NCERT Solutions For Class 11 Physics Chapter 6

NCERT Solutions Class 11 Physics Work, Energy and Power

Ncert Solutions For Class 11 Physics Chapter 6 PDF Free Download

NCERT Solutions for class 11 Physics chapter 6 Work, Energy, and Power are available in PDF format for free download. These books include solutions for all the questions present in the NCERT Physics textbooks and also other important questions from exam point of view. The chapter 6 of class 11 physics covers the following topic

Section Number Topic
6.1 Introduction
6.2 Notions of work and kinetic energy : The work-energy theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable force
6.6 The work-energy theorem for a variable force
6.7 The concept of potential energy
6.8 The conservation of mechanical energy
6.9 The potential energy of a spring
6.10 Various forms of energy : the law of conservation of energy
6.11 Power
6.12 Collisions

Students preparing for their CBSE board exams and other competitive exams can download and refer these materials for having strong foundations about the course. The NCERT Solutions for class 11 Physics chapter 6 pdf available on our website BYJU’S provides detailed solutions and are explained in a logical manner to help students to understand the concepts easily.

 Que.1. It is important to understand the sign of  work done by a force on a body. Find whether the following are positive or negative:

(a) The work done in lifting of a bucket by the use of  a rope.

(b) The work done by the force due to gravity in lifting the bucket.

(c) The work done on the body by friction which is sliding in a plane which is inclined.

(d) The work done on a body which is moving in a constant velocity in a horizontal plane by means of the applied force.

(e) The work done by the resistive force of air on bringing a vibrating bob to rest.

(a) It is clear that the direction of both the force and the displacement are same and thus the work done on it is positive.

(b) It can be noted that the displacement of the object is in upward direction whereas, the force due to gravity is in downward direction. Hence, the word done is negative.

(c) It can be observed that the direction of motion of the object is opposite to the direction of the frictional  force. So, the work done is negative.

(d) The object which is moving in a rough horizontal plane faces the frictional force which is opposite to the direction of the motion. To maintain a uniform velocity, a uniform force is applied on the object. So, the motion of the object and the applied force are in the same direction. Thus, the work done is positive.

(e) It is noted that the direction of the bob and the resistive force of air which is acting on it are in opposite directions. Thus, the work done is negative.


Que.2. A body has a mass of 3 kg which when applied with a force of 8 N moves from rest with a coefficient of kinetic friction = 0.2. Find the following:

(a) When a force is applied for 10 s, what is the work done?

(b) The work done by the friction in 10 s.

(c) When a net force acts on the body for 10 s, what is the work done?

(d) In the time interval of 10 s, the change in the kinetic energy.

Ans. Given: Mass m = 3 kg

Force F = 8 N

Kinetic friction coefficient  \(\mu\) = 0.2

Initial velocity , u = 0

t = 10 s

According to the Newton’s law of motion:

\({a}'\) = \(\frac{F}{m}\) = \(\frac{8}{3}\) = 2.6 m/s2

Friction force = \(\mu\)mg

= 0.2 x 3 x -9.8 = -5.88 N

Acceleration due to friction:

\({a}”\) =  \(\frac{-5.88}{3}\) = -1.96 m/s2

The total acceleration of the body = \({a}'\) + \({a}”\)

= 2.6 + (-1.96) = 0.64 m/s2

According to the equation of the motion

s = ut + \(\frac{1}{2}\) at2

= 0 + \(\frac{1}{2}\) x 0.64 x (10)2

= 32 m

(a). Wa = F x s = 8 x 32 = 256 J

(b). Wt = F x s  = -5.88 x 32 = -188 J

Net force = 8 + (-5.88) = 2.12 N

(c). Wnet = 2.12 x 32 = 68 J

(d) Final velocity v = u + at

= 0 + 0.64 x 10 = 6.4 m/s

Change in kinetic energy = \(\frac{1}{2}\) mv2 -\(\frac{1}{2}\) mu2

= \(\frac{1}{2}\) x 2 (v2 – u2)

= 6.42 – 02

= 41 J


Que.3. One dimensional potential energy functions are given below. A cross indicates the total energy in the ordinary axis. For a given energy, if the particle can’t be found, then specify the region. Also, show the particle’s minimum total energy.


