Two identical blocks  and  each of the mass 2 kg are hanging stationary by a light inextensible flexible string, passing over a light and frictionless pulley, as shown in Fig. A shell of  of mass 1 kg moving vertically upwards with velocity 9 m/s collides with block  and gets stuck to it

Calculate the time after which block  starts moving downwards
a)	0.90 s	b)	1s	c)	0.60 s	d)	0.30 s

(a)

When  strikes with , the combined body starts to rise vertically upward. According to law of conservation of momentum, velocity  of combined body (just after collision) is given by

 or

But  is at rest. Therefore, string becomes slack and bodies move under gravity, combined body upwards and  downwards till string again becomes taut. This happenswhen downward displacement of  becomes equal to upward displacement of combined body. Let it happen at instant  Then,

Or

Displacement of each block at that instant,

And velocity of combined body is and velocity of is  (downward)

Since at this instant, velocities are different, therefore an impulse is developed in the string and magnitude of velocities of bodies becomes equal. Let that velocity magnitude be  and impulse developed be . This impulse acts upwards on both the bodies. Hence for

    (i)

And for combined body,

        (ii)

From above equations,

Now combined body starts to move upwards with velocity  and retardation  where

Combined body comes to an instantaneous rest after time

After this instant, combined body starts moving down. Time interval between this instant and instant of collision is given by

Further height ascended by combined body during time  is

Maximum height raised by combined body is

Loss of energy during collision of  and  is

Loss of energy when impulse is developed in string is

Loss of mechanical energy till  reaches highest point is