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=v2−v1vo=14 4v2−4v1=vo ----(ii)
Solving (i) and (ii) v1+4v2=vo 4v2−4v1=vo–––––––––––––––– 5v1=0 v1=0