A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with a velocity u in a direction perpendicular

Question

A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with a velocity u in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses m M is 1 x . The value of x will be

TestHub · Accepted Answer

From momentum conservation, P i 0 = P f m u = M v … i From angular momentum conservation about O , m u · L 2 = M L 2 12 ω ⇒ ω = 6 m u M L … ( ii ) From e = R . V · S R . V · A 1 = V + ω L 2 u V + ω L 2 = u V + 3 m u M = u m u M + 3 m u M = u 4 m u M = u m M = 1 4 x = 4

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