Work Energy and Power - Result Question 46

49. 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]

(a) 0.5

(b) 0.25

(c) 0.4

(d) 0.8

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Answer:

Correct Answer: 49. (b)

Solution:

(b)

After Collision

According to law of conservation of linear momentum,

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

Coefficient of restitution,

$e=\frac{\text{ Relative velocity of separation }}{\text{ Relative velocity of approach }}$

$ =\frac{\frac{v}{4}}{V} $

or, $\quad e=\frac{1}{4}=0.25$