A) v₂
(B) v₁ + v₂
(C) v₁ × v₂
(D) v₁
Answer:
(B) v₁ + v₂

Question 8.
Two bodies A and B of masses 1 kg and 2 kg moving towards each other with velocities 4 m/s and 1 m/s suffers a head on collision and stick together. The combined mass will
(A) move in direction of motion of lighter mass.
(B) move in direction of motion of heavier mass.
(C) not move.
(D) move in direction perpendicular to the line of motion of two bodies.
Answer:
(A) move in direction of motion of lighter mass.

Question 9.
Which of the following has maximum momentum?
(A) A 100 kg vehicle moving at 0.02 m s^-1.
(B) A 4 g weight moving at 1000 cm s^-1’.
(C) A 200 g weight moving with kinetic energy of 10^-6 J
(D) A 200 g weight after falling through one kilometre.
Answer:
(D) A 200 g weight after falling through one kilometre.

Question 10.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. What is conserved?
(A) Momentum alone.
(B) K.E. alone.
(C) Momentum and K.E. both.
(D) P.E. alone.
Answer:
(A) Momentum alone.

Question 11.
The force exerted by the floor of an elevator on the foot of a person standing there, is more than his weight, if the elevator is
(A) going down and slowing down.
(B) going up and speeding up.
(C) going up and slowing down.
(D) either (A) and (B).
Answer:
(D) either (A) and (B).

Question 12.
If E, G and N represents the magnitudes of electromagnetic, gravitational and nuclear forces between two electrons at a given separation, then,
(A) N = E = G
(B) E < N < G (C) N > G < E (D) E > G > N
Answer:
(D) E > G > N

Qestion 13.
For an inelastic collision, the value of e is
(A) greater than 1
(B) less than 1
(C) equal to 1
(D) none of these
Answer:
(B) less than 1

Question 14.
A perfect inelastic body collides head on with a wall with velocity y. The change in momentum is
(A) mv
(B) 2mv
(C) zero
(D) none of these.
Answer:
(A) mv

Question 15.
Two masses ma and rnb moving with velocitics v_a and v_b in opposite direction collide
elastically and after the collision m_a and m_b move with velocities v_b and v_a respectively. Then the ratio m_am_b is
(A) \(\frac{v_{a}-v_{b}}{v_{a}+v_{b}}\)
(B) \(\frac{\mathrm{m}_{\mathrm{a}}+\mathrm{m}_{\mathrm{b}}}{\mathrm{m}_{\mathrm{a}}}\)
(C) 1
(D) \(\frac{1}{2}\)
Answer:
(C) 1

Question 16.
The frictional force acts _____.
(A) in direction of motion
(B) against the direction of motion
(C) perpendicular to the direction of motion
(D) at any angle to the direction of motion
Answer:
(B) against the direction of motion

Question 17.
A marble of mass X collides with a block of mass Z, with a velocity Y. and sticks to it. The final velocity of the system is
(A) \(\frac{\mathrm{Y}}{\mathrm{X}+\mathrm{Y}} \mathrm{Y}\)
(B) \(\frac{X}{X+Z} Y\)
(C) \(\frac{X+Y}{Z}\)
(D) \(\frac{X+Z}{X}\)
Answer:
(B) \(\frac{X}{X+Z} Y\)

Question 18.
Two balls lying on the same plane collide. Which of the following will be always conserved?
(A) heat
(B) velocity
(C) kinetic energy
(D) linear momentum.
Answer:
(D) linear momentum.

Question 19.
A body is moving with uniform velocity of 50 km h^-1, the force required to keep the body in motion in SI unit is
(A) zero
(B) 10
(C) 25
(D) 50
Answer:
(A) zero

Question 20.
A coolie holding a suitcase on his head of 20 kg and travels on a platform. then work done in joule by the coolie is
(A) 198
(B) 98
(C) 49
(D) zero
Answer:
(D) zero

Question 21.
Out of the following forces, which force is non-conservative?
(A) gravitational
(B) electrostatic
(C) frictional
(D) magnetic
Answer:
(C) frictional

Question 22.
The work done in conservative force is ____.
(A) negative
(B) zero
(C) positive
(D) infinite
Answer:
(B) zero

Question 23.
The angle between the line of action of force and displacement when no work done (in degree) is
(A) zero
(B) 45
(C) 90
(D) 120
Answer:
(C) 90

Question 24.
If the momentum of a body is doubled, its KE. increases by
(A) 50%
(B) 300%
(C) 100%
(L)) 400%
Answer:
(B) 300%

Question 25.
In perfectly inelastic collision, which is conserved?
(A) P.E. only
(B) K.E. only
(C) momentum only
(D) K.E. and momentum
Answer:
(C) momentum only

Question 26.
In case of elaine collision, which s
(A) Momentum and K.E. is conserved.
(B) Momentum conserved and K.E. not conserved,
(C) Momentum not conserved and K.E. conserved.
(D) Momentum and K.E. both not conserved,
Answer:
(A) Momentum and K.E. is conserved.

Question 27.
Pseudo force is true only in
(A) frame of reference which is at rest
(B) inertial frame of reference.
(C) frame of reference moving with constant velocity.
(D) non-inertial frame of reference
Answer:
(D) non-inertial frame of reference

Question 28.
A men weighing 90kg carries a stone of 20 kg to the top of the building 30m high. The work done by hint is (g = 9.8 m/s²)
(A) 80 J
(B) 100 J
(C) 980 J
(D) 29,400 J
Answer:
(D) 29,400 J

Question 29.
A weight lifter is holding a weight of 100 kg on his shoulders for 45 s, the amount of work done by him in joules is
(A) 4500
(B) 100
(C) 45
(D) zero
Answer:
(D) zero

Question 30.
If m is the mass of a body and E its K.E..then its linear momentum is
(A) \(\mathrm{m} \sqrt{\mathrm{E}}\)
(B) \(2 \sqrt{\mathrm{m}} \mathrm{E}\)
(C) \(\sqrt{m} E\)
(D) \(\sqrt{2 \mathrm{mE}}\)
Answer:
(D) \(\sqrt{2 \mathrm{mE}}\)

Question 31.
Torque applied is masimum when the angle between the directions of \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{F}}\) is
(A) 90°
(B) 180°
(C) 0°
(D) 45°
Answer:
(A) 90°

Question 92.
A particle moving with velocity \(\vec{v}\) is acted by three forces shown by the vector triangle PQR. The velocity of the particle will:
(A) remain constant
(B) change according to the smallest force \(\overrightarrow{\mathrm{QR}}\)
(C) increase
(D) decrease

Hint: As the three forces acting on a particle represents a triangle (i.e., a closed loop)
∴ F_net = 0
∴ m \(\vec{a}\) = 0
∴ m\(\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}\)
∴ v remains constant
Answer:
(A) remain constant

Question 93.
A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is:
(A) 25J
(B) 20J
(C) 30J
(D) SJ
Hint: Work done by variable force, W = \(\int_{y_{\text {initial }}}^{y_{\text {final }}} \mathrm{Fdy}\)


Answer:
(A) 25J

Question 94.
Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:
(A) \(\frac{4}{9}\)
(B) \(\frac{5}{9}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{8}{9}\)
Hint:
Fractional loss of K.E. of colliding bodies,

Answer:
(D) \(\frac{8}{9}\)

Question 95.
An object of mass 500 g, initially at rest, is acted upon by a variable force whose X-component varies with X in the manner shown. The velocities of the object at the points X = 8 m and X = 12 m, would have the respective values of

(A) 18m/s and 20.6 m/s
(B) 18 m/s and 24.4 m/s
(C) 23 m/s and 24.4 m/s
(D) 23 m/s and 20.6 m/s
Hint: From work-energy theorem
∆ K.E. = work = area under F-x graph
From x = 0 to x = 8m
\(\frac{1}{2} \mathrm{mv}^{2}\) = (5 × 20) + (3 × 10)
∴ \(\frac{1}{2} \mathrm{mv}^{2}\) = 100 + 30

Answer:
(D) 23 m/s and 20.6 m/s

Question 96.
The centre of mass of two particles system lies
(A) at the midpoint on the line joining the two particles.
(B) at one end of line joining the two particles.
(C) on the line perpendicular in the line joining two particles.
(D) on the line joining the Iwo particles.
Answer:
(D) on the line joining the two particles.

Question 97.
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and θ for the block to remain stationary on the wedge is

Hint:

The mass of block is given to be m. It will remain stationary if forces acting on it are in equilibrium i.e., ma cos θ = mg sin θ
Here, ma = Pseudo force on block.
∴ a = g tan θ
Answer:
a = g tan θ

Question 98.
A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision, When the initial velocity of the lighter block is v, then the value of coefficient of restitution
(e) will
(A) 0.5
(B) 0.25
(C) 0.8
(D) 0.4
Hint:
Given: m₁ = m, m₂ = 4m, u₁ = v, u₂ = 0, v₁ = 0 According to law of conservation of momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
mv + 4m × 0 = m × 0 + 4mv₂

Answer:
(B) 0.25

Question 99.
In a collinear collision, a particle with an initial speed v₀ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:

Hint:
According to law of conservation of momentum,
mv₀ = mv₁ + mv₂

Answer:
(D) \(\sqrt{2} \mathbf{v}_{0}\)

Question 100.
The mass of a hydrogen molecule is 3.32 × 10^-27 kg. If 10²³ hydrogen molecules strike, per second, a fixed wall of area 2 cm² at an angle of 45° to the normal and rebound elastically with a speed of 10³ m/s, then the pressure on the wall is nearly:
(A) 2.35 × 10² N/m²
(B) 4.70 × 10² N/m²
(C) 2.35 × 10³ N/m²
(D) 4.70 × 10³ N/m²
Hint:

Answer:
(C) 2.35 × 10³ N/m²

Question 101.
A bomb at rest explodes into 3 parts of same mass. The momentum of two parts is -3P\(\hat{\mathrm{i}}\) and 2P\(\hat{\mathrm{j}}\) respectively. The magnitude of momentum of the third part is
(A) P
(B) \(\sqrt{5} \mathrm{P}\)
(C) \(\sqrt{11} \mathrm{P}\)
(D) \(\sqrt{13} \mathrm{P}\)

Answer:
(D) \(\sqrt{13} \mathrm{P}\)

Question 102.
A sphere of mass ‘m’ moving with velocity V collides head-on on another sphere of same mass which is at rest. The ratio of final velocity of second sphere to the initial velocity of the first sphere is (e is coefficient of restitution and collision is inelastic)

Hint:
Initial momentum = mv
Final momentum = mv₁ + mv₂

Answer:
(C) \(\frac{\mathrm{e}+1}{2}\)

Question 103.
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively:

Hint:
Tension in spring before cutting the strip,

Answer:
(B) \(\frac{\mathrm{g}}{3}\), g

Question 104.
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be
(A) 9 J
(B) 18 J
(C) 4.5 J
(D) 22 J
Hint:
F = 6t
m = 1 kg
∴ a = 6t

Answer:
(C) 4.5 J

Question 105.
Consider a drop of rain water having mass Ig falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take ‘g’ constant with a value 10 m/s². The work done by the
(i) gravitational force and the
(ii) resistive force of air is:
(A) (i) -10 J
(ii) -8.25 J
(B) (i) 1.25 J
(ii) -8.25 J
(C) (i) 100 J
(ii) 8.75 J
(D) (i) 10 J
(ii) -8.75 J
Hint: Work done by gravitation force is given by (W_g)

Answer:
(D) (i) 10 J
(ii) -8.75 J

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 2 Mathematical Methods Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 2 Mathematical Methods

Question 1.
Explain representation of a vector graphically and symbolically.
Answer:

1. Graphical representation:
   A vector is graphically represented by a directed line segment or an arrow.
   eg.: displacement of a body from P to Q is represented as P → Q.
2. Symbolic representation:
   Symbolically a vector is represented by a single letter with an arrow above it, such as \(\overrightarrow{\mathrm{A}}\). The magnitude of the vector \(\overrightarrow{\mathrm{A}}\) is denoted as |A| or | \(\overrightarrow{\mathrm{A}}\) | or A.

Question 2.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector?
Answer:

1. For a physical quantity, only having magnitude and direction is not a sufficient condition to be a vector.
2. A physical quantity also has to obey vectors law of addition to be termed as vector.
3. Hence, anything that has magnitude and direction is not necessarily a vector.
   Example: Though current has definite magnitude and direction, it is not a vector.

Question 3.
Define and explain the following terms:
i. Zero vector (Null vector)
ii. Resultant vector
iii. Negative vectors
iv. Equal vectors
v. Position vector
Answer:
i. Zero vector (Null vector):
A vector having zero magnitude and arbitrary direction is called zero vector. It is denoted as \(\overrightarrow{0}\).
Example: Velocity vector of stationary particle, acceleration vector of a body moving with uniform velocity.

ii. Resultant vector:
The resultant of two or more vectors is defined as that single vector, which produces the same effect as produced by all the vectors together.

iii. Negative vectors:
A negative vector of a given vector is a vector of the same magnitude but opposite in direction to that of the given vector.
Negative vectors are antiparallel vectors.
In figure, \(\vec{b}\) = – \(\vec{a}\)

iv. Equal vectors:
Two vectors A and B representing same physical quantity are said to be equal if and only if they have the same magnitude and direction.

In the given figure |\(\overrightarrow{\mathrm{P}}\)| = |\(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{R}}\)| = |\(\overrightarrow{\mathrm{S}}\)|

v. Position vector:
A vector which gives the position of a particle at a point with respect to the origin of chosen co-ordinate system is called position vector.

In the given figure \(\overrightarrow{\mathrm{OP}}\) represents position vector of \(\vec{P}\) with respect to O.

Question 4.
Whether the resultant of two vectors of unequal magnitude be zero?
Answer:
The resultant of two vectors of different magnitude cannot give zero resultant.

Question 5.
Define unit vector and give its physical significance.
Answer:
Unit vector: A vector having unit magnitude in a given direction is called a unit vector in that direction.
If \(\vec{p}\) is a non zero vector (P ≠ 0) then the unit vector \(\hat{\mathrm{u}}_{\mathrm{p}}\) in the direction of \(\overrightarrow{\mathrm{P}}\) is given by,
\(\hat{\mathrm{u}}_{\mathrm{p}}\) = \(\frac{\overrightarrow{\mathrm{P}}}{\mathrm{P}}\)
∴ \(\overrightarrow{\mathrm{P}}\) = \(\hat{u}_{p} P\)

Significance of unit vector:

i. The unit vector gives the direction of a given vector.

ii. Unit vector along X, Y and Z direction of a rectangular (three dimensional) coordinate is represented by \(\hat{\mathrm{i}}\), \(\hat{\mathrm{j}}\) and \(\hat{\mathrm{k}}\) respectively Such that \(\hat{\mathbf{u}}_{x}\) = \(\hat{\mathrm{i}}\), \(\hat{\mathbf{u}}_{y}\) = \(\hat{\mathrm{j}}\) and \(\hat{\mathbf{u}}_{z}\) = \(\hat{\mathrm{k}}\)
This gives \(\hat{\mathrm{i}}\) = \(\frac{\overrightarrow{\mathrm{X}}}{\mathrm{X}}\), \(\hat{\mathrm{j}}\) = \(\frac{\overrightarrow{\mathrm{Y}}}{\mathrm{X}}\) and \(\hat{\mathrm{k}}\) = \(\frac{\overrightarrow{\mathrm{Z}}}{\mathrm{Z}}\)

Question 6.
Explain multiplication of a vector by a scalar.
Answer:

1. When a vector \(\overrightarrow{\mathrm{A}}\) is multiplied by a scalar ‘s’, it becomes ‘s\overrightarrow{\mathrm{A}}’ whose magnitude is s times the magnitude of \(\overrightarrow{\mathrm{A}}\).
2. The unit of \(\overrightarrow{\mathrm{A}}\) is different from the unit of ‘s \(\overrightarrow{\mathrm{A}}\)’.
   For example,
   If \(\overrightarrow{\mathrm{A}}\) = 10 newton and s = 5 second, then s\(\overrightarrow{\mathrm{A}}\) = 10 newton × 5 second = 50 Ns.

