A moving block having mass , collides with another stationary block having mass . The lighter block comes to rest after collision. When the initial velocity of the lighter block is , then the value of coefficient of restitution (e) will be

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Solution

Q.
A moving block having mass $m$ , collides with another stationary block having mass $4 m$ . The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$ , then the value of coefficient of restitution (e) will be [2018]

1 0.5

2 0.25

3 0.8

4 0.4

Ans.
(2)

Let final velocity of the block of mass, $4 m = v'$

Initial velocity of block of mass $4 m = 0$

Final velocity of block of mass $m = 0$

According to law of conservation of linear momentum,

$m v + 4 m \times 0 = 4 m v' + 0$ $\Rightarrow v' = \frac{v}{4}$

Coefficient of restitution,

$e = \frac{Relative velocity of separation}{Relative velocity of approach}$ $= \frac{v / 4}{v} = 0.25$

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