JEE Main202120 Jul 2021Morning ShiftPhysicsCenter of Mass, Momentum and CollisionActual
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
Correct answer
0
Step-by-step solution
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