Ans. (a) x > a

The relation which gives the total energy:

E = P.E + K.E

K.E. = E – P.E

K.E of the body is a positive quantity and the region where K.E is negative, the particles does not exist.

For x > a, the potential energy V0 is greater than the total energy E. Hence, the particle does not exist here. The minimum total energy is zero.

(b) All regions

The total energy in all regions is less than the kinetic energy. So the particles do not exist here.


(c) x < b and x > a; =V1

K.E is positive in the region between x > a and x < b. –V1 is the minimum potential energy. K.E = E – (-V1) = E + V1. For the K.E to be positive, total energy must be greater than –V1.

(d) \(\frac{-b}{2}\) < x < \(\frac{a}{2}\); \(\frac{a}{2}\) < x < \(\frac{b}{2}\); -V1

For the given condition, potential energy is greater. So in this region, the particles does not exist. –V1 is the minimum potential energy. K.E = E + V1. For the K.E to be positive, total energy must be greater than –V1. The particle must have a minimum total K.E of –V1.


Que.4. For a particle which executes a linear harmonic motion the potential energy is given by V(x) = Kx2/2, where force constant is K.  K = 0.5 N m-1. A graph is drawn for V(x) verses x. Show that the particle under this potential having an energy of 1 J must return back on reaching x = \(\pm 2\) m

Linear Motion

Ans. Particle energy E = 1 J

K = 0.5 N m-1

K.E = \(\frac{1}{2}\) mv2

Based on law of conservation of energy:

E = V + K

1 = \(\frac{1}{2}\) kx2 + \(\frac{1}{2}\) mv2

Velocity becomes zero when it turns back

1 = \(\frac{1}{2}\) kx2

\(\frac{1}{2}\) x 0.5x2 = 1

X2 = 4

X = \(\pm 2\)

Thus, on reaching x = \(\pm 2\) m, the particle turns back.


Que.5. Explain the following:

(a) When a rocket is air borne , its casing burns up due to friction. The heat energy which is needed is obtained at whose expense? It  is the atmosphere or the rocket?

Ans. When the casing burns up due to the friction, the rocket’s mass gets reduced.

As per the law of conservation of energy:

Total energy = kinetic energy + potential energy

= mgh + \(\frac{1}{2}\) mv2

There will be a drop in total energy due to the reduction in the mass of the rocket. Hence, the energy which is needed for the burning of the casing is obtained from rocket.

(b) Comets travel in an elliptical orbit around the Sun. The gravitational force of the Sun on the comet is not normal to the velocity of the comet but for every complete orbit of the comet, the work done by the gravitational force is zero. Explain.

Ans. The force due to gravity is a conservative force. The work done on a closed path by the conservative force is zero. Hence, for every complete orbit of the comet, the work done by the gravitational force is zero.

(c) An artificial satellite loses its energy when orbiting the Earth irrespective of how thin the atmosphere is, due to atmospheric resistance. Then, how does the speed of the satellite increases when it approaches closer to Earth?

Ans. The potential energy of the satellite revolving the Earth decreases as it approaches the Earth and since the system’s total energy should remain constant, the kinetic energy increases. Thus, the satellite’s velocity increases. In spite of this, the total energy of the system is reduced by a fraction due to the atmospheric friction.

(d) In the given figure, a man carries a mass of 20 kg and walks for 4 m and the same man uses a pulley to pull the same mass and walk the same distance. The work done is greater in which of the above cases ?

Work Done


Scenario I:

m = 20 kg

Displacement of the object, s = 4 m

W = Fs cos \(\theta\) \(\theta\) = It is the angle between the force and displacement

Fs = mgs cos \(\theta\)

W = mgs cos \(\theta\) = 20 x 4 x 9.8 cos \(90^{\circ}\)

= 0                                                                          ( cos \(90^{\circ}\) = 0 )


Scenario II:

Mass = 20 kg

S = 4 m

The applied force direction is same as the direction of the displacement.

\(\theta\) = \(0^{\circ}\)

Cos \(0^{\circ}\) = 1

W = Fs cos \(\theta\)

= mgs \(\theta\)

= 20 x 4 x 9.8 x cos \(0^{\circ}\)

= 784 J

Thus, the work done is more in the second scenario.


