question_answer

A moving block having mass m, collides with another stationary block having mass 4m. 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

A) 0.5                   

B)        0.25

C) 0.4                   

D)        0.8

Correct Answer: B

Solution :

According to law of conservation of linear momentum, \[\operatorname{mv}+ 4m \times  0 = 4\,mv+ 0 \Rightarrow \,\,v'=\frac{\nu }{4}\] \[e=\frac{Relative\text{ }velocity\text{ }of\text{ }separation}{Relative\text{ }velocity\text{ }of\text{ }approach}=\frac{\frac{v}{4}}{v}\] or, \[e=\frac{1}{4}=0.25\]