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Question

A moving block having mass m collides with stationary block of mass 4m, initial velocity of moving block is vo, if coefficient of restitution is 0.25, find speed of lighter block after collision.

A
Zero
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B
vo3
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C
vo
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D
vo3
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Solution

The correct option is A Zero

Conserving the momentum we get,
mvo+0=mv1+4mv2
vo=v1+4v2 -----(i)
Definition of coefficient of restitution e
e=Speed of seperationSpeed of approach=v2v1vo=14
4v24v1=vo ----(ii)

Solving (i) and (ii)
v1+4v2=vo
4v24v1=vo––––––––––––––
5v1=0
v1=0

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