Question 7.
Explain addition of vectors.
Answer:

1. The addition of two or more vectors of same type gives rise to a single vector such that the effect of this single vector is the same as the net effect of the original vectors.
2. It is important to note that only the vectors of the same type (physical quantity) can be added.
3. For example, if two vectors, \(\overrightarrow{\mathrm{P}}\) = 3 unit and \(\overrightarrow{\mathrm{Q}}\) = 4 unit are acting along the same line, then they can be added as, |\(\overrightarrow{\mathrm{R}}\)| = |\(\overrightarrow{\mathrm{P}}\)| + |\(\overrightarrow{\mathrm{Q}}\)|
   |\(\overrightarrow{\mathrm{R}}\)| = 3 + 4 = 7
   
   [Note: When vectors are not in the same direction, then they can be added using triangle law of vector addition.]

Question 8.
State true or false. If false correct the statement and rewrite.
It is possible to add two vectors representing physical quantities having different dimensions.
Answer:
False.
It is not possible to add two vectors representing physical quantities having different dimensions.

Question 9.
Explain subtraction of vectors.
Answer:

1. When two vectors are anti-parallel (in the opposite direction) to each other, the magnitude
2. It is important to note that only vectors of the same type (physical quantity) can be subtracted.
3. For example, if two vectors \(\overrightarrow{\mathrm{P}}\) = 3 unit and \(\overrightarrow{\mathrm{Q}}\) = 4 unit are acting in opposite direction, they are subtracted as, |\(\overrightarrow{\mathrm{R}}\)| = ||\(\overrightarrow{\mathrm{P}}\)| – |\(\overrightarrow{\mathrm{Q}}\)||
   = |3 – 4| = 1 unit, directed along \(\overrightarrow{\mathrm{Q}}\)
   

Question 10.
How can resultant of two vectors of a type inclined with each other be determined?
Answer:
When two vectors of a type are inclined with each other, their resultant can be determined by using triangle law of vector addition.

Question 11.
What is triangle law of vector addition?
Answer:
Triangle law of vector addition:
If two vectors describing the same physical quantity are represented in magnitude and direction, by the two sides of a triangle taken in order, then their resultant is represented in magnitude and direction by the third side of the triangle drawn in the opposite sense, i.e., from the starting point (tail) of the first vector to the end point (head) of the second vector.

Let \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) be the two vectors of same type taken in same order as shown in figure.
∴ Resultant vector will be given by third side taken in opposite order, i.e., \(\overline{\mathrm{OA}}\) + \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{OB}}\)
∴ \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{R}}\)

Question 12.
Using triangle law of vector addition, explain the process of adding two vectors which are not lying in a straight line.
Answer:
i. Two vectors in magnitude and direction are drawn in a plane as shown in figure (a)
Let these vectors be \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\)

ii. Join the tail of \(\overrightarrow{\mathrm{Q}}\) to head of \(\overrightarrow{\mathrm{P}}\) in the given direction. The resultant vector will be the line which is obtained by joining tail of \(\overrightarrow{\mathrm{P}}\) to head of \(\overrightarrow{\mathrm{Q}}\) as shown in figure (b).

iii. If \(\overrightarrow{\mathrm{R}}\) is the resultant vector of \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) then using triangle law of vector addition, we have, \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)

Question 13.
Is it possible to add two velocities using triangle law?
Answer:
Yes, it is possible to add two velocities using triangle law.

Question 14.
Explain, how two vectors are subtracted. Find their resultant by using triangle law of vector addition.
Answer:

1. Let \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) be the two vectors in a plane as shown in figure (a).
   
2. To subtract \(\overrightarrow{\mathrm{Q}}\) from \(\overrightarrow{\mathrm{P}}\), vector \(\overrightarrow{\mathrm{Q}}\) is reversed so that we get the vector –\(\overrightarrow{\mathrm{Q}}\) as shown in figure (b).
3. The resultant vector is obtained by –\(\overrightarrow{\mathrm{R}}\) joining tail of \(\overrightarrow{\mathrm{P}}\) to head of – \(\overrightarrow{\mathrm{Q}}\) as shown in figure (c).
4. From triangle law of vector addition, \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + (-\(\overrightarrow{\mathrm{Q}}\)) = \(\overrightarrow{\mathrm{P}}\) – \(\overrightarrow{\mathrm{Q}}\)

Question 15.
Prove that: Vector addition is commutative.
Answer:
Commutative property of vector addition:
According to commutative property, for two
vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\), \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{p}}\)

Proof:

i. Let two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) be represented in magnitude and direction by two sides \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{AB}}\) respectively.

ii. Complete a parallelogramOABC such that
\(\overrightarrow{\mathrm{OA}}\) = \(\overrightarrow{\mathrm{CB}}\) = \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{OC}}\) = \(\overrightarrow{\mathrm{Q}}\) then join OB.

iii. In △OAB, \(\overrightarrow{\mathrm{OA}}\) + \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{OB}}\)
(By triangle law of vector addition)
∴ \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{R}}\) … (1)
In △OCB, \(\overrightarrow{\mathrm{OC}}\) + \(\overrightarrow{\mathrm{CB}}\) = \(\overrightarrow{\mathrm{OB}}\)
(By triangle law of vector addition)
∴ \(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{P}}\) = \(\overrightarrow{\mathrm{R}}\) … (2)

iv. From equation (1) and (2),
\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{P}}\)
Hence, addition of two vectors obeys commutative law.

Question 16.
Prove that: Vector addition is associative.
Answer:
Associative property of vector addition:
According to associative property, for three vectors \(\overrightarrow{\mathrm{P}}\), \(\overrightarrow{\mathrm{Q}}\) and \(\overrightarrow{\mathrm{R}}\),
(\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)) + \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + (\(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{R}}\))

Proof:


On comparing, equation (2) and (4), we get,
(\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)) + \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + (\(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{R}}\))
Hence, associative law is proved.

Question 17.
State true or false. If false correct the statement and rewrite.
The subtraction of given vectors is neither commutative nor associative.
Answer:
True.

Question 18.
State and prove parallelogram law of vector addition and determine magnitude and direction of resultant vector.
Answer:

i. Parallelogram law of vector add addition;
If two vectors of same type starting from the same point (tails cit the same point), are represented in magnitude and direction by the two adjacent sides of a parallelogram then, their resultant vector is given in magnitude and direction, by the diagonal of the parallelogram starting from the same point.

ii. Proof:

a. Consider two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) of the same type, with their tails at the point O’ and θ’ is the angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) as shown in the figure below.

b. Join BC and AC to complete the parallelogram OACB, with \(\overline{\mathrm{OA}}\) = \(\overrightarrow{\mathrm{P}}\) and \(\overline{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{Q}}\) as the adjacent sides. We have to prove that diagonal \(\overline{\mathrm{OC}}\) = \(\overrightarrow{\mathrm{R}}\), the resultant of sum of the two given vectors.

c. By the triangle law of vector addition, we have,
\(\overrightarrow{\mathrm{OA}}\) + \(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{OC}}\) … (1)
As \(\overrightarrow{\mathrm{AC}}\) is parallel to \(\overrightarrow{\mathrm{OB}}\),
\(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{OB}}\) = \(\overrightarrow{\mathrm{Q}}\)
Substituting \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OC}}\) in equation (1) we have,
\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{R}}\)
Hence proved.

iii. Magnitude of resultant vector:

a. To find the magnitude of resultant vector \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{OC}}\), draw a perpendicular from C to meet OA extended at S.

c. Using Pythagoras theorem in right angled triangle, OSC
(OC)² = (OS)² + (SC)²
= (OA + AS)² + (SC)²
∴ (OC)² = (OA)² + 2(OA).(AS) + (AS²) + (SC)² . . . .(4)

d. From right angle trianle ASC,
(AS)² + (SC)² = (AC)² …. (5)

e. From equation (4) and (5), we get
(OC)² = (OA)² + 2(OA) (AS) + (AC)²
… .(6)

f. Using (2) and (6), we get
(OC)² = (OA)² + (AC)² + 2(OA)(AC) cos θ
∴ R² = P² + Q² + 2 PQ cos θ
∴ R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos \theta}\) ….(7)
Equation (7) gives the magnitude of resultant vector \(\overrightarrow{\mathrm{R}}\).

iv. Direction of resultant vector:
To find the direction of resultant vector \(\overrightarrow{\mathrm{R}}\), let \(\overrightarrow{\mathrm{R}}\) make an angle α with \(\overrightarrow{\mathrm{P}}\).

Equation (9) represents direction of resultant vector.
[Note: If β is the angle between \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{Q}}\), it can be similarly derived that

Question 19.
Complete the table for two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) inclined at angle θ.

Answer:

Question 20.
The diagonal of the parallelogram made by two vectors as adjacent sides is not passing through common point of two vectors. What does it represent?
Answer:
The diagonal of the parallelogram made by two vectors as adjacent sides not passing through common point of two vectors represents triangle law of vector addition.

Question 21.
If | \(\overrightarrow{\mathbf{A}}\) + \(\overrightarrow{\mathbf{B}}\) | = | \(\overrightarrow{\mathbf{A}}\) – \(\overrightarrow{\mathbf{B}}\) | then what can be the
angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) ?
Answer:
Let θ be the angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\), then

Thus, if |\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)| = |\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\) |, then vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) must be at right angles to each other.

Question 22.
Express vector \(\overrightarrow{\mathbf{A C}}\) in terms of vectors \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{C B}}\) shown in the following figure.

Solution:
Using the triangle law of addition of vectors,
\(\overrightarrow{\mathbf{A C}}\) + \(\overrightarrow{\mathbf{C B}}\) = \(\overrightarrow{\mathbf{A B}}\)
∴\(\overrightarrow{\mathbf{A C}}\) = \(\overrightarrow{\mathbf{A B}}\) – \(\overrightarrow{\mathbf{C B}}\)

Question 23.
From the following figure, determine the resultant of four forces \(\overrightarrow{\mathbf{A}}_{1}\), \(\overrightarrow{\mathbf{A}}_{2}\), \(\overrightarrow{\mathbf{A}}_{3}\), \(\overrightarrow{\mathbf{A}}_{4}\).

Solution:
Join \(\overrightarrow{\mathrm{OB}}\) to complete ∆OAB as shown in figure below

Now, using triangle law of vector addition,
\(\overrightarrow{\mathrm{OB}}\) = \(\overrightarrow{\mathrm{OA}}\) + \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{A}}_{1}\) + \(\overrightarrow{\mathrm{A}}_{2}\)
Join \(\overrightarrow{\mathrm{OC}}\) to complete triangle OBC as shown figure below
Similarly, \(\overrightarrow{\mathrm{OC}}\) = \(\overrightarrow{\mathrm{OB}}\) + \(\overrightarrow{\mathrm{BC}}\) = \(\overrightarrow{\mathrm{A}}_{1}\) + \(\overrightarrow{\mathrm{A}}_{2}\) + \(\overrightarrow{\mathrm{A}}_{3}\)

Answer:
\(\overrightarrow{O D}\) is the resultant of the four vectors.

Question 24.
Find the vector that should be added to the sum of (2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\)) and (4\(\hat{\mathbf{i}}\) + 7\(\hat{\mathbf{j}}\) – 4\(\hat{\mathbf{k}}\)) to give a unit vector along the X-axis.
Solution:
Let vector \(\overrightarrow{\mathrm{p}}\) be added to get unit vector (\(\hat{\mathbf{i}}\)) along X-axis.
Sum of given vectors is given as,
(2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\) ) + (4\(\hat{\mathbf{i}}\) + 7\(\hat{\mathbf{j}}\) – 4\(\hat{\mathbf{k}}\)) = 6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)
According to given condition, (6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) + \(\hat{\mathbf{P}}\) = \(\hat{\mathbf{i}}\)
∴ \(\overrightarrow{\mathrm{P}}\) = \(\hat{\mathbf{i}}\) – (6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) = \(\hat{\mathbf{i}}\) – 6\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\) = -5\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\)
Answer:
The required vector is -5\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\).

Question 25.
If \(\overrightarrow{\mathbf{P}}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{Q}}\) = 2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\).Find
i. \(\overrightarrow{\mathbf{P}}\) + \(\overrightarrow{\mathbf{Q}}\)
ii. 3\(\overrightarrow{\mathbf{P}}\) – 2\(\overrightarrow{\mathbf{Q}}\)
Solution:
Given \(\overrightarrow{\mathbf{P}}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\), \(\overrightarrow{\mathbf{Q}}\) = 2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\)
To find:

i. \(\overrightarrow{\mathbf{P}}\) + \(\overrightarrow{\mathbf{Q}}\)
ii. 3\(\overrightarrow{\mathbf{P}}\) – 2\(\overrightarrow{\mathbf{Q}}\)

Calculation:

i. \(\overrightarrow{\mathbf{P}}\) + \(\overrightarrow{\mathbf{Q}}\) = (2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – k) + (2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2k)
= (2 + 2)\(\hat{\mathbf{i}}\) + (3 – 5)\(\hat{\mathbf{j}}\) + (-1 + 2)\(\hat{\mathbf{k}}\)
= 4\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\)

ii. 3\(\overrightarrow{\mathbf{P}}\) = 3(2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) = 6\(\hat{\mathbf{i}}\) + 9\(\hat{\mathbf{j}}\) – 3\(\hat{\mathbf{k}}\)
2\(\overrightarrow{\mathbf{Q}}\) = 2(2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\)) = 4\(\hat{\mathbf{i}}\) – 10\(\hat{\mathbf{j}}\) + 4\(\hat{\mathbf{k}}\)

Question 26.
Find unit vector parallel to the resultant of the vectors \(\overrightarrow{\mathbf{A}}\) = \(\hat{\mathbf{i}}\) + 4\(\hat{\mathbf{j}}\) – 2\(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = 3\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\).
Solution:
The resultant of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is,

Answer:
The required unit vector is \(\frac{1}{3 \sqrt{2}}\)(4\(\hat{\mathbf{i}}\) – \(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\))

Question 27.
Two forces, F₁ and F₂, each of magnitude 5 N are inclined to each other at 60°. Find the magnitude and direction of their resultant force.
Solution:
Given: F₁ = 5 N, F₂ = 5 N, θ = 60°
To find: Magnitude of resultant force (R),
Direction of resultant force (α)

Answer:
i. The magnitude of resultant force is 8.662 N.

ii. The direction of resultant force is 30° w.r.t. \(\overrightarrow{\mathrm{F}_{1}}\).

Question 28.
Water is flowing in a stream with velocity 5 km/hr in an easterly direction relative to the shore. Speed of a boat relative to still water is 20 km/hr. If the boat enters the stream heading north, with what velocity will the boat actually travel?
Answer:
The resultant velocity \(\overrightarrow{\mathrm{R}}\) of the boat can be obtained by adding the two velocities using ∆ OAB shown in the figure.

