First slide

Coefficient of restitution

Question

A block having mass m collides with an another stationary block having mass 2 m. The lighter block comes to rest after collision. If the velocity of first block is v, then the value of coefficient of restitution will must be [AIIMS 2015]

Moderate

A

0.5
B

0.4
C

0.6
D

0.8

Solution

Let the velocity of block of mass 2 m after the collision be v', then from law of conservation of momentum,
$m v = 2 m v' \Rightarrow v' = \frac{v}{2}$
Now, the coefficient of restitution,
$e = \frac{velocity of separation}{velocity of approach} = \frac{v'}{v} = \frac{v / 2}{v} = \frac{1}{2} = 0.5$

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