Que.6. Select the right one for the following:

(a) When a positive work is done on a body by conservative force, its potential energy decreases/ increases/ remains unaltered.

Ans. Decreases

When a body is displaced in the direction of the force, positive work is done on the body by the conservative force due to which the body moves to the center of force. Thus the separation between the two decreases and the potential energy of the body decreases.

 (b) There will be always a loss of kinetic/potential energy when a work is done against the friction.

Ans. Kinetic energy

Velocity of the body is reduced when the work done is in the direction opposite to that of friction. Thus, the kinetic energy decreases.

(c) In a many particle system, the rate of change of total momentum is proportional to the sum of internal force/ external force on the system.

Ans. External force

Change in momentum cannot be produced by internal forces, irrespective of their directions. Thus, the change in total momentum is proportional to the external force on the system.

(d) The quantities which do not change after an inelastic collision of two bodies are the total linear momentum/ total kinetic energy/ total energy of the system of two bodies.

Ans. Total linear momentum

Irrespective of elastic collision or an inelastic collision, the total linear momentum remains the same.


Que.7. State whether the following are true or false and explain.

(a) The momentum and the energy of each body is conserved in an elastic collision.

Ans. False

The momentum and the energy of both the bodies are conserved and mot individually.

(b) Irrespective of internal or external forces on the body, the total energy of the system is always conserved.

Ans. False.

The external forces on the system can do work on the body and are able to change the energy of the system.

(c) Work done is zero for a body which is in motion over a closed loop for all the forces in nature.

Ans. False.

The work done by the conservative force on the moving body in closed loop is zero.

(d) The initial kinetic energy of the system is always higher than the final kinetic energy in an inelastic collision.

Ans. True

The final kinetic energy is always lesser then the initial kinetic energy as there will be always a loss of energy in the form of heat, light, etc.


Que.8. (a) Is the kinetic energy conserved during the elastic collision of two billiard balls?

Ans. The initial and the final kinetic energy is equal in an elastic collision. When the two balls collide, there is no conservation of kinetic energy. It gets converted into potential energy.

(b) In an elastic collision of two balls, is the total linear momentum conserved?

Ans. The total linear momentum of the system is conserved in an elastic collision.

(c) For the above scenarios find the answers for inelastic collision.

Ans. There will be a loss of kinetic energy in an inelastic collision. The K.E after collision is always less than the K.E before collision.

The total linear momentum of the system is conserved in an inelastic collision also.

(d) Is the collision of two billiard balls elastic or inelastic, if the potential energy of the two depends only on the separation distance between their centers?

Ans. It is an elastic collision as the forces involved are conservative forces. It depends on the distance between the centers of the billiard balls.


 Que.9. A body which is at rest initially undergoes an one-dimensional motion with constant acceleration. At the time t, the power delivered to it is proportional to

(i) \(t^{\frac{1}{2}}\)

(ii) \(t^{\frac{3}{2}}\)

(iii) t2

(iv) t

Ans. body mass = m

Acceleration = a

According to the Newton’s second law of motion:

F = ma (constant)

We know that a = \(\frac{dv}{dt}\) = constant

dv = dt x constant

On integrating

v = \(\alpha\)t \(\rightarrow\) 1

Where, \(\alpha\) is also a constant

v \(\propto\) t \(\rightarrow\) 2

The relation of power is given by:

P = F.v

From the equation 1 & 2

P \(\propto\) t

Thus, from the above we conclude that power is proportional to time.


Que.10. Under the influence of a constant power source, a body is moving unidirectionally. The displacement of it in time t is proportional to:

(i) \(t^{\frac{1}{2}}\)

(ii) \(t^{\frac{3}{2}}\)

(iii) t2

(iv) t

Ans.  We know that the power is given by:

P = Fv

= mav = mv\(\frac{dv}{dt}\)

= k (constant)

vdv = \(\frac{k}{m}\) dt

On integration:

\(\frac{v^{2}}{2}\) = \(\frac{k}{m}dt\)

v = \(\sqrt{\frac{2kt}{m}}\)

