Question

A ball of mass m and speed v collides elastically with one end of a uniform rod of mass M which is fixed at the center of mass P and is free to rotate about the axis passing through P. After the collision, the ball is stationary. Find M.

• A. M=m
• B. M=2m
• C. M=$\frac{m}{2}$
• D. M=4m

Answer 1

Answer: (D)

M=4m

Answer 2

Assuming the rod is free to move on a frictionless surface and the collision is elastic, we apply conservation of linear momentum, conservation of angular momentum about the center of mass, and the definition of the coefficient of restitution ($e=1$ for elastic collision).
Let m be the mass of the ball, v its initial speed, M the mass of the rod, and L its length. After the collision, the ball is stationary ($v'=0$). Let $V_{CM}$ be the final velocity of the rod's center of mass and $\omega'$ its final angular velocity.

1. Conservation of linear momentum: $mv = MV_{CM}$ (since $v'=0$).
2. Conservation of angular momentum about the center of mass P: $mv(L/2) = I_P \omega' = \frac{1}{12}ML^2 \omega'$.
3. Coefficient of restitution $e=1$: The relative velocity of separation along the line of impact equals the relative velocity of approach. The velocity of the point of impact Q on the rod is $v_Q = V_{CM} + \omega'(L/2)$. $v_{rel,i} = v - 0 = v$. $v_{rel,f} = v_Q - v' = V_{CM} + \omega'(L/2) - 0$. So, $v = V_{CM} + \omega'(L/2)$.
   From (1), $V_{CM} = mv/M$.
   From (2), $\omega' = \frac{6mv}{ML}$.
   Substitute into (3): $v = \frac{mv}{M} + \frac{6mv}{ML} \frac{L}{2} = \frac{mv}{M} + \frac{3mv}{M} = \frac{4mv}{M}$.
   $v = \frac{4mv}{M} \implies M = 4m$.
   This result is consistent with energy conservation as well.