The direction ot the resultant velocity is

Answer: The velocity of the boat is 20.616 km/hr in a direction 14.04° east of north. .
[Note: tan^-1 (0.25) ≈ 14.04° which equals 14°2]

Question 29.
Rain is falling vertically with a speed of 35 m/s. Wind starts blowing at a speed of 12 m/s in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella? (NCERT)
Solution:
Let the velocity of rain and wind be \(\overrightarrow{\mathbf{V}_{\mathrm{R}}}\) and \(\overrightarrow{\mathbf{V}_{\mathrm{W}}}\), then resultant velocity \(\overrightarrow{\mathrm{v}}\) has magnitude of

If \(\overrightarrow{\mathrm{v}}\) makes an angle θ with vertical then, from the figure

Answer: The boy should hold his umbrella in vertical plane at an angle of about 19° with vertical towards the east.

Question 30.
What are components of a vector?
Answer:

1. The given vector can be written as sum of two or more vectors along certain fixed directions. The vectors into which the given single vector is splitted are called components of the vector.
2. Let \(\overrightarrow{\mathrm{A}}\) = \(\mathrm{A}_{1} \hat{\alpha}\) + \(\mathrm{A}_{2} \hat{\beta}\) + \(\mathrm{A}_{3} \hat{\gamma}\) where, \(\hat{\alpha}\), \(\hat{\beta}\) and \(\hat{\gamma}\) are unit vectors along chosen directions. Then, A₁, A₂ and A₃ are known as components of \(\overrightarrow{\mathrm{A}}\) along three directions \(\hat{\alpha}\), \(\hat{\beta}\) and \(\hat{\gamma}\).
3. It two vectors are equal then, their corresponding components are also equal and vice-versa.
   

[Note: The magnitude of a vector is a scalar while each component of a vector is always a vector.]

Question 31.
What is meant by resolution of vector?
Answer:

1. The process of splitting a given vector into its components is called resolution of the vector.
2. Resolution of vector is equal to replacing the original vector with the sum of the component vectors.

Question 32.
That are rectangular components of vectors? Explain their uses.
Answer:
i. Rectangular components of a vector:
If components of a given vector are mutually perpendicular to each other then they are called rectangular components of that vector.

ii. Consider a vector \(
\overrightarrow{\mathrm{R}}
\) = \(
\overrightarrow{\mathrm{OC}}
\) originating from the origin O’ of a rectangular co-ordinate system as shown in figure.

iii. Draw CA ⊥ OX and CB ⊥ OY.
Let component of \(
\overrightarrow{\mathrm{R}}
\) along X-axis \(
\overrightarrow{\mathrm{R}}_{\mathrm{x}}
\) and component of \(
\overrightarrow{\mathrm{R}}
\) along Y-axis = \(
\overrightarrow{\mathrm{R}}_{\mathrm{y}}
\)
By parallelogram law of vectors,

where, \(
\hat{i}
\) and \(
\hat{j}
\) are unit vectors along positive direction of X and Y axes respectively.

iv. If θ is angle made by \(
\overrightarrow{\mathrm{R}}
\) with X-axis, then

v. Squaring and adding equation (1) and (2) we get,

Equation (3) gives the magnitude of \(
\overrightarrow{\mathrm{R}}
\).

vi. Direction of \(
\overrightarrow{\mathrm{R}}
\) can be found out by dividing equation (2) by (1),

Equation (4) gives direction of \(
\overrightarrow{\mathrm{R}}
\)

vii. When vectors are noncoplanar, it becomes necessary to use the third dimension. If \(
\overrightarrow{\mathrm{R}}_{\mathrm{x}}
\), \(
\overrightarrow{\mathrm{R}}_{\mathrm{y}}
\) and \(
\overrightarrow{\mathrm{R}}_{\mathrm{z}}
\) are three rectangular components of \(
\overrightarrow{\mathrm{R}}
\) along X, Y and Z axes of a three dimensional rectangular cartesian co-ordinate system then.

Question 33.
Find a unit vector in the direction of the vector 3\(
\hat{i}
\) + 4\(
\hat{j}
\).
Solution:

Question 34.
Given \(
\overrightarrow{\mathbf{a}}
\) = \(
\hat{\mathbf{i}}
\) + 2\(
\hat{\mathbf{j}}
\) and \(
\overrightarrow{\mathbf{b}}
\) = 2\(
\hat{\mathbf{i}}
\) + \(
\hat{\mathbf{j}}
\), what are the magnitudes of the two vectors? Are these two vectors equal?
Solution:

The magnitudes of \(
\vec{a}
\) and \(
\vec{b}
\) are equal. However, their corresponding components are not equal, i.e., a_x ≠ b_x and a_y ≠ b_y. Hence, the two vectors are not equal.
Answer:
Magnitudes of two vectors are equal, but vectors are unequal.

Question 35.
Find the vector drawn from the point (-4, 10, 7) to the point (3, -2, 1). Also find its magnitude.
Solution:
If \(
\overrightarrow{\mathrm{A}}
\) is a vector drawn from the point (x₁, y₁, z₁) to the point (x₂, y₂, z₂), then

Question 36.
In a cartesian co-ordinate system, the co-ordinates of two points P and Q are (2, 4, 4) and (-2, -3, 7) respectively, find \(
\overrightarrow{\mathbf{P Q}}
\) and its magnitude.
Solution:
Given: Position vector of P = (2,4,4)

∴ |\(
\overrightarrow{\mathrm{PQ}}
\)| = 8.6 units
Answer: Vector \(
\overrightarrow{\mathrm{PQ}}
\) is -4\(
\hat{\mathbf{i}}
\) – 7\(
\hat{\mathbf{j}}
\) + 3\(
\hat{\mathbf{k}}
\) and its magnitude is 8.6 units.

Question 37.
If \(
\overrightarrow{\mathbf{A}}
\) = 3\(
\hat{i}
\) + 4[/latex] = 3\(
\hat{j}
\) and \(
\overrightarrow{\mathbf{B}}
\) = 7\(
\hat{i}
\) + 24\(
\hat{j}
\), find a vector having the same magnitude as \(
\overrightarrow{\mathbf{B}}
\) and parallel to \(
\overrightarrow{\mathbf{A}}
\).
Solution:
The magnitude of vector \(
\overrightarrow{\mathrm{A}}
\) is | \(
\overrightarrow{\mathrm{A}}
\) |

Answer: The required vector is 15\(
\hat{\mathbf{i}}
\) + 20\(
\hat{\mathbf{j}}
\).

Question 38.
Complete the table.

Answer:

Question 39.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful.
i. Adding any two scalars,
ii. Adding a scalar to a vector of the same dimensions,
iii. Multiplying any vector by any scalar,
iv. Multiplying any two scalars,
v. Adding any two vectors. (NCERT)
Answer:

1. Not any two scalars can be added. To add two scalars it is essential that they represent same physical quantity.
2. This operation is meaningless. Only a vector can be added to another vector.
3. This operation is possible. When a vector is multiplied with a dimensional scalar, the resultant vector will have different dimensions.
   eg.: acceleration vector is multiplied with mass (a dimensional scalar), the resultant vector has the dimensions of force.
   When a vector is multiplied with non – dimensional scalar, it will be a vector having dimensions as that of the given vector.
   eg.: \(
   \overrightarrow{\mathrm{A}}
   \) × 3 = 3\(
   \overrightarrow{\mathrm{A}}
   \)
4. This operation is possible. Multiplication of non-dimensional scalars is simply algebraic multiplication. Multiplication of non dimensional scalars will result into scalar with different dimensions.
   eg.: Volume × density = mass.
5. Not any two vectors can be added. To add two vectors it is essential that they represent same physical quantity.

Question 40.
Explain scalar product of two vectors with the help of suitable examples.
Answer:
Scalar product of two vectors:

1. The scalar product of two non-zero vectors is defined as the product of the magnitude of the two vectors and cosine of the angle θ between the two vectors.
2. The dot sign is used between the two vectors to be multiplied therefore scalar product is also called dot product.
3. The scalar product of two vectors \(
   \overrightarrow{\mathrm{P}}
   \) and \(
   \overrightarrow{\mathrm{Q}}
   \) is given by, \(
   \overrightarrow{\mathrm{P}}
   \) . \(
   \overrightarrow{\mathrm{Q}}
   \) = PQ cos θ
   where, p = magnitude of \(
   \overrightarrow{\mathrm{P}}
   \), Q = magnitude of \(
   \overrightarrow{\mathrm{Q}}
   \)
   θ = angle between \(
   \overrightarrow{\mathrm{P}}
   \) and \(
   \overrightarrow{\mathrm{Q}}
   \)
4. Examples of scalar product:
   1. Power (P) is a scalar product of force (\(
      \overrightarrow{\mathrm{F}}
      \)) and velocity (\(
      \overrightarrow{\mathrm{v}}
      \))
      ∴ P = \(
      \overrightarrow{\mathrm{F}}
      \) . \(
      \overrightarrow{\mathrm{v}}
      \)
   2. Work is a scalar product of force (\(
      \overrightarrow{\mathrm{F}}
      \)) and displacement (\(
      \overrightarrow{\mathrm{s}}
      \)).
      ∴ W = \(
      \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}
      \)

Question 41.
Discuss characteristics of scalar product of two vectors.
Answer:
Characteristics of the scalar product of two vectors:
i. The scalar product of two vectors is equivalent to the product of magnitude of one vector with component of the other in the direction of the first.



vi. Scalar product of two vectors is expressed in terms of rectangular components as
\(
\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}
\) = A_x + B_x + A_yB_y + A_zB_z

vii. For \(
\vec{a} \neq 0, \vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}
\) does not necessarily mean \(
\vec{b}
\) = \(
\vec{c}
\)

Question 42.
Complete the table vector given below:

Answer:

Question 43.
Define and explain vector product of two vectors with suitable examples.
Answer:
i. The vector product of two vectors is a third vector whose magnitude is equal to the product of magnitude of the two vectors and sine of the smaller angle θ between the two vectors.

ii. Vector product is also called cross product of vectors because cross sign is used to represent vector product.

iii. Explanation:

a. The vector product of two vectors \(
\overrightarrow{\mathrm{A}}
\) and \(
\overrightarrow{\mathrm{B}}
\), is a third vector \(
\overrightarrow{\mathrm{R}}
\) and is written as, \(
\overrightarrow{\mathrm{R}}
\) = \(
\overrightarrow{\mathrm{A}}
\) × \(
\overrightarrow{\mathrm{B}}
\) = AB sin θ \(
\hat{\mathrm{u}}_{\mathrm{r}}
\) where, \(
\hat{\mathrm{u}}_{\mathrm{r}}
\) is unit vector in direction of \(
\overrightarrow{\mathrm{R}}
\), i.e., perpendicular to plane containing two vectors. It is given by right handed screw rule.

c. Examples of vector product:

1. Force experienced by a charge q moving with velocity \(\overrightarrow{\mathrm{V}}\) in uniform magnetic field of induction (strength) \(\overrightarrow{\mathrm{B}}\) is given as \(\overrightarrow{\mathrm{F}}\) = q\(\overrightarrow{\mathrm{V}}\) × \(\overrightarrow{\mathrm{B}}\)

2. Moment of a force or torque (\(\begin{aligned}
&\rightarrow\\
&\tau
\end{aligned}\)) is the vector product of the position vector (\(\vec{r}\)) and the force (\(\overrightarrow{\mathrm{F}}\)).
i.e., \(\begin{aligned}
&\rightarrow\\
&\tau
\end{aligned}\) = \(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)

3. The instantaneous velocity (\(\overrightarrow{\mathrm{v}}\)) of a rotating particle is equal to the cross product of its angular velocity (\(\vec{\omega}\)) and its position (\(\overrightarrow{\mathrm{r}}\)) from axis of rotation.
\(\overrightarrow{\mathrm{v}}\) = \(\overrightarrow{\mathrm{r}}\) × \(\vec{\omega}\)

Question 44.
State right handed screw rule.
Answer:
Statement of Right handed screw rule: Hold a right handed screw with its axis perpendicular to the plane containing vectors and the screw rotated from first vector to second vector through a small angle, the direction in which the screw tip would advance is the direction of the vector product of two vectors.

Question 45.
State the characteristics of the vector product (cross product) of two vectors.
Answer:
Characteristics of the vector product (cross product):
i. The vector product of two vectors does not obey the commutative law of multiplication.


vi. The magnitude of cross product of two vectors is numerically equal to the area of a parallelogram whose adjacent sides represent the two vectors.

Question 46.
Derive an expression for cross product of two vectors and express it in determinant form.
Answer:
Expression for cross product of two vectors:
i. Let two vectors \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{Q}}\) be represented in magnitude and direction by,

ii.

iii. Determinant form of cross product of two vectors \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{Q}}\) is given by,

Question 47.
Show that magnitude of vector product of two vectors is numerically equal to the area of a parallelogram formed by the two
vectors.
Answer:
Suppose OACB is a parallelogram of adjacent sides, \(\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{Q}}\).

Question 48.
Distinguish between scalar product (dot product) and vector product (cross product).
Answer:

Question 49.
Given \(\overrightarrow{\mathbf{P}}\) = 4\(\hat{\mathbf{i}}\) – \(\hat{\mathbf{j}}\) + 8\(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{Q}}\) = 2\(\hat{\mathbf{i}}\) – m\(\hat{\mathbf{j}}\) + 4\(\hat{\mathbf{k}}\) find m if \(\overrightarrow{\mathbf{P}}\) and \(\overrightarrow{\mathbf{Q}}\) have the same direction. Solution:
Since \(\overrightarrow{\mathbf{P}}\) and \(\overrightarrow{\mathbf{Q}}\) have the same direction, their corresponding components must be in the same proportion, i.e.,

Question 50.
Find the scalar product of the two vectors \(\overrightarrow{\mathbf{v}}_{1}\) = \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{v}}_{2}\) = \(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\)
Solution:

Answer: Scalar product of two given vectors is – 4.

Question 51.
A force \(\overrightarrow{\mathbf{F}}\) = \(4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) acting on a particle produces a displacement of \(\overrightarrow{\mathbf{S}}\) = \(\overrightarrow{\mathrm{s}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}\) where F is expressed in newton and s in metre. Find the work done by the force.
Solution:

Answer: The work done by the force is 41 J.

Question 52.
Find ‘a’ if \(\overrightarrow{\mathbf{A}}\) = \(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(\mathbf{a} \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) are perpendicular to one another.
Solution:

Question 53.
If \(\overrightarrow{\mathbf{A}}\) = \(5 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) determine the angle between \(\) and \(\). Solution:
Solution:

Question 54.
Find the angle between the vectors
\(\overrightarrow{\mathbf{A}}\) = \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{2} \hat{\mathbf{k}}\).
Solution:
Let angle between the vectors be θ

Answer: The angle between the vectors is 60°.

Question 55.
If \(\overrightarrow{\mathbf{A}}\) = \(2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and \(\vec{B}\) = \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}\), find the component of \(\overrightarrow{\mathbf{A}}\) along \(\overrightarrow{\mathbf{B}}\).
Solution:

Question 56.
\(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) are unit vectors along X-axis and Y-axis respectively. What is the magnitude and direction of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}\)? What are the components of a vector
\(\overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}\) along the directions of \((\hat{\mathbf{i}}+\hat{\mathbf{j}})\) and \((\hat{\mathbf{i}}-\hat{\mathbf{j}})\)? (NCERT)
Answer:

Question 57.
The angular momentum \(\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}\), where \(\overrightarrow{\mathbf{r}}\) is a position vector and \(\overrightarrow{\mathrm{p}}\) is linear momentum of a body.

