Two identical blocks `A` and `B` each of 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 `C` of mass `1 kg` moving vertically upwards with velocity `9 m//s` collides with block `B` and gets stuck to it.

The loss of mechanical energy up to that instant is

Options

A. `32.4J`
B. `40J`
C. `16.5J`
D. `12.5J`

Answer

A

Explanation

When `C` strikes with `B`, the combined body starts to rise vertically upward. According to law of conservation of momentum, velocity `v_(0)` of combined body (just after collision) is given by
`(2+1)v_(0)=1xx9` or `v_(0)=3 m//s`
But `A` is at rest. Therefore, string becomes slack and bodies move under gravity, combined body upwards and `A` downwards till string again becomes taut. This happens when downward displacement of `A` becomes equal to upward displacement of combined body. Let it happen at instant `t_(1)`. Then,
`1/2"gt"_(1)^(2)=3t_(1)-1/2"gt"_(1)^(2)`
or `t_(1)=0.3s` Displacement of each block at that instant,
`h_(1)=1/2"gt"_(1)^(2) 0.45m`
And velocity of combined body is `3-"gt"_(1)=0` and velocity of `A` is `"gt"_(1)=3m//s` (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 `v` and impulse developed be `J`. This impulse acts upwards on both the bodies.
Hence for `A`
`2xx3-J=v’`………i
and for combined body,
`J=3v’`…………..ii
From above equations
`v’=1.2m//s`
Now combined body starts to move upwards with velocity `v’=1.2m//s` and retardation a, where
`a=(3-2)/(3+2)=2m//s^(2)`
Combined body comes to an instantaneous rest after time
`t_(2)=(v^('))/a=0.6s`
After this instant, combined body starts moving down. Time interval between this instant and instant of collision is given by
`t_(1)+t_(2)=0.3+0.6=0.9s`
Further height ascended by combined body during time `t_(2)` is
`h_(2)=v’v_(2)-1/2at_(2)^(2)=0.36m`
Maximum height raised by combined body is
`h=h_(1)+h_(2)=0.81m`
Loss of energy during collision of `B` and `C` is
`E_(1)=1/2xx1xx9^(2)-1/23v_(0)^(2)=27J`
Loss of energy when impulse is developed in string is
`E_(2)=1/2xx2x3^(2)-1/2(2+3)3v_(0)^(2)(2+3)(v’)^(2)=5.4J`
Loss of mechanical energy till `B` reaches highest point is
`E=E_(1)+E_(2)=27+5.4=32.4J`