# NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT solutions for class 11 physics chapter 6 work energy and power: Consider that you are carrying a 10 Kg stationary in your hand and you are not moving. In this situation, you are doing work physically according to you. But in physics, you are doing zero work. Can you find why work is zero here? The solutions of NCERT class 11 physics chapter 6 work energy and power has questions based on the concepts of work and types of work, the relation between work and energy and the concept of power. In the CBSE NCERT solutions for class 11 physics chapter 6 work energy and power, you will study problems on conservation of energy (energy can be neither created nor be destroyed). NCERT solutions for class 11 physics chapter 6 work energy and power also give answers to the questions based on the concepts of the collision, which are important for the exam. Solutions of NCERT are helpful in scoring well in class and board exams. Solutions for both exercise and additional exercise are given.

Exercise

## The main topics of the NCERT  class 11 physics chapter 6 work energy and power are listed below:

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

## Some of the main points on the topic collisions

• For an  elastic collision Law of conservation of momentum and that of Kinetic Energy holds good
• For inelastic collision Law of conservation of momentum hold good but kinetic energy is not conserved
• Coefficient of restitution is the ratio of relative velocity after the collision to relative velocity before the collision
• For a perfectly elastic collision coefficient of restitution is one
• For inelastic collision coefficient of restitution is less than one
• For a perfectly inelastic collision coefficient of restitution is zero

NCERT solutions for class 11 physics chapter 6 work energy and power exercise

work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

In this case, the direction of force and displacement are the same, this work done is positive

In this case, the direction of displacement is upward and the direction of force is downward. Thus work done is negative in nature.

We know that friction acts in the direction opposite to the direction of motion. Hence work done by the frictional force is negative.

## Q 1 (d)  The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: work done by an applied force on a body moving on a rough horizontal plane with uniform velocity

In this case, the applied force supports the motion of the object (balances frictional force). Thus work done by the force is positive.

Work done in this case is negative as the direction of force and motion are not identical.

Using Newton's law we can write :

$a\ =\ \frac{F}{m}$

$=\ \frac{7}{2}\ =\ 3.5\ m/s^2$

The frictional force is given by :

$f\ =\ \mu mg$

$=\ 0.1\times 2\times 9.8\ =\ 1.96\ N$

Its direction will be opposite of the direction of the motion. Thus acceleration produced will be negative.

$a\ =\ \frac{-1.96}{2}\ =\ -0.98\ m/s^2$

Thus the net acceleration is  =   3.5  -   0.98  =   2.52 m/s2.

The total distance travelled is given by :

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or                                                  $=\ 0\ +\ \frac{1}{2}(2.52)10^2\ =\ 126\ m$

Hence the work done by applied force is given by :

$W\ =\ F.s\ =\ 7\times 126\ =\ 882\ J$

The work done by frictional force will be negative as the force opposes the motion.

$W\ =\ f.s\ =\ -1.96\times 126\ =\ -247\ J$

## Q2 (c)  A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the work done by the net force on the body in 10 s,

The net work done will be the sum of work done by applied force and work done by frictional force.

$W\ =\ 882\ -\ 247\ =\ 635\ J$

It is given that initial velocity is zero. The final velocity can be calculated by the equation of motion :

$v\ =\ u\ +\ at$

or                                                       $=\ 0\ +\ (2.52)10$

or                                                       $=\ 25.2\ m/s$

Thus change is kinetic energy is :

$\Delta K\ =\ \frac{1}{2}mv^2\ -\ \frac{1}{2}mu^2$

or                                                           $=\ \frac{1}{2}\times 2(25.2)^2\ -\ 0$

or                                                            $=\ 635\ J$

Total energy = kinetic energy (KE) + potential energy(PE)

KE > 0 since m and v2 is positive. If KE <0 particles cannot be find. If PE>TE, then KE<0 (now in all graph check for this condition)

In case 1 kinetic energy is negative for x<a. So at x<a particle cannot be found.

In case 2 for x<a and for x> b kinetic energy is negative. So the particle cannot be found in these regions.

In the third case, the minimum potential energy is when   $a. At this position, the potential energy is negative (- V1).