Solution:

Question 58.
If \(\overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) are two vectors, find \(|\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}|\)
Answer:

Question 59.
Find unit vectors perpendicular to the plane of the vectors, \(\overrightarrow{\mathbf{A}}\) = \(\) and
\(\overrightarrow{\mathbf{B}}\) = \(2 \hat{\mathbf{i}}-\hat{\mathbf{k}}\)
Solution:
Let required unit vector be \(\hat{\mathrm{u}}\).

Question 60.
\(\overrightarrow{\mathbf{P}}\) = \(\hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{Q}}\) = \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) are two vectors, find the unit vector parallel to \(\overrightarrow{\mathbf{P}} \times \overrightarrow{\mathbf{Q}}\). Also find the vector perpendicular to P and Q of magnitude 6 units.
Solution:

Question 61.
Find the area of a triangle formed by \(\overrightarrow{\mathbf{A}}\) = \(\hat{3} \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\boldsymbol{2} \hat{\mathbf{k}}\) as adjacent sides measure in metre. Solution:
Given: Two adjacent sides of triangle,
\(\overrightarrow{\mathrm{A}}\) = \(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\), \(\overrightarrow{\mathrm{B}}\) = \(\hat{i}+\hat{j}-2 \hat{k}\)
To find: Area of triangle
Formula: Area of triangle =

Answer:
Area of the triangle is 6.1 m².

Question 62.
Find the derivatives of the functions,
i. f(x) = x⁸
ii. f(x) = x³ + sin x
Solution:
i. Using \(\frac{\mathrm{dx}^{\mathrm{n}}}{\mathrm{dx}}\) = nx^n-1,
\(\frac{d\left(x^{8}\right)}{d x}\) = 8x⁷

ii.

Question 63.
Find derivatives of e^2x – tan x
Solution:

Question 64.
Find the derivatives of the functions.
f(x) = x³ sin x
Solution:
Using,

Question 65.
Find derivatives of \(\frac{d}{d x}(x \times \ln x)\)
Solution:
Using,

Question 66.
Evaluate the following integrals.

i. \(\int x^{8} d x\)
Solution:
Using formula \(\int x^{n} d x\) = \(\frac{x^{n+1}}{n+1}\),
\(\int x^{8} d x\) = \(\frac{x^{9}}{9}\)

ii. \(\int_{2}^{5} x^{2} d x\)
Answer:

iii) \(\int(x+\sin x) d x\)
Solution:

iv) \(\int\left(\frac{10}{x}+e^{x}\right) d x\)
Answer:

v) \(\int_{1}^{4}\left(x^{3}-x\right) d x\)
Answer:
Using,
f₁(x) – f₂(x) = \(\int f_{1}(x)-\int f_{2}(x)\)
Here,

Question 67.
A man applies a force of 10 N on a garbage crate. If another man applies a force of 8 N on the same crate at an angle of 60° with respect to previous, then what will be the resultant force and direction of the crate, if crate is stationary.
Answer:

Answer:
A resultant force of 15.62 N is applied on a crate at an angle of 26.56°.

Question 68.
A lady dropped her wallet in the parking lot of a super market. A boy picked the wallet up and ran towards the lady. He set off at 60° to the verge, heading towards the lady with a speed of 10 m s^-1, as shown in the diagram.
Find the component of velocity of boy directly across the parking strip.

Answer:
The angle between velocity vector and the direction of path is 60°.
∴ Component of velocity across the parking strip
= v × cos 60°
= 10^-1 × cos 60°
= 5 m s^-1

Question 69.
On an open ground, a biker follows a track that turns to his left by an angle of 60° after every 600 m. Starting from a given turn, specify the displacement of the biker at the third and sixth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer:
The path followed by the biker will be a closed hexagonal path. Suppose the motorist starts his journey from the point O.

= 1200 m
= 1.2 km
∴ Total path length = \(|\overrightarrow{\mathrm{OA}}|+|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|\)
= 600 + 600 + 600
= 1800 m
= 1.8 km
The ratio of the magnitude of displacement to the total path-length = \(\frac{1.2}{1.8}\) = \(\frac{2}{3}\) = 0.67

ii. The motorist will take the sixth turn at O.
Displacement is zero.
path-length is = 3600 m or 3.6 km.
Ration of magnitude of displacement and path-length is zero.

Question 70.
What is the resultant of vectors shown in the figure below?

Answer:
If number of vectors are represented by the various sides of a closed polygon taken in one order then, their resultant is always zero.

Question 71.
If \(\overrightarrow{\mathbf{P}}\) is moving away from a point and \(\overrightarrow{\mathbf{Q}}\) is moving towards a point then, can their resultant be found using parallelogram law of vector addition?
Answer:
No. Resultant cannot be found by parallelogram law of vector addition because to apply law of parallelogram of vectors the two vectors and should either act towards a point or away from a point.

Question 72.
Which of the throwing is a vector?
(A) speed
(B) displacement
(C) mass
(D) time
Answer:
(B) displacement

Question 73.
The equation \(\vec{a}+\vec{a}=\vec{a}\) is
(A) meaningless
(B) always truc
(C) may he possible for limited values of a’
(D) true only when \(\overrightarrow{\mathrm{a}}=0\)
Answer:
(D) true only when \(\overrightarrow{\mathrm{a}}=0\)

Question 74.
The minimum number of numerically equal vectors whose vector sum can be zero is
(A) 4
(B) 3
(C) 2
(D) 1
Answer:
(C) 2

Question 75.
If \(\vec{A}+\vec{B}=\vec{A}-\vec{B}\) then vector \(\overrightarrow{\mathrm{B}}\) must be
(A) zero vector
(B) unit vector
(C) Non zero vector
(D) equal to \(\overrightarrow{\mathrm{A}}\)
Answer:
(A) zero vector

Question 76.
If \(\hat{\mathrm{n}}\) is the unit vector in the direction of \(\overrightarrow{\mathrm{A}}\), then,
(A) \(\hat{n}=\frac{\vec{A}}{|\vec{A}|}\)
(B) \(\hat{\mathrm{n}}=\overrightarrow{\mathrm{A}}|\overrightarrow{\mathrm{A}}|\)
(C) \(\hat{\mathrm{n}}=\frac{|\overrightarrow{\mathrm{A}}|}{\overrightarrow{\mathrm{A}}}\)
(D) \(\hat{\mathrm{n}}=\hat{\mathrm{n}} \times \overrightarrow{\mathrm{A}}\)
Answer:
(A) \(\hat{n}=\frac{\vec{A}}{|\vec{A}|}\)

Question 77.
Two quantities of 5 and 12 unit when added gives a quantity 13 unit. This quantity is
(A) time
(B) mass
(C) linear momentum
(D) speed
Answer:
(C) linear momentum

Question 78.
A force of 60 N acting perpendicular to a force of 80 N, magnitude of resultant force is
(A) 20N
(B) 70N
(C) 100 N
(D) 140 N
Answer:
(C) 100 N

Question 79.
A river is flowing at the rate of 6 km h^-1. A man swims across it with a velocity of 9 km h^-1. The resultant velocity of the man will be
(A) \(\sqrt{15} \mathrm{~km} \mathrm{~h}^{-1}\)
(B) \(\sqrt{45} \mathrm{~km} \mathrm{~h}^{-1}\)
(C) \(\sqrt{117} \mathrm{~km} \mathrm{~h}^{-1}\)
(D) \(\sqrt{225} \mathrm{~km} \mathrm{~h}^{-1}\)
Answer:
(C) \(\sqrt{117} \mathrm{~km} \mathrm{~h}^{-1}\)

Question 80.
If \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}\) and magnitudes of \(\overrightarrow{\mathrm{A}}\), \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{C}}\) are 5, 4 and 3 unit respectively, then angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is
(A) sin^-1 (3/4)
(B) cos^-1 (4/5)
(C) tan^-1 (5/3)
(D) cos^-1 (3/5)
Answer:
(B) cos^-1 (4/5)

Question 81.
If \(\vec{A}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\overrightarrow{\mathrm{B}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\), then the area of parallelogram formed from these vectors as the adjacent sides will be
(A) 2\(\sqrt{3}\) square units
(B) 4\(\sqrt{3}\) square units
(C) 6\(\sqrt{3}\) square units
(D) 8\(\sqrt{3}\) square units
Answer:
(D) 8\(\sqrt{3}\) square units

Question 82.
A person moves from a point S and walks along the path which is a square of each side 50 m. He runs east, south, then west and finally north. Then the total displacement covered is
(A) 200m
(B) 100 m
(C) 50\(\sqrt{2}\) m
(D) zero
Answer:
(D) zero

Question 83.
The maximum value of magnitude of \((\vec{A}-\vec{B})\) is
(A) A – B
(B) A
(C) A + B
(D) \(\sqrt{\left(A^{2}+B^{2}\right)}\)
Answer:
(C) A + B

Question 84.
The magnitude of the X and Y components of \(\overrightarrow{\mathrm{A}}\) are 7 and 6. Also the magnitudes of the X and Y components of \(\vec{A}+\vec{B}\) are 11 and 9 respectively. What is the magnitude of
(A) 5
(B) 6
(C) 8
(D)
Answer:
(A) 5

Question 85.
What is the maximum n Limber of components into which a force can be resolved?
(A) Two
(B) Three
(C) Four
(D) Any number
Answer:
(D) Any number

Question 86.
The resultant of two vectors of magnitude \(|\overrightarrow{\mathrm{P}}|\) is also \(|\overrightarrow{\mathrm{P}}|\). They act at an angle
(A) 60°
(B) 90°
(C) 120°
(D) 180°
Answer:
(C) 120°

Question 87.
The vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are such that \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{C}}\) and A² + B² = C². Angle θ between positive directions of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is
(A) \(\frac{\pi}{2}\)
(B) 0
(C) π
(D) \(\frac{2 \pi}{3}\)
Answer:
(A) \(\frac{\pi}{2}\)

Question 88.
The expression \(\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) is a
(A) unit vector
(B) null vector
(C) vector of magnitude \(\sqrt{2}\)
(D) scalar
Answer:
(A) unit vector

Question 89.
What is the angle between \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{\mathrm{i}}\)?
(A) 0°
(B) \(\frac{\pi}{6}\)
(C) \(\frac{\pi}{3}\)
(D) None of the above
Answer:
(D) None of the above

Question 90.
\((\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})\) is a unit vector along X-axis. If \(\overrightarrow{\mathrm{P}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\) then \(\overrightarrow{\mathrm{Q}}\) is
(A) \(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\)
(B) \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\)
(C) \(\hat{i}+\hat{j}+\hat{k}\)
(D) \(\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Answer:
(B) \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\)

Question 91.
The magnitude of scalar product of the vectors \(\overrightarrow{\mathrm{A}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{B}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}\) is
(A) 20
(B) 22
(C) 26
(D) 29
Answer:
(C) 26

Question 92.
Three vectors \(\overrightarrow{\mathrm{A}}\), \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{C}}\) satisfy the relation \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) = 0 and \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{C}}\) = 0, then \(\overrightarrow{\mathrm{A}}\) is parallel to
(A) \(\overrightarrow{\mathrm{B}}\)
(B) \(\overrightarrow{\mathrm{C}}\)
(C) \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{C}}\)
(D) \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{C}}\)
Answer:
(C) \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{C}}\)

Question 93.
What vector must be added to the sum of two vectors \(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) so that the resultant is a unit vector along Z axis?
(A) \(5 \hat{\hat{i}}+\hat{\mathrm{k}}\)
(B) \(-5 \hat{i}+3 \hat{j}\)
(C) \(3 \hat{j}+5 \hat{k}\)
(D) \(-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Answer:
(B) \(-5 \hat{i}+3 \hat{j}\)

Question 94.
\(\overrightarrow{\mathrm{A}}=5 \overrightarrow{\mathrm{i}}-2 \overrightarrow{\mathrm{j}}+3 \overrightarrow{\mathrm{k}}\) and \(\overrightarrow{\mathrm{B}}=2 \overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+2 \overrightarrow{\mathrm{k}}\), then component of \(\overrightarrow{\mathrm{B}}\) along \(\overrightarrow{\mathrm{A}}\) is
(A) \(\frac{\sqrt{28}}{38}\)
(B) \(\frac{28}{\sqrt{38}}\)
(C) \(\frac{\sqrt{28}}{48}\)
(D) \(\frac{14}{\sqrt{38}}\)
Answer:
(D) \(\frac{14}{\sqrt{38}}\)

Question 95.
Choose the WRONG statement
(A) The division of vector by scalar is valid.
(B) The multiplication of vector by scalar is valid.
(C) The multiplication of vector by another vector is valid by using vector algebra.
(D) The division of a vector by another vector is valid by using vector algebra.
Answer:
(D) The division of a vector by another vector is valid by using vector algebra.

Question 96.
The resultant of two forces of 3 N and 4 N is 5 N, the angle between the forces is
(A) 30°
(B) 60°
(C) 90°
(D) 120°
Answer:
(C) 90°

Question 97.
The unit vector along \(\hat{\mathrm{i}}+\hat{\mathrm{j}}\) is
(A) \(\hat{\mathrm{k}}\)
(B) \(\hat{\mathrm{i}}+\hat{\mathrm{j}}\)
(C) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)
(D) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\)
Answer:
(C) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 3 Motion in a Plane Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 3 Motion in a Plane

Question 1.
Explain the term: Displacement.
Answer:
Displacement:

1. Displacement of a particle for a time interval is the difference between the position vectors of the object in that time interval.
2. Let \(\overrightarrow{\mathrm{x}_{1}}\) and \(\overrightarrow{\mathrm{x}_{2}}\) be the position vectors of a particle at time t₁ and t₂ respectively. Then the displacement \(\overrightarrow{\mathrm{S}}\) in time ∆t = (t₂ – t\overrightarrow{\mathrm{s}}=\Delta \overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}) is given by \(\overrightarrow{\mathrm{s}}=\Delta \overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}\)
3. Dimensions of displacement are equal to that of length i.e.. [L¹M⁰T⁰].
4. Displacement is a vector quantity.
5. Example:
   • For an object has travelled through 1 m from time t₁ to t₂ along the positive X-direction, the magnitude of its displacement is I m and its direction is along the positive X-axis.
   • On the other hand, for an object has travelled through I m from time t₁ to t₁ along the positive Y-direction, the magnitude of its displacement remains same i.e., I m but the direction of the displacement is along the positive Y-axis.

Question 2.
Explain the term: Path length.
Answer:

1. Path length is the actual distance travelled by the particle during its motion.
2. It is a scalar quantity.
3. Dimensions of path length are equal to that of length i.e.. [L¹M⁰T⁰]
4. Example:
   • If an object travels along the X-axis from x = 3 m to x = 6 m then the distance travelled is 3 m. In this case the displacement is also 3 m and its direction is along the positive X-axis.
   • However, if the object now comes back to x 5, then the distance through which the object has moved increases to 3 + I = 4 m. Its initial position was x 3 m and the final position is now x = 5 m and thus, its displacement is ∆x = 5 – 3 = 2 m, i.e., the magnitude of the displacement is 2 m and its direction is along the positive X-axis.
   • If the object now moves to x =1, then the distance travelled, i.e., the path length increases to 4 + 4 = 8 m while the magnitude of displacement becomes 3 – 1 = 2 m and its direction is along the negative X-axis.

