A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v 0 in a direction perpendicu

Question

A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v 0 in a direction perpendicular to AB . The collision is perfectly elastic and after the collision, the particle comes to rest. The ratio m M is

TestHub · Accepted Answer

Using the principle of conservation of linear momentum

${mv}_{0} = {Mv}_{cm}$

Using the principle of conservation of angular momentum

${mv}_{0} \frac{L}{2} = \frac{{ML}^{2}}{12} ω$

Since the collision is completely elastic, the velocity of approach is equal to the velocity of separation

$v_{0} = v_{cm} + ω \frac{L}{2}$

$\Rightarrow v_{0} = \frac{m}{M} v_{0} + 3 \frac{m}{M} v_{0}$

$\Rightarrow \frac{m}{M} = \frac{1}{4}$

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