A uniform rod A B of length L and mass m is suspended freely at A and hangs vertically at rest when a particle of same mass m is fired horizontally with speed u to strike the rod at its mid point. If the particle is brought to rest after the impact. Then the impulsive reaction at A in horizontal direction is

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(2) m v / 2

(1) m v / 4

(3) m v

(4) 2 m v

answer is A.

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Detailed Solution

From conservation of angular momentum, just before and just after impact about point O we have,

$L_{i} = L_{f} mv \frac{L}{2} = (\frac{{mL}^{2}}{3}) \cdot ω ω = \frac{3 v}{2 L} v_{C} = \frac{L}{2} \cdot ω = \frac{3}{4} v$

Impulse J has changed the momentum of particle from m v to O. Hence,

J=m v

For rod

$J_{H} + J = {mv}_{C} = \frac{3}{4} mv ∴ J_{H} = \frac{3}{4} mv - J = \frac{3}{4} mv - mv = - \frac{mv}{4} or |J_{H}| = \frac{mv}{4}$

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