Question 3.
Explain the terms:
i. Average velocity
ii. Instantaneous velocity
iii. Average speed
iv. Instantaneous speed
Answer:
i) Average velocity:

1. Average velocity (\(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\)) of an object is the displacement (\(\Delta \overrightarrow{\mathrm{x}}\)) of the object during the time interval (∆t) over which average velocity is being calculated, divided by that time interval.
2. Average velocity = (\(\frac{\text { Displacement }}{\text { Time interval }}\))
   \(\overrightarrow{\mathrm{V}_{\mathrm{av}}}=\frac{\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}\)
3. Average velocity is a vector quantity.
4. Its SI unit is m/s and dimensions are [M⁰L¹T^-1]
5. For example, if the positions of an object are x +4 m and x = +6 m at times t = O and t = 1 minute respectively, the magnitude of its average velocity during that time is V_av = (6 – 4)1(1 – 0) = 2 m per minute and its direction will be along the positive X-axis.
   ∴ \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\) = 2 i m/min
   Where, i = unit vector along X-axis.

ii) Instantaneous velocity:

1. The instantaneous velocity (\(\overrightarrow{\mathrm{V}}\)) is the limiting value of ¡he average velocity of the object over a small time interval (∆t) around t when the value of lime interval goes to zero.
2. It is the velocity of an object at a given instant of time.
3. \(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
   where \(\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) derivative of \(\overrightarrow{\mathrm{x}}\) with respect to t.

iii) Average speed:

1. Average speed of an object is the total path length (distance) travelled by the object during the time interval over which average speed is being calculated, divided by that time interval.
2. Average speed = \(\frac{\text { Total path length }}{\text { Total time interval }}\)
3. Average speed is a scalar quantity.
4. Its S.I. unit is m/s and dimensions are [M⁰V¹T^-1].
5. In rectilinear motion;
   • If the motion of the object is only in one direction, then the magnitude of displacement will be equal to the path length and hence the magnitude of average velocity will be equal to the average speed.
   • If the motion of the object reverses its direction, then the magnitude of displacement will be less then the path length and hence the magnitude of average velocity will be less than the average speed.

iv) Instantaneous speed:
The instantaneous speed is the limiting value of the average speed of the object over a small time interval ‘∆t’ around t when the value of time interval goes to zero.

Question 4.
Distinguish between uniform rectilinear motion and non-uniform rectilinear motion.
Answer:

No.	Uniformly rectilinear motion	Non-uniform rectilinear motion
i.	The object is moving with constant velocity.	The object is moving with variable velocity.
ii.	The average and instantaneous velocities are same.	The average and instantaneous velocities are different.
iii.	The average and instantaneous speeds are the same.	The average and instantaneous speeds are different.
iv.	The average and instantaneous speeds are equal to the magnitude of the velocity.	The average speed will be different from the magnitude of average velocity.

Question 5.
Explain the terms:

1. Acceleration
2. Average acceleration
3. Instantaneous acceleration

Answer:

1. Acceleration:
   • Acceleration is the rate of change of velocity with respect to time.
   • It is a vector quantity.
   • Dimension: [M⁰L¹T^-2]
   • If a particle moves with constant velocity, its acceleration is zero.
2. Average acceleration:
   • Average acceleration is the change in velocity divided by the total time required for the change.
   • If \(\overrightarrow{\mathrm{v}_{\mathrm{1}}}\) and \(\overrightarrow{\mathrm{v}_{\mathrm{2}}}\) are the velocities of the T particle at time t₁ and t₂ respectively, then the change in velocity is \(\) and time required for this change is ∆t = t₂ – t₁
     ∴ \(\vec{a}_{a v}=\frac{\vec{v}_{2}-\vec{v}_{1}}{t_{2}-t_{1}}=\frac{\Delta \vec{v}}{\Delta t}\)
3. Instantaneous acceleration:
   • The instantaneous acceleration a is the limiting value of the average acceleration of the object over a small time interval ‘∆t’ around t when the value of time interval goes to zero.
     \(\overrightarrow{\mathrm{a}}_{\text {inst }}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{d} \mathrm{t}}\)
   • Instantaneous acceleration is the slope of the tangent to the velocity-time graph at a position corresponding to given instant of time.
     [Note: Generally, when the term acceleration is used, it is an instantaneous acceleration.]

Question 6.
Draw and explain the position-time graph of:

1. An object at rest.
2. An object moving with uniform velocity along positive x-axis.
3. An object moving with uniform velocity along negative x-axis.
4. An object moving with non-uniform velocity.
5. An object performing oscillatory motion with constant speed.

Answer:

1. The position-time graph of an object at rest:
   • For an object at rest, the position-time graph is a horizontal straight line parallel to time axis.
   • The displacement of the object is zero as there is no change in the object’s position.
   • Slope of the graph is zero, which indicates that velocity of the particle is zero.
     
2. The position-time graph of an object moving with uniform velocity along positive x-axis:
   • When an object moves, the position of the particle changes with respect to time.
   • Since velocity is constant, displacement is proportional to elapsed time.
   • The graph is a straight line with positive slope, showing that the velocity is along the positive x-axis.
   • In this case, as the motion is uniform, the average velocity and instantaneous velocity are equal at all times.
   • Speed is equal to the magnitude of the velocity.
     
3. Position-time graph of an object moving with uniform velocity along negative x- axis:
   • The graph is a straight line with negative slope, showing that the velocity is along the negative x-axis.
   • Displacement decreases with increase in time.
     
4. Position-time graph of a particle moving with non-uniform velocity;
   • When the velocity of an object changes with time, slope of the graph is different at different points. Therefore, the average and instantaneous velocities are different.
   • Average velocity over time interval from t₁ to t₄ around time t₀ = slope of line AB.
   • Average velocity over time interval from t₂ to t₃ = slope of line CD
   • On further reducing the time interval around t₀, it can be deduced that, instantaneous interval at t₀ = the slope of the tangent PQ at t₀.
     
5. Position-time graph of an object performing oscillatory motion with constant speed:
   For an object performing oscillatory motion with constant speed, the direction of velocity changes from positive to negative and vice versa over fixed intervals of time.
   

Question 7.
Explain the velocity-tune graphs of an object:
i) Moving with zero acceleration.
ii) Moving with constant positive acceleration.
iii) Moving with constant negative acceleration.
iv) Moving with non-uniform acceleration.
Answer:
i) Object is moving with zero acceleration:

1. 
2. As the acceleration is zero, the graph will be a straight line parallel to time axis.
3. Velocity of the particle is constant as the acceleration is zero.
4. Magnitude of displacement of object from t₁ to t₂ = v₀ × (t₂ – t₁) shaded area under velocity-time graph.

ii) Object is moving with constant positive acceleration:

1. The velocity-time graph is linear.
2. Velocity increases with increase in time. as acceleration is positive (along the direction of velocity).
3. The area under the velocity-time graph between two instants of time t₁ and t₂ gives the displacement of the object during that time interval.
4. Slope of the graph is \(\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}\) = positive acceleration
   

iii) Object is moving with constant negative acceleration:

1. The velocity-time graph is linear.
2. Velocity decreases with increase in time as acceleration is negative (opposite to the direction of velocities).
3. The area tinder the velocity-time graph between two instants of time t₁ and t₂ gives the displacement 0f the object during that time interval.
4. Slope of the graph is \(\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}\) negative acceleration
   

iv. Object is moving with non-uniform acceleration:

1. 
2. Velocity-time graph is non-linear.
3. The area under the velocity-time graph between two instants of time t₁ and t₂ gives the displacement of the object during that time interval area under the velocity-time curve =
   
   = x(t₂) – x(t₁)
   = displacement of the object from t₁ to t₂.

Question 8.
A ball thrown vertically upwards from a point P on earth reaches a point Q and returns back to earth striking at a point R. Draw speed-time graph to depict the motion of the ball (Neglect air resistance).
Answer:

• 
• A ball which is thrown up with a certain initial speed goes up with decreasing speed to a certain height where its speed becomes zero.
• Now, during its downward motion, the speed goes on increasing from zero and reaches its initial value when it strikes the ground.
• The speed-time graph for the motion of a ball is as shown in the figure.

Question 9.
Figure shows velocity-time graph for various situations. What does each graph indicate?

Answer:

1. Initial velocity, u > 0. Also, velocity is constant with time. Hence, acceleration is zero.
2. As finite initial velocity is increasing with time, acceleration, a > 0 and is constant.
3. Initial velocity, u = 0. Velocity is increasing with time so, acceleration a is positive. But it is decreasing in magnitude with time.
4. Initial velocity, u = 0. Velocity is linearly increasing with time. Hence, starting from rest acceleration is constant.
5. Initial velocity, u = 0. Acceleration and velocity is increasing with time.
6. Initial velocity u > 0. Velocity decreases and ultimately comes to rest. Hence, acceleration a < 0.

Question 10.
‘The distances travelled by an object starting from rest and having a positive uniform acceleration in successive seconds are in the ratio 1:3:5:7….’ Prove it.
Answer:

1. Consider an object under free fall, Initial velocity u = 0, acceleration a = g
2. The distance travelled by the object in equal time intervals t₀ can be given by the second law of motion as,
   s = ut₀ + \(\frac{1}{2}\) gt₀²
3. Distance travelled in the first time interval to,
   s₁ = 0 + \(\frac{1}{2}\)gt₀² = \(\frac{1}{2}\) gt₀²
   Substituting \(\frac{\mathrm{g}}{2}\) = A, we have s₁ = At₀²
4. Distance travelled in the time interval 2t₀ = A (2t₀)²
   ∴ The distance travelled in the second t₀ interval, s₂ = A(4t₀² – t₀²) = 3At₀\frac{\mathrm{g}}{2} = 3s₁
5. Distance travelled in the time interval 3t₀ = A(3t₀)²
   ∴ The distance travelled in the third to interval,
   s₃ = A (9t₀² – 4t₀²) = 5 At₀² = 5s₁
6. On continuing, it can be seen that the distances travelled, (s₁ : s₂ : s₃ ….) are in the ratio (1 : 3 : 5 :….)

Question 11.
Explain the concept of relative velocity along a straight line with the help of an example.
Answer:

1. Consider two trains A and B moving on two parallel tracks in the same direction.
2. Case 1: Train B overtakes train A.
   For a passenger in train A, train B appears to be moving slower than train A. This happens because the passenger in train A perceives the velocity of train B with respect to him/her i.e., the difference in the velocities of the two trains which is much smaller than the velocity of train A.
3. Case 2: Train A overtakes train B.
   For a passenger in train A, train B appears to be moving faster than train A. This happens because the passenger in train A perceives the velocity of the train B w.r.t. to him/her i.e., the difference in the velocities of the two trains which is larger than the velocity of train A.
4. If \(\vec{v}_{\mathrm{A}}\) and \(\vec{v}_{\mathrm{B}}\) be the velocities of two bodies then relative velocity of A with respect to B is given by \(\vec{v}_{A B}=\vec{v}_{A}-\vec{v}_{B}\).
5. Similarly the velocity of B with respect to A is given by, \(\vec{v}_{A B}=\vec{v}_{B}-\vec{v}_{A}\).
   Thus, relative velocity of an object w.r.t. another object is the difference in their velocities
6. If two objects start form the same point at t = 0, with different velocities, distance between them increases with time in direct proportion to the relative velocity between them.

Solved Problems

Question 12.
A person walks from point P to point Q along a straight road ¡n 10 minutes, then turns back and returns to point R which ¡s midway between P and Q after further 4 minutes. If PQ is 1 km, find the average speed and velocity of the person in going from P to R.
Solution:
Given: time taken (t) = 10 + 4 = 14 minutes,
distance (s) = PQ + QR = 1 + 0.5 = 1.5 km,
displacement = PQ – QR = 1 – 0.5 = 0.5km
To find: Average speed, average velocity (v)

The average speed and average velocity of the person is 6.42 km/hr and 2.142 km/hr respectively.

Question 13.
A car moves at a constant speed of 60 km/hr for 1 km and 40 km/hr for next 1 km. What ¡s the average speed of the car?
Solution:
Given. v₁ = 60 km/hr, x₁ = 1 km,
v₂ = 40 km/hr, x₂ = 1 km
To find: Average speed of car (V_av)
Formula: v_av = \(\frac{\text { total path length }}{\text { total time interval }}\)
Calculation: From given data,

∴ Average speed of car = 48 km/hr
The average speed of the car is 48 km/hr

Question 14.
A stone is thrown vertically upwards from the ground with a velocity 15 m/s. At the same instant a ball is dropped from a point directly above the stone from a height of 30 m. At what height from the ground will the stone and the ball meet and after how much time? (Use g = 10 m/s² for ease of calculation).
Solution:
Let the stone and the ball meet after time t₀. From second equation of motion, the distances travelled by the stone and the ball in that time is given as,
S_stone = 15 t₀ – \(\frac{1}{2}\) gt₀²
S_ball = \(\frac{1}{2}\) gt₀²
When they meet. S_stone + S_ball = 30
∴ 15t₀ – \(\frac{1}{2}\) gt₀² + \(\frac{1}{2}\) gt₀² = 30
t₀ = \(\frac{30}{15}\) = 2 s
∴ S_stone = 15 (2) – \(\frac{1}{2}\) (10) (2)² = 30 – 20 = 10 m
The stone and the ball meet at a height of 10 m after time 2s.

Question 15.
A ball is dropped from the top of a building 122.5 m high. How long will it take to reach the ground? What wilt be its velocity when it strikes the ground?
Solution:
Given: s = h = 122.5 m, u = 0,
a = g = 9.8 ms²
To find: i) Time taken to reach the ground (t)
ii) Velocity of ball when it strikes ground (v)

Formulae: i) s = ut + \(\frac{1}{2}\) at²
ii) v = u + gt
Calculation: From formula (i),
122.5 = 0 + \(\frac{1}{2}\) × 9.8 t²
t² = \(\frac{122.5}{4.9}\) = 25
t = \(\sqrt {25}\) = 5 second
From formula (ii),
v = u + gt
v = 0 + 9.8 × 5 = 49 m/s

i) Time taken to reach the ground is 5 s.
ii) Velocity of the ball when it strikes the ground is 49 m/s.

Question 16.
The position vectors of three particles are given by
\(\overrightarrow{\mathrm{x}}_{1}=(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \mathrm{m}\), \(\overrightarrow{\mathrm{x}}_{2}=(5 \mathrm{t} \hat{\mathrm{i}}+5 \mathrm{t} \hat{\mathrm{j}}) \mathrm{m}\) and \(\overrightarrow{\mathrm{x}}_{3}=\left(5 \mathrm{t} \hat{\mathrm{i}}+10 \mathrm{t}^{2} \hat{\mathrm{j}}\right) \mathrm{m}\) as a function of time t.
Determine the velocity and acceleration for each, in SI units.
Solution:

v₂ = \(\sqrt{5^{2}+5^{2}}\) = 5\(\sqrt{2}\) m/s
tan θ = \(\frac{5}{5}\) = 1
∴ θ = 45°
Direction of v₂ makes an angle of 45° to the horizontal.

Thus, third particle is getting accelerated along the y-axis at 20 m/s².

