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