To get the displacement:

v = \(\frac{dx}{dt}\) = \(\sqrt{\frac{2k}{m}}\) \(t^{\frac{1}{2}}\)

dx = \(k{}'\) \(t^{\frac{1}{2}}\) dt

where \(k{}'\) = \(\sqrt{\frac{2k}{3}}\)

x = \(\frac{2}{3}\) \(k{}'\) \(t^{\frac{2}{3}}\)

Hence, from the above equation it is shown that x \(\propto\) \(t^{\frac{3}{2}}\)


Que.11. A constant force F is applied to a body which is constrained to move in    z-axis is given by F = -\(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) N

Unit vectors are \(\hat{i}\), \(\hat{j}\), \(\hat{z}\) which are along the x, y and z axis of the system. If a body is moved to a distance of 6m along the z-axis, find the work done.

Ans. F = -\(\hat{i}\) +2\(\hat{j}\) + 3\(\hat{k}\) N

Displacement s = 6 \(\hat{k}\) m

Work done, W = F.s

= (-\(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\)). (4\(\hat{k}\))

= 0 + 0 – 6 x 4

= 24 J

Thus, the work done on the body is 24 J


Que.12. An electron is detected with a kinetic energy of 20 keV and also a proton with a kinetic energy of 200 keV. Find out which is faster and also find their speed ratios.

Ans. Electron mass, m­e = 9.11 x 10-31 kg

Proton mass, mp = 1.67 x 10-27 kg

Electron’s kinetic energy

ke = 20 keV = 20 x 103 eV

= 20 x 103 x 1.60 x 10-19

= 3.2 x 10-15 J

Proton’s kinetic energy,

kp = 200 kev = 2 x 105 eV

= 3.2 x 10-14 J

To find the velocity of electron ve, the kinetic energy is used.

Eke = \(\frac{1}{2}\) m\(v_{e}^{2}\)

Ve = \(\sqrt{\frac{2\times E_{ke}}{m}}\)

= \(\sqrt{\frac{2\times 3.2\times 10^{-15}}{9.11\times 10^{-31}}}\)

= 8.38 x 107 m/s

To find the velocity of proton vp, the kinetic energy is used.

Ekp = \(\frac{1}{2}\) m \(v_{p}^{2}\)

vp =  \(\sqrt{\frac{2\times E_{kp}}{m}}\)

vp = \(\sqrt{\frac{2\times 3.2\times 10^{-15}}{1.67\times 10^{-27}}}\)

= 6.19 x 106 m/s

Thus, electron moves faster when compared with proton.

The speed ratios are:

\(\frac{v_{e}}{v_{p}}\) = \(\frac{8.38\times 10^{7}}{6.19\times 10^{6}}\) = \(\frac{13.53}{1}\)


Que.13. From a height of 1000 m above the ground, a water drop of 4 mm falls. Till the half way, the speed decreases due to the air resistance and after the half mark, it attains its maximum speed and moves in a constant speed after it. Give the work done by the gravity in the first and the last phase of the journey. If the speed of the drop on reaching the ground is 20 m s-1, find the work done by the resistive force for the entire journey.

Ans. Radius of the water drop, r = 4 mm = 4 x 10-3 m

Volume V = \(\frac{4}{3}\) \(\pi\) \(r^{3}\)

= \(\frac{4}{3}\) x 3.14 x ( 4 x 10-3 )3 m-3

Density of water, \(\rho\) = 103 kg m-3

Mass m = \(\rho\)V

m = \(\frac{4}{3}\) x 3.14 x ( 4 x 10-3 )3 x 103 kg

Force due to gravity, F = mg

F = \(\frac{4}{3}\) x 3.14 x ( 4 x 10-3 )3 x 103 x 9.8 N

The work done on the water drop by the force of gravity in the first half of the journey:

W1 = Fs

= \(\frac{4}{3}\) x 3.14 x (4 x 10-3) x 103 x 9.8 x 500

= 1.31 J

The work done on the water drop by the force of gravity in the second half of the journey:

W2 = 1.31 J

According to the law of conservation of energy, the total energy of the water drop remains the same if there is no resistive force.

Therefore, the total energy at the top:

ET = mgh + 0

= \(\frac{4}{3}\) x 3.14 x (4 x 10-3)3 x 103 x 9.8 x 1000 x 10-5

= 2.62 J

The drop reaches the ground at a velocity of 20 m/s.