Question 17.
The initial velocity of an object is \(\overrightarrow{\mathrm{u}}=5 \hat{\mathrm{i}}+10 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\). Its constant acceleration is \(\vec{a}=2 \hat{i}+3 \hat{j} \mathrm{~m} / \mathrm{s}^{2}\). Determine the velocity and the displacement after 5 s.
Solution:

Question 18.
An aeroplane A, is travelling in a straight line with a velocity of 300 km/hr with respect to Earth. Another aeroplane B, is travelling in the opposite direction with a velocity of 350 km/hr with respect to Earth. What is the relative velocity of A with respect to B? What should be the velocity of a third aeroplane C moving parallel to A, relative to the Earth if it has a relative velocity of 100 km/hr with respect to A?
Solution:
Given: v_A = 300 km/hr, v_B = 350 km/hr,
v_CA = 100 km/hr
To find: i) Velocity of plane A relative to B (v_A – v_B)
ii) Velocity of aeroplane C (v_C)

Formula: i) v_AB = v_A – v_B
ii) v_CA = v_C – v_A

Calculations: From formula (i),
v_AB = v_A – v_B = 300 – (-350)
∴ v_AB = 650 km/hr
From formula (ii),
v_C = v_CA + v_A = 100 + 300 = 400 km/hr

i) The relative velocity of A with respect to B is 650 km/hr.
ii) The velocity of plane C relative to Earth is 400 km/hr.

Question 19.
A car moving at a speed 10 m/s on a straight road is ahead of car B moving in the same direction at 6 m/s. Find the velocity of A relative to B and vice-versa.
Solution:
Given: v_A = 10 m/s, v_B = 6 m/s,
To find: i) Velocity of A relative to B (v_A – v_B)
ii) Velocity of B relative to A (v_B – v_A)

Formulae: i) v_AB = v_A – v_B
ii) v_BA = v_B – v_A

Calculation: From formula (i),
v_AB = 10 – 6 = 4 m/s
From formula (ii),
v_BA = 6 – 10 = -4 m/s
-ve sign indicates that driver of car A sees the car B lagging behind at the rate of 4 m/s.
∴ v_AB = 4 m/s, v_BA = -4 m/s

i) Velocity of A relative to B is 4 m/s.
ii) Velocity of B relative to A is -4 m/s.

Question 20.
Two trains 120 m and 80 m in length are running in opposite directions with velocities 42 km/h and 30 km/h respectively. In what time will they completely cross each other?
Solution:
Given: l₁ = 120 m, l₂ = 80 m,
v_A = 42 km/h = 42 × \(\frac{5}{18}\) = \(\frac{35}{3}\) m/s,
v_B = -30km/h= -30 × \(\frac{5}{18}\) = \(\frac{-25}{3}\) m/s
To find: Time taken by trains to cross each other (t)
Formula: Time = \(\frac{\text { Distance }}{\text { speed }}\)

Calculation :
Total distance to be travelled
= sum of lengths of two trains
= 120 + 80 = 200m
Relative velocity of A with respect to B is v_AB,
v_AB = v_A – v_B
= \(\frac{35}{3}\) – (\(\frac{-25}{3}\))
= \(\frac{60}{3}\)
∴ v_AB = 20m/S
From formula,
∴ Time taken to cross each other (t) = \(\frac{\text { Distance }}{\text { speed }}\)
= \(\frac{200}{20}\)
= 10 s
Time taken by the two trains to cross each other is 10 s.

Question 21.
A jet aeroplane travelling at the speed of 500 km/hr ejects its products of combustion at speed of 1500 km/hr relative to jet plane. What is the relative velocity of the latter with respect to an observer on the ground?
Solution:
Let us consider the positive direction of motion towards the observer on the ground.
Suppose \(\vec{v}_{\mathrm{a}}\) and \(\vec{v}_{\mathrm{cj}}\) be the velocities of the aeroplane and relative velocity of combustion products w.r.t. aeroplane respectively.

∴ \(\vec{v}_{\mathrm{cj}}\) = 1,500 km/hr (towards the observer on the ground) and \(\vec{v}_{\mathrm{a}}\) = 500 km/hr (away from the observer on the ground)
∴ – \(\vec{v}_{\mathrm{a}}\) = -500 km/ hr (towards the observer on the ground)

Let \(\vec{v}_{\mathrm{c}}\) be the velocity of the combustion products towards the observer on ground then,
\(\vec{v}_{\mathrm{c} j}=\vec{v}_{\mathrm{c}}-\vec{v}_{\mathrm{a}}\)
∴ \(\vec{v}_{\mathrm{c} }=\vec{v}_{\mathrm{cj}}-\vec{v}_{\mathrm{a}}\)
= 1500 + (-500)
= 1000 km/hr
∴ \(\vec{v}_{\mathrm{c}}\) = 1000 km/hr
The relative velocity of the combustion products w.r.t. the observer is 1000 km/hr.

Question 22.
Derive the expression for average velocity and instantaneous velocity for the motion of an object in x-y plane.
Answer:
i) Consider an object to be at point A at time t₁ in an x—y plane.

ii) At time t₁, the position vector of the object is given as,

vii) The instantaneous velocity of the object at point A along the trajectory is along the tangent to the curve at A. This is shown by the vector AB.

Equation (3) is the slope of the tangent to the curve at the point at which we are calculating the instantaneous velocity.

Question 23.
Derive the expression for average acceleration and instantaneous acceleration for the motion of an object in x-y plane.
Answer:
i) Consider an object moving in an x-y plane.
Let the velocity of the particle be \(\vec{v}_{1}\) and \(\vec{v}_{2}\) at time t₁ and t₂ respectively.
ii) The average acceleration (\(\overrightarrow{\mathrm{a}}_{\mathrm{av}}\)) of the particle between t₁ and t₂ is given as,


Equation (1) is the slope of the tangent to the curve at the point at which we are calculating the instantaneous acceleration.

Question 24.
Explain relative velocity between two objects moving in a plane.
Answer:

1. If \(\vec{v}_{A}\) and \(\vec{v}_{B}\) be the velocities of two bodies then relative velocity of A with respect to B is given by, \(\vec{v}_{A B}=\vec{v}_{A}-\vec{v}_{B}\)
2. Similarly, the velocity of B with respect to A is given by, \(\vec{v}_{\text {BA }}=\vec{v}_{\text {B }}-\overrightarrow{v_{A}}\)
3. Thus, the magnitudes of the two relative velocities are equal and their directions are opposite.
4. For a number of objects A, B, C, D—Y, Z, moving with respect to the other. The velocity of A relative to Z can be given as, \(\overrightarrow{\mathrm{v}}_{\mathrm{AZ}}=\overrightarrow{\mathrm{v}}_{\mathrm{AB}}+\overrightarrow{\mathrm{v}}_{\mathrm{BC}}+\overrightarrow{\mathrm{v}}_{\mathrm{CD}}+\ldots+\overrightarrow{\mathrm{v}}_{\mathrm{XY}}+\overrightarrow{\mathrm{v}}_{\mathrm{YZ}}\)
   The order of subscripts is: A → B → C → D → Z

Question 25.
Write a note on projectile motion.
Answer:

1. An object in flight after being thrown with some velocity is called a projectile and its motion is called projectile motion.
2. Example: A bullet fired from a gun, football kicked in air, a stone thrown obliquely in air etc.
3. In projectile motion, the object is moving freely under the influence of Earth’s gravitational field.
4. The projectile has two components of velocity, one in the horizontal i.e., along the x- direction and the other in the vertical i.e., along the y-direction.
5. As acceleration due to gravity acts only along the vertically downward direction, the vertical component changes in accordance with the laws of motion with a_x = 0 and a_y = -g.
6. As no force is acting in the horizontal direction, the horizontal component of velocity remains unchanged.
   [Note: Retarding forces like air resistance etc. are neglected in projectile motion unless otherwise stated.]

Question 26.
Obtain an expression for the time of flight of a projectile.
Answer:
Expression for time of flight:

1. Consider a body projected with velocity \(\vec{u}\), at an angle θ of projection from point O in the co-ordinate system of the X-Y plane, as shown in figure.
   
2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
   u_x = u cos θ (Horizontal component)
   u_y = u sin θ (Vertical component)
3. Thus, the horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to, v_y = u_y + a_y t with a_y = – g and u_y = u sin θ
4. The components of velocity of the projectile at time t are given by, v_x = u_x = u cos θ
   v_y = u_y – gt = u sin θ – gt ………….. (1)
5. At maximum height.
   v_y = 0, t = t_A = time of ascent = time taken to reach maximum height.
   ∴ 0 = u sin θ – gt_A ……..[From(l)]
   u sin θ = gt_A
   t_A = \(\frac{\mathrm{u} \sin \theta}{\mathrm{g}}\) ………….. (2)
   This is time of ascent of projectile.
6. The total time in air i.e., time of flight T is given as,
   T = 2t_A ………… [From(2)]
   = \(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) ………… (3)
   Equation (3) represents time of flight of projectile.

Question 27.
Define Horizontal range of projectile:
Answer:
The maximum horizontal distance travelled by the projectile is called the horizontal range (R) of the projectile.

Solved Examples

Question 28.
An aeroplane Is travelling northward with a velocity of 300 km/hr with respect to the Earth. Wind is blowing from east to west at a speed of 100 km/hr. What is the velocity of the aeroplane with respect to the wind?
Solution:
Given:
velocity of aeroplane w.r.t Earth,
\(\vec{v}_{A E}=300 \hat{j}\)
velocity of wind w.r.t Earth,
\(\vec{v}_{w E}=-100 \hat{i}\)

Question 29.
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms^-1. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. Take = 9.8 m/s².
Solution:
Given: h = 490m, u_x = 15 ms^-1, a_y = 9.8 ms^-1,
a_x = 0
To find: i) Time taken (t)
ii) Velocity (v)

Formulae: i) t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
ii) v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)

Calculation: t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 490}{9.8}}\) = 10 s
v_x = u_x + a_xt= 15 + 0 × 10 = 15 m/s
u_y = u_y + a_yt = 0 + 9.8 × 10 = 98 m/s
∴ v = \(\sqrt{\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}}=\sqrt{15^{2}+98^{2}}\)
= 99.1 m/s
The stone taken 10 s to reach the ground and hits the ground with 99.1 m/s.

Question 30.
A body is projected with a velocity of 40 ms^-1. After 2 s it crosses a vertical pole of height 20.4 m. Find the angle of projection and horizontal range of projectile, (g = 9.8 ms^-2).
Solution:
Given: u = 40 ms^-1, t = 2 s, y = 20.4 m,
a_y = -9.8 m/s²

To find: i) Angle of projection (θ)
ii) Horizontal range of projectile (R)

Formulae: i) y = u_y t + \(\frac{1}{2}\) a_y t²
ii) R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)

Calculation: Taking vertical upward motion of the projectile from point of projection up to the top of vertical pole we have
u_y = 40 sinθ,
From formula (i),
∴ 20.4 = 40 sinθ × 2 + \(\frac{1}{2}\) (-9.8) × 2²
∴ 20.4 = 80 sinθ – 19.6
or sinθ = \(\frac{(20.4+19.6)}{80}=\frac{1}{2}\)
or θ = 30°.
From formula (ii),
Horizontal range = \(\frac{40^{2}}{9.8}\) sin 2 × 30°
= 141.4 m
The angle of projection is 30°. The horizontal range of projection is 141.4m

Question 31.
A stone is thrown with an initial velocity components of 20 m/s along the vertical, and 15 m/s along the horizontal direction. Determine the position and velocity of the stone after 3 s. Determine the maximum height that it will reach and the total distance travelled along the horizontal on reaching the ground. (Assume g = 10 m/s²)
Solution:
The initial velocity of the stone in x-direction = u cos θ = 15 m/s and in y-direction = u sin θ = 20 m/s
After 3 s, v_x = u cos θ = 15 m/s
v_y = u sin θ – gt
= 20 – 10(3)
= -10 m/s
10 m/s downwards.
∴ v = \(\sqrt{\mathrm{v}_{x}^{2}+\mathrm{v}_{\mathrm{y}}^{2}}=\sqrt{15^{2}+10^{2}}=\sqrt{225+100}=\sqrt{325}\)
∴ v = 18.03m/s
tan α = v_y/v_x = 10/15 = 2/3
∴ α = tan^-1 (2/3) = 33° 41’ with the horizontal.
S_x = (u cos θ)t = 15 × 3 = 45m,
S_y = (u sin θ)t – \(\frac{1}{2}\)gt² = 2o × 3 – 5(3)²
∴ S_y = 15m
The maximum vertical distance travelled is given by,
H = \(\frac{(\mathrm{u} \sin \theta)^{2}}{(2 \mathrm{~g})}=\frac{20^{2}}{(2 \times 10)}\)
∴ H = 20m
Maximum horizontal distance travelled
R = \(\frac{2 \cdot u_{x} \cdot u_{y}}{g}=\frac{2(15)(20)}{10}\) = 60 m

Question 32.
A body is projected with a velocity of 30 ms^-1 at an angle of 300 with the vertical.
Find
i) the maximum height
ii) time of flight and
iii) the horizontal range
Solution:
Given:
30 ms^-1, θ = 90° – 30° = 60°

To find: i) The maximum height reached (H)
ii) Time of flight (T)
iii) The horizontal range (R)

i) The maximum height reached by the body is 34.44 m.
ii) The time of flight of the body is 5.3 s.
iii) The horizontal range of the body is 79.53 m.

Question 33.
A projectile has a range of 50 m and reaches a maximum height of 10 m. What is the e1eation of the projectile?
Solution:
Given: R = 50m, H = 10 m
To find: Elevation of the projectile (θ)

∴ θ = tan^-1 (0.8)
∴ θ = 38.66°
The elevation of the projectile is 38.66°

Question 34.
A bullet fired at an angle of 300 with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit a target 5 km away? Assume the muzzle speed to be fixed and neglect air resistance.
Solution:
R = 3km = 3000m, θ = 30°,
Distance of target R’ = 5km
Horizontal range, R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)
∴ 3000 = \(\frac{\mathrm{u}^{2} \sin 60^{\circ}}{\mathrm{g}}\)
∴ \(\frac{\mathrm{u}^{2}}{\mathrm{~g}}=\frac{3000}{\sin 60^{\circ}}=\frac{3000 \times 2}{\sqrt{3}}\) = 2000\(\sqrt {3}\)
Maximum horizontal range,
R_max = \(\frac{\mathrm{u}^{2}}{\mathrm{~g}}\) = 2000 \(\sqrt {3}\) m
= 2000 × 1.732 = 3464m = 3.46km
Since R’ > R_max, Target cannot be hit.

Question 35.
Q.54. A ball is thrown at an angle θ and another ball ¡s thrown at an angle (90° – θ) with the horizontal direction from the same point with velocity 39.2 ms^-1. The second ball reaches 50 m higher than the first balL find their individual heights. [Take g = 9.8 ms^-2]
Solution:
For the first ball: Angle of projection = θ,
u = 39.2 ms^-1
H = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\)
H = \(\frac{(39.2)^{2} \sin ^{2} \theta}{2 \times 9.8}\) …………… (i)
For the second ball: Angle of projection
= 90° – θ,
u = 39.2 ms^-1,
maximum height reached = (H + 50) m

or 2H = 78.4 – 50 = 28.4
∴ H = 14.2 m
∴ Height of first ball = H = 14.2 m
Height of second ball = H + 50 = 14.2 + 50 = 64.2 m

i) Height reached by the first ball is 14.2 m.
ii) Height reached by the second ball is 64.2m.

