Question

# 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.

Solution

## Suppose after collision the heavier mass get speed $$v_1$$ then applying momentum conservation we get $$mv +0 = 0 + 2mv_1$$ or $$v_1 = \dfrac {v}{2}$$ coefficient of restitution is $$\dfrac { \text{relative velocity of separation}} {\text{ relative velocity of approach}}$$ = $$\dfrac{v_1 -0}{v}$$  put value of $$v_1$$ as $$\dfrac{v}{2}$$ we get the coefficient as $$0.5$$Physics

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