Total energy at the ground:

EG = \(\frac{1}{2}\)mv2 + 0

= \(\frac{1}{2}\) x \(\frac{4}{3}\) x 3.14 x (4 x 10-3)3 x 103 x 9.8 x (20)2

= 0.525 J

Resistive force = EG – ET = –2.095 J


Que.14. A molecule with a speed of 300 m s-1 hits the wall of the container at an angle \(40^{\circ}\) with the normal and rebounds with the same speed. During the collision is the momentum conserved?  Is it an elastic or an inelastic collision?

Ans. The collision is an elastic collision.

Whether the collision is an elastic or an inelastic collision, the momentum gets conserved. The molecule travels at a speed of 300 m/s and strikes the wall and rebounds with the same speed. Thus, the wall’s rebound velocity is zero. During the collision, the total kinetic energy gets conserved.


Que.15. A water pump takes 20 minutes to pump water to fill a tank of volume 40 m3. If the efficiency of pump is 40 % and the tank is at a height 50 m, how much electric power is needed for the pump?

Ans. Tank volume V = 40 m3

Operation time, t = 20 min = 20 x 60 = 1200 s

Height of the tank, h = 50 m

Efficiency, \(\eta\) = 40 %

Water density, \(\rho\) = 103 kg/m3

Mass of water, m = \(\rho\)V = 40 x 103 kg

Output power, P0 = \(\frac{Work\; done}{Time\;taken}\) = \(\frac{mgh}{t}\)

= \(\frac{40\times 10^{3}\times 9.8\times 50}{1200}\)

= 16.333 x 103 W

To find input power:

\(\eta\) = \(\frac{P_{0}}{P_{i}}\) = 40 %

i = \(\frac{16.333}{40}\times 100\times 10^{3}\)

= 40.8 KW


Que.16. On a frictionless table, two ball bearing which are identical are in contact with each other and they are hit by an another ball of same mass head-on with the initial speed V. Which of the following are correct, if the collision is an elastic collision.

Friction less Table

Ans. In each case, the total momentum before and after collision is same. So, the kinetic energy should be conserved before and after collision.

Before collision:

Kinetic energy = \(\frac{1}{2}\)mV2 + \(\frac{1}{2}\)(2m)0

= \(\frac{1}{2}\)mV2                                      

Scenario I:

After collision, the kinetic energy = \(\frac{1}{2}\)m x 0 + \(\frac{1}{2}\)(2m) x \(\frac{V}{4}^{2}\)

= \(\frac{1}{4}\) mV2

Thus, the K.E is not conserved.

Scenario II:

Kinetic energy after collision =  \(\frac{1}{2}\)(2m) x 0 + \(\frac{1}{2}\)mV2

= \(\frac{1}{2}\)mV2

The kinetic energy is conserved in this case

Scenario III:

Kinetic energy after collision = \(\frac{1}{2}\)(3m) \(\frac{V}{9}^{2}\)

= \(\frac{1}{6}\)mV2

The kinetic energy is not conserved in this case.

Hence, the scenario II is the correct answer.


Que.17. A ball A which is at an angle \(30^{\circ}\) to the vertical is released and it hits a ball B of same mass which is at rest. Does the ball A rises after collision? The collision is an elastic collision.

Elastic Collission

Ans. In an elastic collision when the ball A hits the ball B which is stationary, the ball B Acquires the velocity of the ball A while the ball A comes to rest immediately after collision. There is transfer of momentum to the moving body from the stationary body. Thus, the ball A comes to rest after collision and ball B moves with velocity of ball A.


Que.18. From a horizontal position, the bob of the pendulum is released. The pendulum’s length is 2.5 m. At what speed does the bob arrive at the lowest point? 10 % of the initial energy is dissipated due to the air resistance.

Ans. Length, l = 2.5 m

Mass = m

Energy dissipated due to air resistance = 10 %

The total energy of the system remains constant due to the law of conservation of energy.

At horizontal position:

Potential energy Ep = mgl

Kinetic energy Ek = 0

Total energy = mgl

At the lowest point:

Potential energy, Ep = mgl

Kinetic energy, Ek = \(\frac{1}{2}\)mv2

Total energy E= \(\frac{1}{2}\)mv2

10% energy is dissipated when the bob moves from the horizontal position. The total energy at the lowest point is 90 % of the energy at the horizontal point.

