# NCERT Physics Class 11 Exemplar Ch 6 Work Energy and Power Part 6

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## SA

Q 34. A graph of potential energy verses x is shown in Fig. A particle of energy is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

Ans:

Q 35. A ball of mass m, moving with a speed, collides in elastically with an identical ball at rest. Show that

(a) For head-on collision, both the balls move forward.

(b) For a general collision, the angle between the two velocities of scattered balls is less than

Ans:

(a) For head on collision:

Conservation of momentum

Or

And

Since has the same sign as, therefore the ball moves on after collision.

(b) Conservation of momentum

But KE is lost

Thus and are related as shown in the figure.

is acute (less than would give)

Q 36. Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C and D in which the relation between potential energy V, kinetic energy (K) and total energy E is as given below:

Region A:

Region B:

Region C:

Region D:

State with reason in each case whether a particle can be found in the given region or not.

Ans:

Region A: No, as KE will become negative.

Region B: Yes, total energy can be greater than PE for non-zero K.E.

Region C: Yes, KE can be greater than total energy if its PE is negative.

Region D: Yes, as PE can be greater than KE.

Q 37. The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in Fig.

If the length of the pendulum is 1m, calculate

(a) The height to which bob A will rise after collision.

(b) The speed with which bob B starts moving. Neglect the size of the bobs and assume the collision to be elastic.

Ans:

(a) Ball A transfers its entire momentum to the ball on the table and does not rise at all.

(b)

Q 38. A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of. Calculate

(a) The loss of P.E. of the drop.

(b) The gain in K.E. of the drop.

(c) Is the gain in K.E. equal to loss of P.E.? If not why.

Take

Ans:

(a) Loss of

(b) Gain in

(c) No, because a part of PE is used up in doing work against the viscous drag of air.

Q 39. Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact. One of the bobs is released after being displaced by so that it collides elastically head-on with the other bob.

(a) Describe the motion of two bobs.

(b) Draw a graph showing variation in energy of either pendulum with time, for , where T is the period of each pendulum.

Ans:

(b)

Q 40. Suppose the average mass of raindrops is and their average terminal velocity. Calculate the energy transferred by rain to each square metre of the surface at a place which receives of rain in a year.

Ans:

Q 41. An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of kg is moving with a speed of 36 km h-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If of energy of the wagon is lost due to friction, calculate the spring constant.

Ans:

10% of this is stored in the spring.

Q 42. An adult weighing raises the centre of gravity of his body by while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.

Ans:

In 6 km there are 6000 steps.

This is 10 % of intake.

.

Q 43. On complete combustion a litre of petrol gives off heat equivalent to. In a test drive a car weighing kg. Including the mass of driver, runs 15 km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.

Ans:

With 0.5 efficiency, 1 litre generates, which is used for 15 km drive.

. With

: force of friction.