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.

- Zero
- vo3
- vo
- 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=v2−v1vo=14

4v2−4v1=vo ----(ii)

Solving (i) and (ii)

v1+4v2=vo

4v2−4v1=vo––––––––––––––––

5v1=0

v1=0

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

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