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Question

A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet in m/s after it emerges from the block?

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Solution

Suppose the velocity of the bullet of mass m is v and it strikes the block of mass M. After collision, the linear velocity of the block is V and that of the bullet is v'.
Applying law of conservation of linear momentum, we get
mv=MV+mv
or 500×0.01=2V+0.01v
or 5 = 2V + 0.01 v' (i)
Now, the kinetic energy gained by the block 12MV2 raises it to a height h where it gains gravitational potential energy Mgh. By conservation of energy, we get
12MV2=Mgh
or V=2gh=2×9.8×0.1=1.4 m/s
Putting the value of V in Eq. (i), we get
5=2×1.4+0.01v or v' = 220 m/s

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