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

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Solution

Suppose after collision the heavier mass get speed $v_{1}$ then applying momentum conservation we get $m v + 0 = 0 + 2 m v_{1}$ or $v_{1} = \frac{v}{2}$
coefficient of restitution is $\frac{relative velocity of separation}{relative velocity of approach}$ = $\frac{v_{1} - 0}{v}$ put value of $v_{1}$ as $\frac{v}{2}$ we get the coefficient as $0.5$

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

Suppose after collision the heavier mass get speed v1 then applying momentum conservation we get mv+0=0+2mv1 or v1=v2 coefficient of restitution is relative velocity of separation relative velocity of approach = v1−0v put value of v1 as v2 we get the coefficient as 0.5

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