The kinetic energy in this case is given by :

$K.E.\ =\ E\ -\ (-V_1)\ =\ E\ +\ V_1$

And the minimum energy of particle is  - V1.

In the fourth case,  the particle will not exist in the states which will have potential energy greater than the total energy.

Thus particle will not exist in    $\frac{-b}{2}   and   $\frac{-a}{2} .

The minimum energy of particle will be - V1 as it is the minimum potential energy.

## Q4  The potential energy function for a  particle executing linear simple harmonic motion is given by $V ( x ) = k x^2 /2$, where k is the force constant of the oscillator. For $k = 0.5 Nm^{-1}$ the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of  total energy 1 J moving under this potential must ‘turn back’ when it reaches $x = \pm 2$

The total energy of the particle is given by :

$E\ =\ K.E\ +\ P.E$

or                                               $=\ \frac{1}{2}mv^2\ +\ \frac{1}{2} kx^2$

At the extreme position, the velocity of the object is zero thus its kinetic energy at that point is zero.

$E\ =\ \frac{1}{2} kx^2$

or                                      $1\ =\ \frac{1}{2} (0.5)x^2$

or                                      $x^2\ =\ 4$

or                                      $x\ =\ \pm \ 2$

Hence the extreme position are $\pm \ 2\ m$

The total energy is given by :

$E\ =\ K.E\ +\ P.E$

or                                           $=\ \frac{1}{2}mv^2\ +\ mgh$

The burning of casing results in a reduction in the mass of the rocket. This leads to a lowering in the total energy.

Thus heat required for burning is obtained from the expenses of the rocket.

This is because the gravitational force is a conservative force. And we know that the work done by a conservative force in a closed path is always zero. That's why the work done by the gravitational force is zero in a complete orbit revolution.

The total energy of artificial satellite remains constant. Thus when it approaches towards the earth the distance between them decreases. This results in a decrease in the potential energy of the satellite. By energy conservation, the kinetic energy of satellite increases and so does the velocity.

In the first case,

Work done is  :

$W\ =\ F.s\ =\ Fs \cos \Theta$

or                                                                           $=\ mgs \cos \Theta$

or                                                                           $=\ 15\times 9.8\times2\times \cos 90^{\circ}$

or                                                                           $=\ 0$

In the second case :

$W\ =\ Fs \cos \Theta$

or                                                     $=\ mgs \cos 0^{\circ}$

or                                                     $=\ 15\times 9.8\times 2\ =\ 294\ J$

Thus work done in the second case is greater than the first case.

It is given that work done by the conservative force is positive, thus the force acts in the direction of the motion. This results in a decrease in distance between the bodies. Thus it's potential energy decreases

Work done by the body against friction results in a decrease in the velocity of the body. Thus the kinetic energy of the body decreases.

The internal force cannot produce a change in the total momentum as no external force is acting. Thus the change in total momentum is proportional to the external forces acting on the body.

The conservation of total linear momentum doesn't depend upon the fact whether it is an elastic collision or an inelastic collision.

In an elastic collision of two bodies, the momentum and energy of each body is conserved.

False:- The linear momentum and energy will be conserved if both are considered in a system. But for individual bodies, this conservation of momentum and energy doesn't hold. This is because the impact during the collision may transfer energy/momentum of one ball to the other ball.

## Q7 (b)  State if each of the following statements is true or false. Give reasons for your answer.  (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

False:- Internal forces will not change the energy of the system but external forces can change the total energy by changing their magnitude or direction.

False:-   This is true only for conservative forces e.g. gravitational force.  For e.g in case of frictional force (non-conservative force), the work done in a closed-loop cannot be zero as energy is wasted throughout.

True but not always:-   In the case of inelastic collisions, few amounts of energy is converted into other forms of energy such as sound or in deformation. Thus final kinetic energy is always less as compared to initial kinetic energy. But in case of the explosion of a bomb final kinetic energy is greater than the initial kinetic energy

No, because at the time of the collision, the kinetic energy is converted to the potential energy. Thus total kinetic energy is not constant at the collision.

Yes, in case of elastic collision the total linear momentum of the system remains conserved as no external force is acting on the system of balls.

