A small cube of mass m slides along a horizontal frictionless surface to collide
ID: 1618386 • Letter: A
Question
A small cube of mass m slides along a horizontal frictionless surface to collide with the bottom end of a rod. The rod is uniform with length L and mass M and is supported by a frictionless pivot through its center and perpendicular to its long axis. The cube has an elastic collision with the rod. Of course, if the cube comes in from the left, the rod will leave the collision spinning ccw. But:
(a) what is the spped of the small cube after the collision? Does it keep going in its initial direction or does it bounce back?
(b) Also, what is the angular speed of the rod after the collision?
Explanation / Answer
(A) for elastic collision,
velocity of approach = velocity of separation
v0 = wL/2 - v
Applying angular momentum conservation,
m v0 L/2 = I w + m v L/2
m v0 L/2 = (M L^2 / 12)(2/L)(v0 + v) + m v L / 2
m v0 L /2 - M v0 L / 6 = M v L/6 + m v L / 2
( v0 L / 6) [ 3 m - M ] = ( v L /6) (M + 3m)
v = (3 m - M) v0 / (M + 3m) ........Speed of cube after colllision
its direction depends on M/m ratio.
if M > 3m then it will go back
if M < 3 m then it will keep moving in original direction.
(b) w = (v0 + v) 2 / L
= v0 (M + 3m + 3m - M)/ (M + 3m) ] (2 / L)
= 12 m v0 / (M + 3m)L