First slide

Types of collision

Question

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

Moderate

A

0.5
B

0.25
C

0.8
D

0.4

Solution

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,
$\begin{array}{l} m v + 4 m \times 0 = 4 m v^{'} + 0 \\ v^{'} = \frac{v}{4} \end{array}$
Coefficient of restitution,
$e = \frac{Relative velocity of separation}{Relative velocity of approach} = \frac{v / 4}{v}$

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