## Q8 (c) What are the answers to (a) and (b) for an inelastic collision?

The total kinetic energy of the system cannot be conserved in case of inelastic collision as there is loss of energy in the form of deformation. But the total linear momentum of the system remains constant even in the case of inelastic collision as no external force is acting.

Since the potential energy of the system depends upon the separation between the bodies thus the forces acting on the body are conservative in nature. We know that conservative forces produce elastic collisions.

It is given that acceleration is constant thus force will also be constant (by Newton's law of motion   F = ma).

Also,

$a\ =\ \frac{dv}{dt}\ =\ constant$

or                                        $dv\ =\ C\ dt$

Thus                                    $v\ \propto \ t$

Now, the work done by the force is given by :

$P\ =\ F.v$

Hence power is directly proportional to the time.

## Q10  A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

$(i) t^{1/2}\: \: (ii) t \: \: (iii) t^{3/2}\: \: (iv) t^2$

We know that the power is given by :

$P\ =\ F.v$

or                                                    $=\ ma.v$

or                                                   $=\ m\frac{dv}{dt}v$

It is given that power is constant, thus :

$mv\frac{dv}{dt}\ =\ constant$

or                                              $vdv\ =\ \frac{C}{m}dt$

By integrating both sides, we get

$v\ =\ \left ( \sqrt{\frac{2Ct}{m}} \right )$

Also, we can write :

$v\ =\ \frac{dx}{dt}$

or                                              $\frac{dx}{dt}\ =\ \sqrt{\frac{2C}{m}}t^\frac{1}{2}$

By integrating we get the relation :

$x\ \propto \ \ t^\frac{3}{2}$

## Q11  A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by$F = - \hat i + 2 \hat j + 3 \hat k N$ where$\hat i , \hat j , \hat k$ 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?

Force is given to be :

$F = - \hat i + 2 \hat j + 3 \hat k N$

And the displacement is :

$s\ =\ 4 \hat k\ m$

Thus the work done is given by :

$W\ =\ F.s$

or                                                              $=\left ( - \hat i + 2 \hat j + 3 \hat k \right ). \left ( 4 \hat k \right )$

or                                                               $=0 +0 \ -\ 3(4)\ =\ \ 12\ J$

## Q12  An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass  $= 9.11 \times 10 ^{-31} Kg$ , proton mass        $1.67 \times 10 ^{-27} Kg , 1 e V = 1.60 \times 10 ^{-19} J$)

The kinetic energy of the electron is given by :

$K_e\ =\ \frac{1}{2}mv_e^2$

or                                             $1.6\times 10^{-15}\ J\ =\ \frac{1}{2}\times 9.11\times 10^{-31}\times v_e^2$

Thus velocity is obtained as :

$v_e\ =\ \sqrt{\frac{2\times 1.6\times 10^{-15}}{9.11\times 10^{-31}}}$

or                                                               $=\ 5.93\times 10^7\ m/s$

Similarly, we can find the velocity of the proton :

$K_p\ =\ \frac{1}{2}mv_p^2$

$1.6\times 10^{-14}\ J\ =\ \frac{1}{2}\times 1.67\times 10^{-27}\times v_p^2$

Thus velocity is obtained as :

$v_p\ =\ \sqrt{\frac{2\times 1.6\times 10^{-14}}{1.67\times 10^{-27}}}$

or                                                               $=\ 4.38\times 10^6\ m/s$

Thus the ratio of their velocities is :

$\frac{v_e}{v_p}\ =\ \frac{5.93\times 10^{7}}{4.38\times 10^6}\ =\ 13.54$

The volume of the drop is :

$V\ =\ \frac{4}{3}\pi r^3\ =\ \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3$

Thus the mass of raindrop is :

$m\ =\ \rho v$

or                                                 $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ Kg$

Thus the work done is given by :

$W\ =\ F.s$

or                                                 $=\ mgs$

or                                                 $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3 \times 9.8\ \times 250$

or                                                  $=\ 0.082\ J$

Now the total energy at the peak point is :

$E_p\ =\ mgh\ +\ 0\ =\ mgh$

or                                                                                $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ \times 9.8 \times 500$

or                                                                                 $=\ 0.146\ J$

And the energy at the ground is :

$E_b\ =\ 0\ +\ \frac{1}{2}mv^2\ =\ \frac{1}{2}mv^2$

or                                                                                     $=\ \frac{1}{2}\times 10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ \times (10)^2$

or                                                                                      $=\ 1.67\times 10^{-3} \ J$

Thus work done by the resistive force is :

$=\ 1.67\times 10^{-3} \ J\ -\ 0.164\ J\ =\ -\ 0.162\ J$

The momentum is conserved in the collision as no external force is acting on the system. In the given case the rebound velocity is the same as the initial velocity thus the kinetic energy of the molecule initially and finally are same. Hence this is an elastic collision.