\(\frac{1}{2}\)mv2 = \(\frac{90}{100}\) x mgl

v = \(\sqrt{\frac{2\times 90\times 2.5\times 9.8}{100}}\)

= 6.64 m/s


Que.19. A sandbag of 30 kg is carried by a trolley of mass 250 Kg which moves at a speed of 24 km/h on a frictionless track. The sand starts to leak through a holes at a rate of 0.06 kg s-1. After the sand bag gets empty, find the speed of the trolley.

Ans. The trolley with a sand bag on it moves with the speed of 24 km/h. The external force is zero on the system. There is no velocity change even when the sand starts leaking from the bag as there are no external force produced on the system due to the leaking action. Hence, the speed remains same.


Que.20. A body travels in a straight line whose velocity and mass of v = a\(x^{\frac{3}{2}}\) and 0.6 kg where a = 5 \(m^{\frac{-1}{2}}\) s-1. Find the work done when it is displaced from x = 0 to x = 2 m.

Ans. Mass = 0.6 kg

v = a \(x^{\frac{3}{2}}\) where a = 5 \(m^{\frac{-1}{2}}\)  s-1

Initial velocity u = 0

Final velocity v = 10\(\sqrt{2}\) m/s

Work done W = \(\frac{1}{2}\)m (v2 – u2)

= \(\frac{1}{2}\) x 0.5 x { (10\(\sqrt{2}\))2 – (0)2}

= \(\frac{1}{2}\) x 0.5 x 10 x 10 x 2

= 50 J.


Que.21. The windmill sweeps a circle of area A with their blades. If the velocity of the wind is perpendicular to the circle, find the air passing through it in time t and also the kinetic energy of the air. 25 % of the wind energy is converted into electrical energy and v = 36 km/h, A = 30 m2 and the density of the air is 1.2 kg m-3. What is the electrical power produced?

Area = A

Velocity = V

Density = \(\rho\)

(a) Volume of the wind through the windmill per sec = Av

Mass = \(\rho\)Av

Mass m through the windmill in time t = \(\rho\)Avt

(b) kinetic energy = \(\frac{1}{2}\)mv2

= \(\frac{1}{2}\) (\(\rho\)Avt)v2 = \(\frac{1}{2}\) \(\rho\)Av3t

(c) Area = 30 m2

Velocity = 36 km/h

Density of air \(\rho\) = 1.2 kg m-3

Electric energy = 25 % of wind energy

= \(\frac{25}{100}\) x kinetic energy

= \(\frac{1}{8}\) \(\rho\)Av3t

Power = \(\frac{Electric\;energy}{Time}\)

= \(\frac{1}{8}\) \(\frac{\rho Av^{3}t}{t}\) = \(\frac{1}{8}\) \(\rho\)Av3

= \(\frac{1}{8}\) x 1.2 x 30 x 103

= 4.5 x 103 W =4.5 kW


Que.22. A person who wants to lose weight lifts a mass of 20 kg to a height of 0.6 m a thousand times. Each time she lower the mass, the potential energy lost is dissipated. (a) What is the work done against gravitational force? The fat supplies 3.8 x 107 J of energy which is converted with an efficiency of 20 % into mechanical energy. How much fat is used up? 

Ans. Mass m = 20 kg

Height h = 0.6 m

No. of times weight lifted n = 1000

Work done against gravitational force = n(mgh)

= 1000 x 20 x 9.8 x 0.6 = 117.6 KJ

Energy equivalent to 1 Kg of fat = 3.8 x 107 J

Efficiency = 20 %

Mechanical energy = \(\frac{20}{100}\) x 3.8 x 107 J

= \(\frac{1}{5}\) x 3.8 x 107 J

Mass of fat lost = \(\frac{1}{\frac{1}{5}\times 3.8\times 10^{7}}\times 117.6\times 10^{3}\)

=1.5 x 10-2 kg


Que.23. 10 KW of power is used by a family.

(a) On the horizontal surface, the average solar energy generated is 300 W per square meter. If the electrical energy conversion is 25%, then to supply 10 KW, how  large should the area be?

