**1** **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:$\\$ (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.$\\$ (b) work done by gravitational force in the above case,$\\$ (c) work done by friction on a body sliding down an inclined plane,$\\$ (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,$\\$ (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.**

(a) Positive$\\$ In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.$\\$ (b) Negative$\\$ In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative. $\\$ (c) Negative$\\$ Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.$\\$ (d) Positive$\\$ Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.$\\$ (e) Negative$\\$ The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

**2** **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.$$\\$ Compute the$\\$ (a) work done by the applied force in $10 s,$$\\$ (b) work done by friction in $10 s,$$\\$ (c) work done by the net force on the body in $10 s,$$\\$ (d) change in kinetic energy of the body in $10 s,$$\\$ and interpret your results.**

Mass of the body, $m = 2 kg$$\\$ Applied force, $F = 7 N$$\\$ Coefficient of kinetic friction, $\mu = 0.1$$\\$ Initial velocity,$ u = 0$$\\$ Time, $t = 10 s$$\\$ The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:$\\$ $a ' =\dfrac{F}{m}= \dfrac{7 }{ 2} = 3.5 ms^{-2}$ Frictional force is given as:$\\$ $f = \mu mg\\ = 0.1 × 2 × 9.8 = - 1.96 N$$\\$ The acceleration produced by the frictional force:$\\$ $a "=\dfrac{- 1.96 }{ 2}=-0.98 ms^2$$\\$ Total acceleration of the body: $a ' + a "\\ = 3.5+(-0.98)=2.52 ms^{-2}$ The distance travelled by the body is given by the equation of motion: $s=ut +(\dfrac{1}{2}) at^2\\ =0 +(\dfrac{1}{2})* 2.52*(10)^2=126 m$$\\$ (a) Work done by the applied force, $W_a =F*s =7 *126 =882 J$$\\$ (b) Work done by the frictional force, $W_f =F *s =-1.96 * 126 =- 247 J$$\\$ (c) Net force = $7 + (-1.96) = 5.04 N$$\\$ Work done by the net force, $W_{net}= 5.04*126 = 635 J$$\\$ (d) From the first equation of motion, final velocity can be calculated as: $v = u + at\\ = 0 + 2.52 × 10 = 25.2 m/s$ Change in kinetic energy$=(\dfrac{1}{2})mv^2-(\dfrac{1}{2})mu^2\\ =(\dfrac{1}{2})*2(v^2-u^2)\\ =(25.2)^2-0^2=635 J$

**3** **Given in figure are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.**

Total energy of a system is given by the relation:$\\$ $E = P.E. + K. E.\\ \therefore K.E. = E - P.E.$$\\$ Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where $K.E.$ becomes negative.$\\$ (i)For $ x> a,P.E. (V_0) > E$$\\$ $\therefore K.E.$ becomes negative. Hence, the object cannot exist in the region $x > a.$$\\$ (ii)For $x < a$ and $ x> b,P.E.(V_0) >E.$$\\$ $\therefore K.E. $becomes negative. Hence the object cannot be present in the region $x < a$ and $x > b.$$\\$ (iii)$x > a$ and $x < b ; - V_1$$\\$ In the given case, the condition regarding the positivity of $K.E.$ is satisfied only in the region between $x > a$ and $x < b.$$\\$ The minimum potential energy in this case is $- V_1$ . Therefore, $K.E. = E - (- V_1)=E +V_1 .$ Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $-V_1 $. So, the minimum total energy the particle must have is $- V_1 .$ $\\$ (iv)$-\dfrac{b}{2} < x< \dfrac{a}{2};$$\\$ $\dfrac{a}{2}< x< \dfrac{b}{2};-V_1$$\\$ In the given case, the potential energy $(V_0)$ of the particle becomes greater than the total energy$\\$ (E) For $-\dfrac{b}{2} < x < \dfrac{b}{2}$ and $-\dfrac{a}{2} < x < \dfrac{a}{2}. $\\$ Therefore,$ K . E . = E -(- V_1)= E +V_1 .$\\$ Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $-V_1$ . So, the minimum total energy the particle must have is $-V_1 $.