Mass of the water is :

$m\ =\ \rho v$

or                                               $=\ 30\times10^3 \ Kg$

Thus the output power is given by :

$Power\ =\ \frac{Work\ done}{Time}$

or                                                         $=\ \frac{mgh}{t}$

or                                                         $=\ \frac{30\times 10^3\times 9.8\times 40}{900}$

or                                                         $=\ 13.067\times 10^3\ W$

Also, we are given that efficiency is 30 per cent.

Thus the input power is :

$P_i\ =\ \frac{13.067}{\frac{30}{100}}\times 10^3$

or                                                     $=\ 43.6\ KW$

The initial kinetic energy of the system is given by :

$=\ \frac{1}{2}mv^2\ -\ \frac{1}{2}2m(0)$

or                                                   $=\ \frac{1}{2}mv^2$

Case (i):-      The final kinetic energy is :

$=\ \frac{1}{2}m.0\ -\ \frac{1}{2}2m(\frac{v}{2})^2\ =\ \frac{1}{4}mv^2$

Thus the kinetic energy is not conserved in this case.

Case (ii):-      The final kinetic energy is :

$=\ \frac{1}{2}2m.0\ -\ \frac{1}{2}mv^2\ =\ \frac{1}{2}mv^2$

Thus kinetic energy is conserved in this case.

Case (iii):-    The final kinetic energy is:-

$=\ \frac{1}{2}\times 3m\times \left ( \frac{v}{3} \right )^2$

or                                                       $=\ \frac{1}{2}m v^2$

Thus the kinetic energy is not conserved in this case.

## Q17  The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

This is an elastic collision thus the transfer of momentum will take place. It is given that bob B is at rest and bob A has some velocity. So in momentum transfer, bob B will gain the velocity in the left direction whereas bob A will come to rest (complete momentum transfer takes place).

Hence bob A will not rise.

## Q18  The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Consider the extreme position (horizintal) :-

The kinetic energy at this position is zero as velocity is zero.

Thus total energy is given by :    $=\ mgl\ +\ 0 \ =\ mgl$

Now consider the mean position (lowermost point) :

Here the potential energy of bob is zero.

Whereas   kinetic energy is  :

$=\ \frac{1}{2}mv^2$

Further, it is given that 5 per cent of energy is dissipated due to air resistance while coming down.

Thus energy equation becomes (conservation of energy):-

$\frac{1}{2}mv^2\ =\ \frac{95}{100}\times mgl$

or

$v\ =\ \sqrt{\frac{2\times 95\times1.5\times 9.8}{100}}\ =\ 5.28\ m/s$

Since the sand is falling in the trolley thus the force generated on the system (trolley and sandbag) is an internal force. There is no external force thus momentum of the system doesn't change. Hence speed remains the same i.e.,  27 Km/hr.

The relation between work done and the kinetic energy is given by :

$Work\ =\ \frac{1}{2}mv^2\ -\ \frac{1}{2}mu^2$

Using the relation   $v = ax ^{3/2}$  we can write :

Initial velocity  =   0    (at x  =  0 )

And          the final velocity    =   $10\sqrt{2}\ m/s$   (at  x  = 2).

Thus work done is :

$Work\ =\ \frac{1}{2}m(v^2\ -\ u^2)$

or                                                 $=\ \frac{1}{2}\times 0.5\times (10\sqrt{2})^2$

or                                                  $=\ 50\ J$

The volume of wind   =   $Av$$\rhov$ here  $A$ is the swept circle and $v$ is the velocity.

Thus the mass of the wind is  : -    $\rho Av$  ,  $\rho$   is the density of the air.

