Question

A block of mass m moving with speed v collides with another block of mass 2 m at rest. The lighter block comes to rest after collision. The value of coefficient of restitution is

• A. $\frac{1}{2}$
  
• B. $\frac{1}{2}$
  
• C. $\frac{3}{4}$
  
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{2}$


Suppose the second block moves at a speed v after collision
mv = 2mv’ or v’ =$\frac{v}{2}$
Velocity of separation = $\frac{v}{2}$
Velocity of approach = v
By definition, $e=\frac{Velocityofseperation}{velocityofapproach}=\frac{1}{2}$