A tennis ball with (small) mass m2 sits on top of a basketball with (large) mass

Question

A tennis ball with (small) mass m2 sits on top of a basketball with (large) mass m1. The bottom of the basketball is a height h above the ground, and the bottom of the tennis ball is a height h + d above the ground. The balls are dropped. Working in the approximation where m1 is much larger than m2, and assuming the balls bounce elastically, (a) to what height does the tennis ball bounce? Now consider n balls, B1,..,Bn having masses m1,m2,..,mn (such that m1 >> m2 >> .. >> mn) sitting in a vertical stack. The bottom of B1 is a height h above the ground, the bottom of Bn is a height h + L above the ground. The balls are dropped. With the same assumptions, to what height does the top ball bounce? (c) What is the minimum number of balls needed for the top one to reach escape velocity? (for part c you may take L to be vanishingly small)

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Explanation / Answer

Basketball and tennis ball (a) For simplicity, assume that the balls are separated by a very small distance, so that the relevant bounces happen a short time apart. This assumption isn’t necessary, but it makes for a slightly cleaner solution. Just before the basketball hits the ground, both balls are moving downward with speed (using mv^2/2 = mgh) v= (2gh)^1/2. Just after the basketball bounces off the ground, it moves upward with speed v, while the tennis ball still moves downward with speed v. The relative speed is therefore 2v. After the balls bounce off each other, the relative speed is still 2v. (This is clear if you look at things in the frame of the basketball, which is essentially a brick wall.) Since the upward speed of the basketball essentially stays equal to v, the upward speed of the tennis ball is 2v + v = 3v. By conservation of energy, it will therefore rise to a height of H = d + (3v)^2/(2g). But v^2 = 2gh, so we have H = d + 9h. (b) Just before B1 hits the ground, all of the balls are moving downward with speed v = v2gh. We will inductively determine the speed of each ball after it bounces off the one below it. If Bi achieves a speed of vi after bouncing off Bi-1, then what is the speed of Bi+1 after it bounces off Bi? The relative speed of Bi+1 and Bi (right before they bounce) is v + vi. This is also the relative speed after they bounce. Since Bi is still moving upwards at essentially speed vi, the final upward speed of Bi+1 is therefore (v + vi) + vi. Thus, vi+1 = 2vi + v. (3) Since v1 = v, we obtain v2 = 3v (in agreement with part (a)), v3 = 7v, v4 = 15v, etc. In general, which is easily seen to satisfy eq. (3), with the initial value v1 = v. vn = (2^n - 1)v, from conservation of energy, Bn will bounce to a height of H=l+[((2^n-1)v)^2]/2g =l+(2^n-1)^2h It turns out that the relative speed is the same before and after any elastic collision, independent of what the masses are. This is easily seen by working in the center-of-mass frame, where the masses simply reverse their velocities.

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