Hence mass of wind flowing through windmill in time t is     $=\ \rho Avt$.

The kinetic energy is given by :

$=\ \frac{1}{2}mv^2$

or                                                $=\ \frac{1}{2}\ \rho Avt\ v^2$

or                                              $=\ \frac{1}{2}\ \rho At\ v^3$

Thus the kinetic energy of wind is      $\frac{1}{2}\ \rho At\ v^3$  J.

It is given that 25 per cent of wind energy is converted into electrical energy.

Thus electric energy produced is :

$=\ \frac{25}{100} \times \frac{1}{2}\ \rho At v^3$

or                                                $=\ \frac{1}{8}\ \rho At v^3$

Now the electric power is given by  :

$Power\ =\ \frac{Energy}{Time}$

or                                                                $=\ \frac{\frac{1}{8}\ \rho At v^3}{t}\ =\ {\frac{1}{8}\ \rho A v^3}$

or                                                                 $=\ {\frac{1}{8}\ \times 1.2\times 30 \times (10)^3}$

or                                                                 $=\ 4.5\ KW$

## Q22 (a)  A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.  How much work does she do against the gravitational force?

The work done against the gravitational force is given by :

=   Number of times the weight is lifted $\times$ work done in 1 time.

$=\ 1000\times mgh$

or                                          $=\ 1000\times 10\times 9.8\times 0.5$

or                                          $=\ 49\ KJ$

Efficiency is given to be 20 per cent.

Thus energy supplied by the person :

$=\ \frac{20}{100}\times 3.8\times 10^7$

Thus the amount of fat lost is :

$=\ \frac{49\times 10^3}{\frac{20}{100}\times 3.8\times 10^7}$

or                                                 $=\ 6.45\times 10^{-3}\ Kg$

It is given that the efficiency of energy conversion is 20 per cent.

According to question, we can write (equating power used by family) :

$8\times 10^3\ =\ \frac{20}{100}\times A\times 200$                           (Here A is the area required.)

or                                                   $A\ =\ \frac{8\times 10^3}{40}$

or                                                          $=\ 200\ m^2$

Thus required area is 200 m2.

A typical has dimensions of  $14 \times 14\ m^2$.

The area of the roof of the house is $225\ m^2$.

This is nearly equal to the area required for the production of the given amount of electricity.

## NCERT solutions for class 11 physics chapter 6 work energy and power additional exercise

We are given :

Mass of the bullet m: 0.012 Kg

Mass of the block M: 0.4  Kg

The initial velocity of the bullet u: 70  m/s

The initial velocity of the block     :     0

The final velocity of the system (bullet + block): v

For finding the final speed of system we will apply the law of conservation of momentum :

$mu_b\ +\ M(0)\ =\ (m\ +\ M)v$

or                                             $v\ =\ \frac{0.84}{0.412}\ =\ 2.04\ m/s$

Now for the system, we will apply the law of conservation of energy :

The potential energy at the highest point  =   Kinetic energy at the lowest point

$(m+M)gh\ =\ \frac{1}{2}(m+M)v^2$

or                                                         $h\ =\ \frac{1}{2}\times \frac{v^2}{g}$

or                                                               $=\ \frac{1}{2}\times \frac{2.04^2}{9.8}$

or                                                               $=\ 0.2123\ m$

Hence heat produced is :

$=\ \frac{1}{2}mu^2\ -\ \frac{1}{2}(m\ +\ M)v^2$

or                                      $=\ \frac{1}{2}(0.012)(70)^2\ -\ \frac{1}{2}(0.412)(2.04)^2$

or                                      $=\ 29.4\ -\ 0.857\ =\ 28.54\ J$

The FBD of the track is shown in the figure below :

Using the law of conservation of energy we have :

$\frac{1}{2}mv_1^2\ =\ \frac{1}{2}mv_2^2$

or                                            $v_1\ =\ v_2$

Hence both stones will reach the bottom with the same speed.

For stone 1 we can write :

$F\ =\ mg\sin \Theta_1$

or                                $a_1\ =\ g\sin \Theta_1$

For stone 2 we have :

$a_2\ =\ g\sin \Theta_2$

Also, using the equation of motion,

$v\ =\ u\ +\ at$

or                              $t\ =\ \frac{v}{a}$

It is given that             $\Theta _2 > \Theta_1$

or                               $a _2 > a_1$

Thus                           $t _1 > t_2$

Hence, the stone travelling on the steep plane will reach before.