(b) Compare the typical house with this area?

Ans. Power P = 10 x 103 W

Solar energy = 300W

Conversion efficiency = 25%

Area required = A

10 x 103 = 25 % x (A x 300)

= \(\frac{25}{100}\) x A x 300

A = \(\frac{10\times 10^{3}}{75}\) =133 m2

This area is almost equal to the roof having dimensions 11 m x 11 m.


Que.24. A bullet travels at a speed of 80 m s-1 whose mass is 0.018 kg and strikes the piece of wood whose mass is 0.6 kg and comes to rest instantly. By means of a thin wire, the block is suspended from the ceiling. How high does  the block rise and also find the heat produced.

Ans. Bullet mass = 0.018 Kg

Initial bullet speed, u­b = 80 m/s

Mass of wood, M = 0.6 Kg

Initial speed of wood uB = 0

Final speed = v

According to law of conservation of momentum:

Mub + MuB = (m + M)v

0.018 x 80 + 0.6 x 0 = (0.018 + 0.6)v

v = \(\frac{1.44}{0.618}\) = 2.33 m/s

System mass m’ = 0.618 kg

System velocity = 2.33 m/s

Height = h

According to law of conservation of energy

Potential energy at highest point = kinetic energy at lowest point

m’gh = \(\frac{1}{2}\)m’v2

h = \(\frac{1}{2}\)( \(\frac{v^{2}}{g}\))

= \(\frac{1}{2}\) x \(\frac{2.33^{2}}{9.8}\)

h= 0.276 m

Heat produced = kinetic energy of the bullet – system

= \(\frac{1}{2}\)mu2 – \(\frac{1}{2}\)m’v2

= \(\frac{1}{2}\) x 0.018 x 802 – \(\frac{1}{2}\) x 0.618 x 2.332

= 57.6 – 1.67

= 55.93 J


Que.25. Two stones are allowed to slide down from point A without any friction as shown. Does the stones reach down at the same time and is the speed remains same till they reach there? \(\theta _{1}\) = \(30^{\circ}\), \(\theta _{2}\) = \(60^{\circ}\) and height = 10 m. Find the speed and the time taken by the stones.




The potential energy at point A is same as the initial height (h) is same

According to law of conservation of energy, at point B and C, the kinetic energy will be same for the stones.

\(\frac{1}{2}\)m\(v_{1}^{2}\) = \(\frac{1}{2}\)m\(v_{2}^{2}\)

v1 = v2 = v

m = mass of each stone

v = Speed of each stone at B and C

They reach at the same speed v.

For stone I:

Net force on the stone:

Fnet = ma1 = mg sin\(\theta _{1}\)

a1  g sin\(\theta _{1}\)

Stone II:

a2 = g sin\(\theta _{2}\) \(\theta _{2}\) > \(\theta _{1}\)

sin \(\theta _{2}\) > sin \(\theta _{1}\)

a2 > a1

Time can be found using motion equation

v = u + at

t1 = \(\frac{v}{a}\) (u = 0)

For stone I:

t1 = \(\frac{v}{a_{1}}\)

For stone II

t2 = \(\frac{v}{a_{2}}\)

a2 > a1

t2 < t1

So, the stone moving in a steep plane reaches first.

By law of conservation of energy:

Mgh = \(\frac{1}{2}\)mv2

v = \(\sqrt{2gh}\)

= \(\sqrt{2\times 9.8\times 10}\)

= \(\sqrt{196}\) = 14 m/s

t1 = \(\frac{v}{a_{1}}\) = \(\frac{v}{g\; sin\theta _{1}}\) = \(\frac{14}{9.8\times sin30}\) = \(\frac{14}{9.8\times \frac{1}{2}}\) = 2.86 s

t2 = \(\frac{v}{a_{2}}\) =\(\frac{v}{g\; sin\theta _{2}}\) =\(\frac{14}{9.8\times sin30}\) =\(\frac{14}{9.8\times \frac{\sqrt{3}}{2}}\) = 1.65 s.


Que.26. A spring is attached to a 1 kg block which is inclined on a rough surface having a spring constant of 100 N m-1. The block moves 10 cm with the string in the unstretched position before coming to rest. Find the coefficient of friction between the block and the incline.