**4** **The potential energy function for a particle executing linear simple harmonic motion is given by$V(x)=kx^2/2,$ where $k$ is the force constant of the oscillator. For $k=0.5 m^{-1},$ 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 $x =\pm 2 m .$**

Total energy of the particle, $E = 1 J$$\\$ Force constant, $k = 0.5 N m^{-1}$$\\$ Kinetic energy of the particle, $K =(\dfrac{1}{2})mv^2$$\\$ According to the conservation law:$\\$ $E = V + K\\ 1 =(\dfrac{1}{2}) k x^2 +(\dfrac{1}{2} )m v^2$$\\$ At the moment of ‘turn back’, velocity (and hence K) becomes zero.$\\$ $\therefore 1=(\dfrac{1}{2})kx^2\\ (\dfrac{1}{2})*0.5x^2=1\\ x^2=4\\ x=\pm2 $$\\$Hence, the particle turns back when it reaches $x =\pm 2 m .$

**5** **Answer the following:$\\$ (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?$\\$ (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?$\\$ (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?$\\$ (d) In Fig. $6.13 $ (i) the man walks $2 m$ carrying a mass of $15 kg $ on his hands. In Fig. $6.13$ (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of $15 kg$ hangs at its other end. In which case is the work done greater?**

(a) Rocket$\\$ The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket. According to the conservation of energy:$\\$ Total energy = Potential energy + Kinetic energy$\\$ $=mgh +(\dfrac{1}{2}) =mv ^2$$\\$ The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.$\\$ (b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero.$\\$ (c) When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in $P.E.$ results in an increase in $K.E.$ Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.$\\$ (d) Work done in fig $6.13$ (i) Mass,$m=15 kg $$\\$ Displacement , $s=2 m $$\\$ Work done, $W=Fs \cos \theta $$\\$ Where, $\theta=$ Angle between force and displacement $\\$ $=mgs \cos \theta =15*2*9.8\cos 90^o=0$$\\$ Work done in fig$6.13$(ii) Mass, $m=15 kg$$\\$ Displacement, $s=2 m $$\\$ Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.$\\$ Therefore, the angle between them, $ \theta =0^o$$\\$ Since $\cos 0^o=1$$\\$ Work done ,$W =F s \cos \theta =m gs$$\\$ $=15*9.8*2=294 J$$\\$ Hence, more work is done in the fig.$6.13$(ii).

**6** **$\textbf{Underline the correct alternative}:$$\\$ (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.$\\$ (b) Work done by a body against friction always results in a loss of its kinetic/potential energy.$\\$ (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.$\\$ (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.**

(a)$ \underline{Decreases}$$\\$ A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.$\\$ (b) $\underline{Kinetic energy}$$\\$ The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.$\\$ (c) $\underline{External force}$$\\$ Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system.$\\$ (d) $\underline{Total linear momentum}$$\\$ The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

**7** **$\textbf{State if each of the following statements is true or false. Give reasons for your answer.}$$\\$ (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.$\\$ (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.$\\$ (c) Work done in the motion of a body over a closed loop is zero for every force in nature.$\\$ (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.**

(a) False $\\$ In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.$\\$ (b) False$\\$ The external forces on the body may change the total energy of the body.$\\$ (c) False$\\$ The work done in the motion of a body over a closed loop is zero for a conservation force only.$\\$ (d) True$\\$ In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

**8** **Answer carefully, with reasons:$\\$ (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?$\\$ (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?$\\$ (c) What are the answers to (a) and (b) for an inelastic collision?$\\$ (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).**

(a) No$\\$ $K.E.$ is not conserved during the given elastic collision, $K.E.$ before and after collision is the same. In fact during collision, $K.E.$ of the balls gets converted into potential energy.$\\$ (b) Yes$\\$ In an elastic collision, the total linear momentum of the system always remains conserved.$\\$ (c) No; Yes$\\$ In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.$\\$ (d) Elastic$\\$ In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.