Question 36.
A body is thrown with a velocity of 49 m/s at an angle of 30° with the horizontal. Find
i) the maximum height attained by it
ii) the time of flight and
iii) the horizontal range.
Solution:
Given: u = 49 m/s. θ = 30°
To find: i) The maximum height attained (H)
ii) The time of flight (T)
iii) The horizontal range (R)

i) The maximum height attained by the body is 30.625 m
ii) The time of flight of the body is 5 s.
iii) The horizontal range of the body is 212.2 m.

Question 37.
A fighter plane flying horizontally at a altitude of 1.5 km with a speed of 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with a muzzle velocity of 600 m/s to hit the plane? At what minimum altitude should the pilot fly to avoid being hit? [Take g = 10 m /s²]
Solution:
Given: h = 1.5 km = 1500 m,
u = 600 m/s,
y = 720 km/h = 720 × \(\frac{5}{18}\) = 200 m/s

To find: i) Angle of firing (θ)
ii) Minimum altitude (H)

Formula: H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Calculation:
Let the shell hit the plane t seconds after firing,
∴ 600 cos(90 – θ) × t = 200 t
∴ cos(90 – θ) = \(\frac{1}{6}\)
∴ 90° – θ = cos^-1(\(\frac{1}{3}\))
cos ^-1(\(\frac{1}{3}\)) = 90° – θ
∴ 70°28’ = 90° – θ
∴ θ = 90° – cos^-1 (\(\frac{1}{3}\))
∴ θ = 19°47’ with vertical
To avoid being hit, the plane must be at a minimum height, i.e., maximum height reached by the shell.
From formula,

∴ H = 15.9 km

i) Angle made by gun with the vertical is 19°47′.
ii) Minimum altitude at which the pilot should fly is 15.9 km.

Question 38.
A both is thrown with a velocity of 40 m/s in a direction making an angle of 30° with the horizontal. Calculate
i) Horizontal range
ii) Maximum height and
iii) Time taken to reach the maximum height.
Solution:
Given: u = 40 m/s, θ = 30°
To find: i) Horizontal range (R)
ii) Maximum height (H_max)
iii) Time to reach max. height (t_A)

i) Horizontal range of the body is 141.4 m.
ii) Maximum height reached by the body is 20.41 m.
iii) Time taken by the body to reach the maximum height is 2.041 s.

Question 39.
If a child launches paper plane with a velocity of 6 m/s² at an angle θ with vertical.
i) What will be the maximum range of the projectile?
ii) What will be the maximum height of the projectile?
iii) Will the plane hit a lady standing at a distance of 6m?
Solution:
i) Range of projectile to given by.

H_max = 1.63 m
iii) As maximum range of projectile is 6.53 m and lady is standing 6m away, plane will hit the lady.

Question 40.
Explain the term uniform circular motion.
Answer:

1. The motion of a body along the circumference of a circle with constant speed is called uniform circular motion.
2. The magnitude of velocity remains constant and its direction is tangential to its circular path.
3. The acceleration is of constant magnitude and it is perpendicular to the tangential velocity. It is always directed towards the centre of the circular path. This acceleration is called centripetal acceleration.
4. The centripetal force provides the necessary centripetal acceleration.
5. Examples of U.C.M:
   • Motion of the earth around the sun.
   • Motion of the moon around the earth.
   • Revolution of electron around the nucleus of atom.

Question 41.
What is meant by period of revolution of U.C.M. Obtain an expression for the period of revolution of a particle performing uniform circular motion.
Answer:
The time taken by a particle performing uniform circular motion to complete one revolution is called as period of revolution. It is denoted by T.

Expression for time period:
During period T, particle covers a distance equal to circumference 2πr of circle with uniform speed v.

Question 42.
For a particle performing uniform circular motion, derive an expression for angular speed and state its unit.
Answer:

1. Consider an object of mass m, moving with a uniform speed v, along a circle of radius r. Let T be the time period of revolution of the object, i.e., the time taken by the object to complete one revolution or to travel a distance of 2πr.
   Thus, T = 2πr/v
   ∴ Speed, v = \(\frac{\text { Distance }}{\text { Time }}=\frac{2 \pi \mathrm{r}}{\mathrm{T}}\) …………….. (1)
2. During circular motion of a point object, the position vector of the object from centre of the circle is the radius vector r. Its magnitude is radius r and it is directed away from the centre to the particle, i.e., away from the centre of the circle.
3. As the particle performs UCM, this radius vector describes equal angles in equal intervals of time.
4. The angular speed gives the angle described by the radius vector.
5. During one complete revolution, the angle described is 2π and the time taken is period T. Hence, the angular speed ω is given as, ….[From (1)]
   ω = \(\frac{\text { Angle }}{\text { time }}=\frac{2 \pi}{\mathrm{T}}=\frac{\left(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\right)}{\mathrm{r}}\) …………….. [From (1)]
   = \(\frac{\mathrm{v}}{\mathrm{r}}\)
6. The unit of angular speed is radian/second.

Question 43.
Derive an expression for centripetal acceleration of a particle performing uniform circular motion.
Answer:
Expression for centripetal acceleration by calculus method:
i) Suppose a particle is performing U.C.M in anticlockwise direction.
The co-ordinate axes are chosen as shown in the figure.
Let,
A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
\(\overrightarrow{\mathrm{r}}\) = instantaneous position vector at time t

ii) From the figure,
\(\vec{r}=\hat{i} x+\hat{j} y\)
where, \(\hat{i}\) and \(\hat{j}\) are unit vectors along X-axis and Y-axis respectively.

iii) Also, x = r cos θ and y = r sin θ

iv) Velocity of the particle is given as rate of change of position vector.

vi) From equation (1) and (2),
\(\overrightarrow{\mathrm{a}}=-\omega^{2} \overrightarrow{\mathrm{r}}\) ……….. (3)
Negative sign shows that direction of acceleration is opposite to the direction of position vector. Equation (3) is the centripetal acceleration.

vii) Magnitude of centripetal acceleration is given by,
a = ω²r

viii) The force providing this acceleration should also be along the same direction, hence centripetal.
∴ \(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}=-\mathrm{m} \omega^{2} \overrightarrow{\mathrm{r}}\)
This is the expression for the centripetal force on a particle undergoing uniform circular motion.

ix) Magnitude of F = mω²r = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mωv

Question 44.
Discuss the factors on which time period of conical pendulum depends.
Answer:
Time period of conical pendulum is given by,
T = 2π \(\sqrt{\frac{\cos \theta}{\mathrm{g}}}\) …………. (i)
where, l = length of the string
g = acceleration due to gravity
θ = angle of inclination
From equation (i), it is observed that period of conical pendulum depends on following factors.
i) Length of pendulum (l): Time period of conical pendulum increases with increase in length of pendulum, i.e., T ∝ \(\sqrt {l}\)
ii) Acceleration due to gravity (g): Time period of conical pendulum decreases with increase in g. i.e., T ∝ \(\frac{1}{\sqrt{g}}\)
iii) Angle of inclination (θ): As θ increases, cos θ decreases, hence, time period of conical pendulum decreases with increase in θ. (For 0 < θ < π) i.e., T ∝ \(\sqrt{\cos \theta}\)

Question 45.
Is there any limitation on semi vertical angle in conical pendulum? Give reason.
Answer:
Yes.

1. For a conical pendulum, Period T ∝ \(\sqrt{\cos \theta}\)
   ∴ Tension F ∝ \(\frac{1}{\cos \theta}\)
   Speed v ∝ \(\sqrt{\tan \theta}\)
   With increase in angle θ, cos θ decreases and tan θ increases. For θ = 90°, T = 0, F = ∞ and v = ∞ which cannot be possible.
2. Also, θ depends upon breaking tension of string, and a body tied to a string cannot be resolved in a horizontal circle such that the string is horizontal. Hence, there is limitation of semi vertical angle in conical pendulum.

Solved Examples

Question 46.
An object of mass 50 g moves uniformly along a circular orbit with an angular speed of 5 rad/s. If the linear speed of the particle is 25 m/s, ¡s the radius of the circle? Calculate the centripetal force acting on the particle.
Solution:
Given: ω = 5 rad/s, v = 25 m/s,
m = 50 g = 0.05 kg
To find: radius (r), centipetal force (F)
Formula: i) v = ωr
ii) F = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)

Calculation: From formula (i),
r = v/ω = 25/5 m = 5 m.
From formula (ii),
F = \(\frac{0.05 \times 25^{2}}{5}\) = 6.25 N.

i) Radius of the circle is 5 m.
ii) Centripetal force acting on the particle is 6.25 N.

Question 47.
An object is travelling in a horizontal circle with uniform speed. At t = 0, the velocity is given by \(\overrightarrow{\mathbf{u}}=20 \hat{\mathbf{i}}+35 \hat{\mathbf{j}}\) km/s. After one minute the velocity becomes \(\overrightarrow{\mathbf{v}}=-20 \hat{\mathbf{i}}-35 \hat{\mathbf{j}}\). What is the magnitude of the acceleration?
Solution:
Magnitude of initial and final velocities,
u= \(\sqrt{(20)^{2}+(35)^{2}}\) m/s
∴ u = \(\sqrt{1625}\) m/s
∴ u = 40.3 m/s
As the velocity reverses in 1 minute, the time period of revolution is 2 minutes.
T = \(\frac{2 \pi \mathrm{r}}{\mathrm{u}}\), giving r = \(\frac{\text { uT }}{2 \pi}\)
Now,
a = \(\frac{\mathrm{u}^{2}}{\mathrm{r}}=\frac{\mathrm{u}^{2} 2 \pi}{\mathrm{uT}}=\frac{2 \pi \mathrm{u}}{\mathrm{T}}=\frac{2 \times 3.142 \times 40.3}{2 \times 60}\)
= {antilog[log(3.142) + log(40.3) – log(60)]}
= {antilog(0.4972 + 1.6053 – 1.7782)}
= {antilog(0.3243)}
= 2.110.
∴ a = 2.11 m/s²
The magnitude of acceleration is 2.11 m/s².

Question 48.
A racing car completes 5 rounds of a circular track in 2 minutes. Find the radius of the track if the car has uniform centripetal acceleration of π²/s².
Solution:
Given: 5 rounds = 2πr(5),
t = 2minutes = 120s
To find: Radius (r)
Formula: a_cp = ω²r
Calculation: From formula,
a_cp = ω²r
∴ π² = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
But v = \(\frac{2 \pi r(5)}{t}=\frac{10 \pi r}{t}\)
∴ π² = \(\frac{100 \pi^{2} \mathrm{r}^{2}}{\mathrm{rt}^{2}}\)
∴ r = \(\frac{120 \times 120}{100}\) =144m
The radius of the track is 144 m.

Question 49.
A car of mass 1500 kg rounds a curve of radius 250m at 90 km/hour. Calculate the centripetal force acting on it.
Solution:
Given: m= 1500 kg, r = 250m,
v = 90 km/h = 90 × \(\frac{5}{18}\) = 25m/s
To find: Centripetal force (F_CP)
Formula: F_CP = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
Calculation: From formula,
F_CP = \(\frac{1500 \times(25)^{2}}{250}\)
∴ F_CP = 3750 N
The centripetal force acting on the car is 3750 N.

Question 50.
A one kg mass tied at the end of the string 0.5 m long is whirled ¡n a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass.
Solution:
Given: Breaking tension, F = 50 N,
m = 1 kg, r = 0.5m
To find: Maximum speed (v_max)
Formula: B.T (F) = max. C.F \(\frac{m v_{\max }^{2}}{r}\)
Calculation: From formula,
v²_max = \(\frac{F \times r}{m}\)
∴ v²_max = \(\frac{50 \times 0.5}{1}\)
∴ v_max = \(\sqrt{50 \times 0.5}\) = 5 m/s
The greatest speed that can be given to the mass is 5 m/s.

Question 51.
A mass of 5 kg is tied at the end of a string 1.2 m long revolving in a horizontal circle. If the breaking tension in the string is 300 N, find the maximum number of revolutions per minute the mass can make.
Solution:
Given: Length of the string, r = 1.2 m,
Mass attached. m = 5 kg,
Breaking tension, T = 300 N
To find: Maximum number of revolutions per minute (n_max)
Formula: T_max = F_max = mrω²_max
Calculation: From formula,
∴ 5 × 1.2 × (2πn)² = 300
∴ 5 × 1.2 × 4π²n² = 300
∴ n²_max = \(\frac{300}{4 \times(3.142)^{2} \times 6.0}\) = 1.26618
∴n_max = \(\sqrt{1.26618}\) = 1.125 rev/s
∴ n_max = 1.125 × 60
∴ n_max = 67.5 rev/min
The maximum number of revolutions per minute made by the mass is 67.5 rev /min.

Question 52.
A coin placed on a revolving disc, with its centre at a distance of 6 cm from the axis of rotation just slips off when the speed of the revolving disc exceeds 45 r.p.m. What should be the maximum angular speed of the disc, so that when the coin is at a distance of 12 cm from the axis of rotation, it does not slip?
Solution:
Given. r₁ = 6cm, r₂ = 12cm, n₁ = 45 r.p.m
To Find: Maximum angular speed (n₂)
Formula: Max. C.F = mrω²
Calculation: Since, mr₁ω₁² mr₂ω₂² [As mass is constant]

The maximum angular speed of the disc should be 31.8 r.p.m.

Question 53.
A stone of mass 0.25 kg tied to the end of a string is whirled in a circle of radius 1.5 m with a speed of 40 revolutions/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Given: m = 0.25 kg, r.= 1.5 m, T_max = 200 N,
n = 40 rev. min^-1 = \(\frac{40}{60}\) rev s^-1
To find: i) Tension (T)
ii) Maximum speed (v_max)

i) The tension in the string is 6.55 N.
ii) The maximum speed with which the stone can be whirled around is 34.64 m s^-1.

Question 54.
In a conical pendulum, a string of length 120 cm is fixed at rigid support and carries a mass of 150 g at its free end. If the mass is revolved in a horizontal circle of radius 0.2 m around a vertical axis, calculate tension in the string. (g = 9.8 m/s²)
Solution:
Given: l = 120 cm = 1.2rn, r = 0.2m,
m = 150 g = 0.15 kg
To find: Tension in the string (T)

∴ Substituting in formula,

Tension in the string is 1.491 N.

