Read · step-by-step explanationHarvard “Problem of the Week”

Why do two stacked balls bounce so high?

Drop a tennis ball stacked on a basketball and the tennis ball bounces to about nine times the height it fell from. Step by step, we check which conditions that answer depends on.

2026-10-14　Domain: Physics

Analysis data (cue → response texts, stage analyses, explanation steps) is shown in the original Japanese. English rendering of the data layer is pending; titles, summaries and observations are in English.

PROBLEM

The problem

SubjectA light tennis ball is placed on top of a heavy basketball and the pair is dropped. Assuming elastic collisions and a sufficiently large mass ratio, consider the height the upper ball reaches. An original summary, not a reproduction of the text or figure.

Analysed dialogue (excerpt, 10 messages)

Learner

How do you solve this?

Here, look at it from a frame moving with the basketball. The tennis ball approaches at speed 2v. Since the basketball is heavy enough, it can be treated as a moving wall. The tennis ball reflects off it and ends up at 2v upward relative to the basketball.

Returning to the speed seen from the ground, the basketball's own upward speed v is added, giving

2v+v=\boxed{3v}

Height risen is proportional to the square of the speed, so the height risen from the point of collision is

\frac{(3v)^2}{2g}=9h

If the bottom of the tennis ball is at about d at the moment of collision, the height of its bottom at the top is d+9h. Ignoring the size of the balls, it rises to 9 times the drop height.

Learner

So we ignore the time taken to reverse during the collision, and any energy loss?

Yes. In this problem we treat the collision time as negligible and the energy loss in the collision as zero.

Specifically, these are the idealisations:

- The collision with the floor and the subsequent collision between the balls are treated as instantaneous and sequential.
- Deformation during the collision and energy loss to heat and sound are ignored.
- The basketball is heavy enough that its velocity barely changes when it collides with the tennis ball.

Learner

Does the problem statement give a reason for that?

On the other hand, that the collision time is zero, and that the two collisions happen one after the other, is not stated. Those are idealisations made in the solution earlier. “Elastic” does not by itself imply a collision time of zero.

So “no loss” is a condition of the problem, while “instantaneous, sequential collisions” is an assumption on the solution side. That distinction should have been made clear earlier.

Learner

What happens to the basketball's motion after the collision?

It hands energy to the tennis ball, and its upward speed becomes slightly smaller. It then slows under gravity, reaches its highest point and falls.

Let the basketball's mass be M and the tennis ball's m, with velocities just before the collision of v upward and v downward respectively. The basketball's velocity after an elastic collision is

V=\frac{M-3m}{M+m}v
=v-\frac{4m}{M+m}v

(taking upward as positive).

In this problem M\gg m, so the slowdown is slight and we treat it as still about v upward. But the velocity is not entirely unchanged. The energy lost through this slight slowdown is what passes to the tennis ball.

Learner

So this uses conservation of momentum?

Yes. For the collision between the balls, we use conservation of momentum and the elastic-collision condition.

Taking upward as positive, with velocities before the collision of v for the basketball and -v for the tennis ball, and V,u afterwards,

from conservation of momentum,

Mv-mv=MV+mu

In an elastic collision the relative speed of approach equals the relative speed of separation, so

u-V=2v

Solving these two equations,

V=\frac{M-3m}{M+m}v,\qquad
u=\frac{3M-m}{M+m}v

If M\gg m, then V\approx v,\ u\approx3v.

The earlier explanation of “treating it as a moving wall” is a shortcut to this result using relative velocity. Conservation of momentum alone does not fix the velocities after the collision; the elastic-collision condition is also needed.

David Morin, Harvard University, Problem of the Week 1 (2002-09-16), first half. Our own summary, not a reproduction of the original text or figure. The analysed object is a learner–AI dialogue about this problem (excerpt of 10 messages). · Source (original PDF)

OUTPUT

Step-by-step explanation

Cues and operations one step at a time. Figures and equations come from this problem’s judgment data.

1. First, under what conditions

A light tennis ball is placed on top of a heavy basketball and dropped. Why does the small ball on top bounce so high?

Here we consider perfectly elastic collisions, in which no kinetic energy is lost. We further approximate the collision with the floor and the subsequent collision between the two balls as happening one after the other in a short time. Motion during which the floor and both balls push on each other at once is not included in this calculation.

Let the mass of the large ball be M and that of the small ball m. We first keep the mass difference in the calculation, and at the end consider the case where M is much larger than m.

2. Just after the bounce off the floor

Let the distance fallen before hitting the floor be h and the acceleration due to gravity g. Released gently with air resistance ignored, the speed v just before hitting the floor satisfies

v² = 2gh

The large ball bounces elastically off the floor, so immediately afterwards it moves upward at speed v. The small ball, meanwhile, is still moving downward at speed v.

We take the two balls to collide immediately after this. Taking upward as positive, the velocities before the collision are +v for the large ball and −v for the small ball.

3. Writing the two-ball collision as equations

Let the velocity of the large ball after the collision be V and that of the small ball u.

First, view the two balls as one system. Once off the floor, if the impulse of gravity during the short collision can be ignored, the total momentum of the two balls is conserved. Therefore

Mv − mv = MV + mu

However, with two unknown velocities, this equation alone does not determine them.

The other condition we use is that of a perfectly elastic collision. In one dimension, conservation of momentum and of kinetic energy together mean the relative velocity reverses in the collision. Here the balls approach at speed 2v before the collision, so they separate at the same speed afterwards.

u − V = 2v

Momentum is conserved because the impulse of external forces can be ignored; kinetic energy is conserved because the collision is perfectly elastic. We use these two conditions separately.

4. The small ball's speed, and the motion of the large ball

From the relative-velocity equation, u = V + 2v; substitute into the momentum equation.

(M − m)v = MV + m(V + 2v)

(M − 3m)v = (M + m)V

Therefore

V = (M − 3m)v / (M + m)

u = (3M − m)v / (M + m)

If M is much larger than m, the large ball rises at almost speed v and the small ball at about 3v.

The large ball's velocity is not strictly unchanged, however. The reduction is v − V = 4mv / (M + m). The kinetic energy the large ball loses is received by the small ball, which is why the small ball bounces high.

5. How high it goes

Let H be the distance the small ball rises from the point of collision. On the way up, kinetic energy turns into potential energy and the velocity is 0 at the top, so

H = u² / (2g)

Using u ≈ 3v and v² = 2gh,

H ≈ 9h

That is, the small ball rises from the point of collision by about 9 times the distance it fell. Without approximating the mass ratio, H = [(3M − m)/(M + m)]²h.

This 9h is not a height measured from the floor. If the diameter of the large ball is d and the bottom of the small ball collides at roughly d above the floor, the height of the small ball's bottom at the top is about d + 9h.

The order of thinking is: the fall, the bounce off the floor, the two-ball collision, the rise. Choosing the conservation law that applies in each scene, the whole motion can be solved in sequence.