Ans. Mass m = 1 Kg

Spring constant, k = 100 N m-1

Displacement = 0.1 m

At equilibrium:

R = mg cos \(37^{\circ}\)

f = \(\mu\) R = mg sin \(37^{\circ}\)

coefficient of friction = \(\mu\)

Net force = mg sin \(37^{\circ}\) – f

= mg sin \(37^{\circ}\) – \(\mu\) mg cos\(37^{\circ}\)

= mg(sin \(37^{\circ}\) – \(\mu\) cos\(37^{\circ}\)

Work done is equal to potential energy at equilibrium.

mg(sin \(37^{\circ}\) – \(\mu\) cos\(37^{\circ}\))x = \(\frac{1}{2}\)kx2

1 x 9.8 (sin \(37^{\circ}\) – \(\mu\) cos\(37^{\circ}\)) = \(\frac{1}{2}\) x 100 x 0.1

0.602 –  \(\mu\) x 0.799 = 0.510

\(\mu\) = \(\frac{0.092}{0.799}\) = 0.115


Que.27. An elevator is moving down with an uniform speed of 8 m s-1 and a bolt of mass 0.4 kg falls from the ceiling. It hits the elevator’s floor and does not rebound. The length of the elevator is 3 m. Find the heat produced during the impact. What will be the answer if the elevator is stationary?

Ans. Mass m = 0.4 kg

Speed of the elevator = 8 m/s

Height h = 3 m

The potential energy gets converted into heat energy as the relative velocity is zero.

Heat produced = loss of potential energy

= mgh = 0.4 x 9.8 x 3

= 11.76 J

Even if the elevator is stationary, the heat produced is still the same.


Que.28. On a frictionless track, a trolley moves with a speed of 36 km/h with a mass of 200 Kg. A child whose mass is 20 kg runs on the trolley with a speed of 4 m s1 from one end to other which is 20 m. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the  trolley moved from the time the child began to run.

Ans. Mass m = 200 Kg

Speed v = 36 km/h = 10 m/s

Mass of boy = 20 Kg

Initial momentum = (M + m)v

= (200 +20) x 10

= 2200 kg m/s

\(v{}'\) is final velocity of the trolley

Final velocity of boy = M\(v{}'\) +m(\(v{}'\) – 4 )

= 200\(v{}'\) + 20\(v{}'\) – 80

= 220\(v{}'\) – 80

According to law of conservation of energy:

Initial momentum = final momentum

2200 = 220\(v{}'\) – 80

\(v{}'\) = \(\frac{2280}{220}\) = 10.36 m/s

Length l = 20 m/s

\(v{}”\) = 4 m/s

t = \(\frac{20}{4}\) = 5 s

Distance moved by the trolley = \(v{}”\) x t = 10.36 x 5 = 51.8 m


Que.29. Which of the following does not describe the elastic collision of two billiard balls? Distance between the centers of the balls is r.

Elastic Collision

Ans. (i), (ii), (iii), (iv) and (vi).

The potential energy of two masses in a system is inversely proportional to the distance between them. The potential energy of the system of two balls will decrease as they get closer to each other. When the balls touch each other, the potential energy becomes zero, i.e. at r = 2R. The potential energy curve in (i), (ii), (iii), (iv) and (vi) do not satisfy these conditions. So, there is no elastic collision.


Que.30. The decay of free neutrons at rest: n\(\rightarrow\)p + e

Show that the two body decay must give an electron of fixed energy and therefore, can’t account for continuous energy distribution in \(\beta\)-decay of a neutron or a nucleus.


Ans. The decay process of free neutron at rest:

n\(\rightarrow\) p + e

From Einstein’s mass energy relation

Electron energy = \(\Delta\)mc2

\(\Delta\)m = mass defect = mass of neutron – (mass of proton and electron)

c = Speed of light

The presence of neutrino on the LHS of decay explains the continuous energy distribution.

A few of the important topics in this chapter are given below

  • Law of conservation of energy
  • Work Energy theorem
  • Kinetic energy and potential energy
  • Potential energy of a spring
  • Collisions

Note that none of the topics in the chapter should be skipped.

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