Question 55.
A conical pendulum has length 50 cm. Its bob of mass 100 g performs uniform circular motion in horizontal plane, so as to have radius of path 30 cm. Find
i) The angle made by the string with vertical
ii) The tension in the supporting thread and
iii) The speed of bob.
Solution:

Given: l = 150 cm = 0.5 m, r = 30 cm = 0.3 m,
m = 100 g = 100 × 10^-3 kg = 0.1 kg
To find: i) Angle made by the string with vertical (θ)
ii) Tension in the supporting thread (T)
iii) Speed of bob (y)

Formulae: i) tan θ = –\(\frac{r}{\mathrm{~h}}\)
ii) tan θ = \(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\)

Calculation: By Pythagoras theorem, l² = r² + h²
h² = l² – r²
h² = 0.25 – 0.09 = 0.16
h = 0.4m
i) From formula (1),
tan θ = \(\frac{0.3}{0.4}\) = 0.75
∴ θ = tan^-1 (0.75)
θ = 36°52’

ii) The weight of bob is balanced by vertical component of tension T
∴ T cos θ = mg
cos θ = \(\frac{\mathrm{h}}{l}=\frac{0.4}{0.5}\) = 0.8
∴ T = \(\frac{\mathrm{mg}}{\cos \theta}=\frac{0.1 \times 9.8}{0.8}\)
∴ T = 1.225 N

iii) From formula (2),
v² = rg tan θ
∴ v² = 0.3 × 9.8 × 0.75 = 2.205
∴ v = 1.485 m/s

i) Angle made by the string with vertical is 36°52′. ‘
ii) Tension in the supporting thread is 1.225 N.
iii) Speed of the bob is 1.485 m/s

Apply Your Knowledge

Question 56.
Explain the variation of acceleration, velocity and distance with time for an object under free fall.
Answer:

1. For a free falling object, considering the downward direction as negative, the object is released from rest.
   ∴ initial velocity u = 0 and a = -g = -9.8 m/s²
   ∴ The kinematical equations become,
   v = u + at = 0 – gt = -gt = -9.8t
   s = ut + \(\frac{1}{2}\)at2 = o + \(\frac{1}{2}\)(-g)t² = –\(\frac{1}{2}\) 9.8t²
   = -4.9t²
   v² = u² + 2as = 0 + 2(-g)s
   = -2gs = -2 × 9.8s
   = -19.6s
2. These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance.
3. The variation of acceleration, velocity and distance with the time is as shown in figure a, b and c respectively.

Question 57.
The position-time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below.
i) (A/B) lives closer to the school than (B/A)
ii) (A/B) starts from the school earlier than (B/A)
iii) (A/B) walks faster than (B/A)
iv) A and B reach home at the (same/different) time
v) (A/B) overtakes (B/A) on the road (once/twice)

Answer:

1. A lives closer to the school than B. This is because, OQ > OP, hence B has to cover larger distance than A.
2. A starts from the school earlier than B. This is because, A starts at t = 0 whereas B starts at some finite time greater than zero.
3. As slope of B is greater than that of A, hence B walks faster than A. iv. A and B reach home at different times.
4. This is because the value of ‘t’ corresponding to P and Q for A and B respectively is different.
5. B overtakes A on the road once. This is because A and B meet each other only once on their way back home. As B starts from school later than A and walks faster than A, hence B overtakes A once on his way home.

Question 58.
A bowler throws the ball up to correct distance by controlling his velocity and angle of throw, as shown in the figure given below

i) What will be the range of the projectile?
ii) What will be the height of the projectile from ground?
Answer:

1. Range of projectile is given by,
   R = \(\frac{u^{2} \sin 2 \theta}{g}=\frac{6^{2} \times \sin (2 \times 30)}{9.8}\)
   R = 3.18 m
2. Height of projectile is given by,
   H = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{6^{2} \times \sin ^{2} 30}{2 \times 9.8}\) = 0.46m
   Height achieved by ball from ground is
   H = 0.46 + 1 = 1.46m

i) The range of the projectile is 3.18 m.
ii) The height of the projectile is 1.46 m.

Question 59.
A child takes reading of two cars running on highway, for his school project. He draws a position-time graph of the two cars as shown in the figure

i) What is the velocity of two cars when they meet together?
ii) What is the difference in velocities of the two cars when they cover their maximum distance?
iii) What will be acceleration of the two cars in first 20 s?
Solution:
i) According to graph, the velocity of two cars when they meet each other are,
x = 70m
t = 308
v = \(\frac{x}{t}=\frac{70}{30}\) = 2.33 m/s

ii) According to graph, for maximum distance.
For 1^st car,
x₁ = 120m
t₁ = 50 s
v₁ = \(\frac{x_{1}}{t_{1}}=\frac{120}{50}\)
v₁ = 2.4 m/s

For 2^nd car,
x₂ = 90 m
t₂ = 60 s
v₂ = \(\frac{x_{2}}{t_{2}}=\frac{90}{60}\)
v₂ = 1.5 m/s
Difference in velocities is given by,
v₁ – v₂ = 2.4 – 1.5 = 0.9 m/s

iii) According to graph,
Acceleration of 1^st car in first 20 s
v₁ = \(\frac{\mathrm{x}_{1}}{\mathrm{t}}\)
v₁ = \(\frac{60}{20}\)
v₁ = 3 m/s
a₁ = \(\frac{\mathrm{v}_{1}}{\mathrm{t}}=\frac{3}{20}\)
a₁ = 0.15 m/s²
Acceleration of 2^nd car in first 20 s
v₂ = \(\frac{\mathrm{x}_{2}}{\mathrm{t}}\)
v₂ = \(\frac{40}{20}\)
v₂ = 2 m/s
a₂ = \(\frac{\mathrm{v}_{2}}{\mathrm{t}}=\frac{2}{20}\)
a₂ = 0.1 m/s²
Now,
a₁ – a₂ = 0.15 – 0.1
= 0.05 m/s²

i) The velocity of two cars when they meet together is 2.33 m/s.
ii) The difference in velocities of two cars when they cover maximum distance is 0.9 m/s.
iii) The accelerator of two cars in 20 s is 0.05 m/s².

Question 60.
The speed of a projectile u reduces by 50% on reaching maximum height. What is the range on the horizontal plane?
Solution:
If θ is the angle of projection, then velocity of projectile at height point = u cos θ
u cos θ = \(\frac{50}{100}\) u = \(\frac{1}{2}\) u
or cos θ = \(\frac{1}{2}\) cos 60°
or θ = 60°
Horizontal range,

Question 61.
In projectile motion, vertical motion and horizontal motion are dependent of each other. Yes or No? Justify your answer.
Answer:
No. In projectile motion, the horizontal and vertical motion are independent of each other because both motions don’t affect each other.

Question 62.
In angular projection of a projectile, at highest point, what will be the components of horizontal and vertical velocities?
Answer:
At highest point of angular projection of a projectile, the horizontal component of its velocity is non zero and the vertical component of its velocity is momentarily zero.

Question 63.
What angle will be described between velocity and acceleration at highest point of projectile path?
Answer:
At highest point of projectile path, the angle between velocity and acceleration is 90°.

Quick Review

Multiple Choice Questions

Question 1.
The velocity-time relation of a particle starting from rest is given by v = kt where k = 2 m/s². The distance travelled in 3 sec is
(A) 9 m
(B) 16 m
(C) 27 m
(D) 36 m
Answer:
(A) 9 m

Question 2.
If the particle is at rest, then the x – t graph can be only
(A) parallel to position – axis
(B) parallel to time – axis
(C) inclined with acute angle
(D) inclined with obtuse angle
Answer:
(B) parallel to time – axis

Question 3.
A body is thrown vertically upwards, maximum height is reached, then it will have
(A) zero velocity and zero acceleration.
(B) zero velocity and finite acceleration.
(C) finite velocity and zero acceleration.
(D) finite velocity and finite acceleration.
Answer:
(B) zero velocity and finite acceleration.

Question 4.
Which of the following is NOT a projectile?
(A) A bullet fired from gun.
(B) A shell fired from cannon.
(C) A hammer thrown by athlete.
(D) An aeroplane in flight.
Answer:
(D) An aeroplane in flight.

Question 5.
The range of projectile is 1 .5 km when it is projected at an angle of 15° with horizontal. What will be its range when it is projected at an angle of 45° with the horizontal?
(A) 0.75 km
(B) 1.5 km
(C) 3 km
(D) 6 km
Answer:
(C) 3 km

Question 6.
Which of the following remains constant for a projectile fired from the earth?
(A) Momentum
(B) Kinetic energy
(C) Vertical component of velocity
(D) Horizontal component of velocity
Answer:
(D) Horizontal component of velocity

Question 7.
In case of a projectile, what is the angle between the instantaneous velocity and acceleration at the highest point?
(A) 45°
(B) 1800
(C) 90°
(D) 0°
Answer:
(C) 90°

Question 8.
A player kicks up a ball at an angle θ with the horizontal. The horizontal range is maximum when θ is equal to
(A) 30°
(B) 45°
(C) 60°
(D) 90°
Answer:
(B) 45°

Question 9.
The greatest height to which a man can throw a stone is h. The greatest distance to which he can throw it will be
(A) h/2
(B) 2h
(C) h
(D) 3h
Answer:
(B) 2h

Question 10.
Two balls are projected at an angle θ and (90° – θ) to the horizontal with the same speed. The ratio of their maximum vertical
heights is
(A) 1 : 1
(B) tan θ : 1
(C) 1 : tan θ
(D) tan² θ : 1
Answer:
(D) tan² θ : 1

Question 11.
When air resistance is taken into account while dealing with the motion of the projectile, to achieve maximum horizontal range, the angle of projection should be,
(A) equal to 45°
(B) less than 45°
(C) greater than 90°
(D) greater than 45°
Answer:
(D) greater than 45°

Question 12.
The maximum height attained by projectile is found to be equal to 0.433 of the horizontal range. The angle of projection of this projectile is
(A) 30°
(B) 45°
(C) 60°
(D) 75°
Answer:
(C) 60°

Question 13.
A projectile is thrown with an initial velocity of 50 m/s. The maximum horizontal distance which this projectile can travel is
(A) 64m
(B) 128m
(C) 5m
(D) 255m
Answer:
(D) 255m

Question 14.
A jet airplane travelling at the speed of 500 kmh^-1 ejects the burnt gases at the speed of 1400 kmh^-1 relative to the jet airplane. The speed of burnt gases relative to stationary observer on the earth is
(A) 2.8 kmh^-1
(B) 190 kmh^-1
(C) 700 kmh^-1
(D) 900 kmh^-1
Answer:
(D) 900 kmh^-1

Question 15.
A projectile projected with certain angle reaches ground with
(A) double angle
(B) same angle
(C) greater than 90°
(D) angle between 90° and 180°
Answer:
(B) same angle

Question 16.
The time period of conical pendulum is _________.

Answer:
(C) \(2 \pi \sqrt{\frac{l \cos \theta}{\mathrm{g}}}\)

Question 17.
A projectile projected with certain velocity reaches ground with (magnitude)
(A) zero velocity
(B) smaller velocity
(C) same velocity
(D) greater velocity
Answer:
(C) same velocity

Question 18.
The period of a conical pendulum in terms of its length (l), semivertical angle (θ) and acceleration due to gravity (g) is:

Answer:
(C) \(4 \pi \sqrt{\frac{l \cos \theta}{4 \mathrm{~g}}}\)

Question 19.
Consider a simple pendulum of length 1 m. Its bob performs a circular motion in horizontal plane with its string making an angle 600 with the vertical. The period of rotation of the bob is(Take g = 10 m/s²)
(A) 2s
(B) 1.4s
(C) 1.98 s
(D) none of these
Answer:
(B) 1.4s

Question 20.
The period of a conical pendulum is
(A) equal to that of a simple pendulum of same length l.
(B) more than that of a simple pendulum of same length l.
(C) less than that of a simple pendulum of same length l.
(D) independent of length of pendulum.
Answer:
(C) less than that of a simple pendulum of same length l.

Competitive Corner

Question 1.
Two particles A and B are moving in uniform circular motion in concentric circles of radii r_A and r_B with speed v_A and v_B respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be:
(A) r_B : r_A
(B) 1 : 1
(C) r_A : r_B
(D) v_A : v_B
Answer:
(B) 1 : 1
Hint:
Time period of rotation (A and B) is same

Question 2.
When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x₁ along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x₂ distance. Then x₁ : x₂ will be:
(A) 1 : \(\sqrt {3}\)
(B) 1 : 2\(\sqrt {3}\)
(C) 1 : \(\sqrt {2}\)
(D) \(\sqrt {3}\) : 1
Answer:
(A) 1 : \(\sqrt {3}\)
Hint:
v² = u² + 2as
∴ v² = u² + 2 g sin θ x
sin θ. x = constant
∴ x ∝ \(\frac{1}{\sin \theta}\)
∴ \(\frac{x_{1}}{x_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{1 / 2}{\sqrt{3} / 2}\) = 1 : \(\sqrt {3}\)

Question 3.
A person travelling in a straight line moves with a constant velocity v₁ for certain distance x’ and with a constant velocity v₂ for next equal distance. The average velocity y is given by the relation

Answer:
(C) \(\frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}\)
Hint:
Let, t’ be the time taken to travel distance ‘x’ with constant velocity ‘v₁’
∴ t₁ = \(\frac{\mathrm{x}}{\mathrm{v}_{2}}\)
Let ‘t₂’ be the time taken to travel equal distance ‘x’ with constant velocity ‘v₂‘

Question 4.
Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings 100 m apart and of same height of 200 m, with the same velocity of 25 m/s. When and where will the two bullets collide? (g = 10 m/s²)
(A) They will not collide
(B) After 2 s at a height of 180 m
(C) After 2 s at a height of 20 m
(D) After 4 s at a height of 120 m
Answer:
(B) After 2 s at a height of 180 m
Hint:

Let the bullets collide at time t
The horizontal displacement x₁ and x₂ is given by the equation
x₁ = ut and x₂ = ut
∴ x₁ + x₂ = 100
∴ 25t + 25t = 100
∴ t = 2s
Vertical displacement ‘y’ is given by
y = \(\frac{1}{2}\) gt² = \(\frac{1}{2}\) × 10 × 2² = 20m
∴ h = 200 – 20= 180m

Question 5.
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field \(\vec{E}\). Due to the force q\(\vec{E}\), its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
(A) 2 m/s, 4 m/s
(B) 1 m/s, 3 m/s
(C) 1 m/s, 3.5 m/s
(D) 1.5 m/s, 3 m/s
Answer:
(B) 1 m/s, 3 m/s
Hint:
Car at rest attains velocity of 6 m/s in t₁ = 1 s.
Now as direction of field is reversed, velocity of car will reduce to 0 m/s in next 1 s. i.e., at t₂ = 2 s. But, it continues to move for next one second. This will give velocity of -6 m/s to car at t₃ = 3 s.
Using this data, plot of velocity versus time will be

Question 6.
All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Answer:
(D)
Hint:
The graphs (A), (B) and (C) represent the uniformly retarded motion, i.e., velocity decreases uniformly. However, the slope of the curve in graph (D), indicates increasing velocity. Hence, graph (D) is incorrect.

Question 7.
A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n^th power of R. If the period of rotation of the particle is T, then:
(A) T ∝ R^(n+1)/2
(B) T ∝ R^n/2
(C) T ∝ R^3/2 for any n
(D) T ∝ R^{\(\frac{n}{2}\)+1}
Answer:
(A) T ∝ R^(n+1)/2
Hint:
The centripetal force acting on the particle is provided by the central force,

Question 8.
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be:
(A) \(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{2}\)
(B) \(\frac{t_{1} t_{2}}{t_{2}-t_{1}}\)
(C) \(\frac{\mathbf{t}_{1} t_{2}}{\mathbf{t}_{2}+t_{1}}\)
(D) t₁ – t₂
Answer:
(C) \(\frac{\mathbf{t}_{1} t_{2}}{\mathbf{t}_{2}+t_{1}}\)
Hint:
Let velocity of Preeti be v₁, velocity of escalator be v₂ and distance travelled be L.

Question 9.
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

Answer:
(A)
Hint:
If a body is projected in vertically upward direction, then its acceleration is constant and negative. If direction of motion is positive i.e.. vertically up) and initial position of body is taken as origin, then the velocity decreases uniformly. At highest point its velocity is equal to zero and then it accelerates uniformly downwards returning to its reference position.