Question ID - 101269 | SaraNextGen Top Answer
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
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|
Calculate the time after which block |
|||||||
|
a) |
0.90 s |
b) |
1s |
c) |
0.60 s |
d) |
0.30 s |
starts moving downwards(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
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