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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.
  1. Zero
  2. vo3
  3. vo
  4. vo3


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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