For finding speed and time we can use conservation of energy.

$mgh\ =\ \frac{1}{2}mv^2$

or                                      $v\ =\ \sqrt{2gh}$

or                                           $=\ \sqrt{2\times 9.8\times 10}$

or                                           $=\ 14\ m/s$

And the time is given by :

$t_1\ =\ \frac{v}{a_1}\ =\ \frac{14}{9.8\times \sin 30^{\circ}}\ =\ 2.86\ s$

and                                $t_2\ =\ \frac{v}{a_2}\ =\ \frac{14}{9.8\times \sin 60^{\circ}}\ =\ 1.65\ s$

Displacement (x) of the block is given as :   =  0.1 m.

Using equilibrium conditions we can write :

$R\ =\ mg \cos 37^{\circ}$

and                                          $\mu R\ =\ mg \sin 37^{\circ}$                                 ( $\mu R$ is the frictional force).

We can write work done in terms of potential energy as :

$mg\ \left ( \sin 37^{\circ}\ -\ \mu \cos 37^{\circ} \right )x\ =\ \frac{1}{2}kx^2$

or                                       $1\times g\ \left ( \sin 37^{\circ}\ -\ \mu \cos 37^{\circ} \right )x\ =\ \frac{1}{2}100\times (0.1)^2$

or                                                     $\mu\ =\ 0.125$.

Thus the coefficient of friction is 0.125.

## Q27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of $7 m s ^{-1}$. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

In this case, the heat produced is the loss in the potential energy.

Thus,

heat produced  =  mg h

or                                            $=\ 0.3\times 9.8\times 3$

or                                            $=\ 8.82\ J$

The heat produced (when the lift is stationary) will remain the same as the relative velocity of the bolt with respect lift still remains zero.

The initial momentum of the system (boy + trolley) is given as :

$=\ (m\ +\ M)V$

$=\ (200\ +\ 20)10\ =\ 2200\ Kg\ m/s$

Now assume v' is the final velocity of the trolley with respect to the ground.

Then the final momentum will be :

$=\ Mv'\ +\ m(v'\ -\ 4)\ =\ 220v'\ -\ 80$

Conserving momentum :

$220v'\ -\ 80\ =\ 2200$

or                                                      $v'\ =\ 10.36\ m/s$

The time taken by the boy is :

$=\ \frac{10}{4}\ =\ 2.5\ s$

Hence the distance moved by the trolley is :

$=\ 10.36\times 2.5 \ =\ 25.9\ m$

The potential energy of the system depends inversely on the separation between the balls. Thus the potential energy will decrease as the balls will come closer and will become zero as they touch each other.

Thus elastic collision is best described only by the graph (v).

[Note: The simple result of this exercise was one among the several arguments advanced by W.
Pauli to predict the existence of a third particle in the decay products of $\beta -decay$ This
particle is known as the neutrino. We now know that it is a particle of intrinsic spin ½ (like
e—, p or n), but is neutral, and either massless or having an extremely small mass
(compared to the mass of an electron) and which interacts very weakly with matter. The
correct decay process of the neutron is :$n \rightarrow p+ e ^-$ ]

By Einstein’s mass-energy relation we can write :

$Energy\ =\ \Delta mc^2$

Here $\Delta m$ and C are constant thus two-body decay is unable to explain (or account for) the continuous energy distribution in the β-decay of a neutron.

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## NCERT Solutions for Class 11 Subject wise

 NCERT solutions for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry CBSE NCERT solutions for class 11 physics

## Importance of NCERT solutions for class 11 physics chapter 6 work energy and power:

• Many questions were done in the solutions of NCERT class 11 physics chapter 6 work energy and power based on the concept of the work-energy theorem which is important throughout the physics course. The concept will be used in class 12 also.
• CBSE NCERT solutions for class 11 physics chapter 6 work energy and power will help to score well in-class exams and also competitive exams like NEET.
• Learning NCERT solutions will help to solve problems from other reference books also.