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

Question

A rod $ \mathrm{AB} $ of mass $ \mathrm{M} $ and length $ \mathrm{L} $ is lying on a horizontal frictionless surface. A particle of mass $ \mathrm{m} $ travelling along the surface hits the end $ \mathrm{A} $ of the rod with a velocity $ \mathrm{v}_{0} $ in a direction perpendicular to $ \mathrm{AB} $. The collision is perfectly elastic and after the collision, the particle comes to rest. The ratio $ \frac{m}{M} $ is, $ \frac{1}{4} $, $ \frac{1}{3} $, $ \frac{1}{2} $, $ 1 $

MARKS App · Accepted Answer

Using the principle of conservation of linear momentum mv 0 = Mv cm Using the principle of conservation of angular momentum mv 0 L 2 = 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 L 2 ⇒ v 0 = m M v 0 + 3 m M v 0 ⇒ m M = 1 4

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