A small ball of mass m is aligned above a larger ball of mass M = 1.9 kg (with a
ID: 1550791 • Letter: A
Question
A small ball of mass m is aligned above a larger ball of mass M = 1.9 kg (with a slight separation, as with the baseball and basketball of Figure (a)), and the two are dropped simultaneously from height h = 2.5 m. (Assume the radius of each ball is negligible compared to h.) (a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball? (b) What height does the small ball then reach (see Figure (b))?
Explanation / Answer
(a) Using conservation of mechanical energy during fall, Mgh = Mv2/2, the speed of the larger ball just before
bounce is equal to, vi =
2gh. After elastic bounce from the floor, the larger ball moves upwards with speed vi
,
while the small ball at that moment has the same speed and moves downward.
NOTE: We are assuming that the collision with the floor is instantaneous t 0, and, therefore, can be elastic,
because the work done by the mg, that is external force for the ball-floor system, can be neglected.
After the collision with the floor, the large ball moves upward with the speed vi and collides with the small ball.
That collision is assumed to be elastic, with the condition that the larger ball stops after it.
NOTE: This collision is elastic if we assume that it is instantaneous (like we did for the collision with the floor).
Therefore,
Vf (=2m/m + M)(vi) +( M m/m + M) vi = 0 which gives: M = 3m = 3*1.9 = 5.7kg
(b) Using eqution we can find speed of the smaller ball after the collision:
vf = m M/m + M) (vi) + 2M/m + M)vi =3M m/m + M vi = 2vi = 2root2gh
The final height is, using conservation of mechanical energy:
hf =v^2f/2g= 4h = 4 (2.5 m) = 10 m