A tennis ball of mass mt is held just above a basketball of mass mb, as shown in
ID: 2205866 • Letter: A
Question
A tennis ball of mass mt is held just above a basketball of mass mb, as shown in the figure below. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height h and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling.
(a) The two balls meet in an elastic collision. To what height does the tennis ball rebound? (Use any variable or symbol stated above along with the following as necessary: g.) ONLY in Symbols
ht =
Explanation / Answer
Just after the basketball has hit the ground it rebounds with a speed of V the tennis ball is still moving downwards with a speed of -V so the relative speed of them is 2V That will also be the relative speed of the tennis ball relative to the basketball both before and after the collision. Now IF the basketball had an infinite mass then the result is simple. The basketball is going up at V and the tennis ball is going up at 2V relative to it. So the speed from the earth is 3V and it has 9 times the energy and will go to 9 times the height. As it is NOT an infinite mass the problem as viewed from the earth is complex. You need to find the total momentum. immediately after the basketball has bounced. MV - mv and the total energy 1/2 Mv^2 + 1/2 m v^2 Neither of these values may change after the second collision. So you can set up a simultaneous equation for the energy and momentum after the collision and equate them to the energy and momentum prior to the collision, ie. MV2+mv3 = Mv- mv and 1/2 M(v2)^2 + 1/2 m (v3)^2 = 1/2 M v^2 + 1/2 m v^2 or you can use my technique of calculating things relative to the centre of mass. Then convert that back to the earth frame of reference.
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