Classical-Mechanics · Unit 19 · Video 4 · Interactive Practice

Bouncing Superballs: Why the Small Ball Flies Nine Times Higher

IKey Formulas

Formula	Name	What it gives
$v_a = \sqrt{2gh_i}$	Fall from $h_i$	The speed both balls carry into the floor
$v' = v - u$	Change of inertial frame	Velocities seen by an observer rising at $u$ — every difference $v_1 - v_2$ is untouched
$v_b = 2v_a + v_a = 3v_a$	Bounce off a rising wall	Ball 1's speed just after the two-ball collision
$h_f = \dfrac{v_b^2}{2g} = 9h_i$	Rise after the bounce	Height grows as the square of the speed

Key Insight: Relative velocity is the same in every inertial frame, so ride upward with ball 2: the heavy ball becomes a wall at rest, the elastic bounce is a pure reversal of $2v_a$ , and adding the frame's own $v_a$ back gives $3v_a$ .

IIVisualization 1 — Five States, One Drop

Five states carry the pair from rest at $h_i$ to ball 1 alone, at rest at $9h_i$ .

State initial

IIIVisualization 2 — Choosing the Observer

The relative velocity is the same in every frame — one choice turns ball 2 into a wall at rest.

Observer rises at u 0.00 v_a

💡 The wall picture needs $m_2 \gg m_1$ : for a finite mass ratio the collision changes ball 2's velocity too, so the frame that starts at ball 2's rest is no longer its rest frame afterwards.

IVVisualization 3 — Exact Result and the Heavy-Ball Limit

How light must ball 1 be before the exact elastic result delivers the full factor of nine?

Mass ratio m1/m2 0.010

Exact one-dimensional elastic result, with $v_{1i} = -v_a$ and $v_{2i} = +v_a$

v_b = \frac{m_1 - m_2}{m_1 + m_2}\,(-v_a) + \frac{2m_2}{m_1 + m_2}\,(v_a) = \frac{3m_2 - m_1}{m_1 + m_2}\,v_a

Heavy-ball limit

\frac{h_f}{h_i} = \left(\frac{v_b}{v_a}\right)^{2} = \left(\frac{3 - m_1/m_2}{1 + m_1/m_2}\right)^{2} \quad\text{and}\quad \frac{m_1}{m_2} \to 0 \;\Longrightarrow\; \frac{h_f}{h_i} \to 3^2 = 9

💡 Real superballs also lose a few percent of their energy at every bounce, so a $m_1/m_2 = 1/10$ pair falls short even of the $6.95\,h_i$ this exact curve promises.

VQuiz Questions

Problem 1 · Speed Out, Height Up

Given: the same stacked pair dropped from $h_i = 0.80\ \text{m}$ , with $g = 9.8\ \text{m/s}^2$ , $m_2 \gg m_1$ and every collision elastic — find ball 1's speed just after the two-ball collision and the height it reaches.

✅ Correct!

v_a = \sqrt{2(9.8)(0.80)} = 3.96\ \text{m/s}

, ball 1 leaves at

3v_a = 11.9\ \text{m/s}

, and

h_f = v_b^2/2g = 7.2\ \text{m} = 9h_i

❌ That is the fixed-floor answer. Ball 1 does not bounce off the ground; it bounces off ball 2, which is already moving up at

v_a

when they meet.

❌ $2v_a$ is ball 1's speed in the rising frame. That frame is itself climbing at

v_a

, so the ground observer adds it back:

2v_a + v_a = 3v_a

❌ The speed is right, the height is not.

h_f = v_b^2/2g

— tripling the speed multiplies the height by

3^2 = 9

, not by

3

Show solution

Step 1 — the fall sets the scale. Mechanical energy is constant from the release to the floor, so $\tfrac{1}{2}mv_a^2 = mgh_i$ :

v_a = \sqrt{2gh_i} = \sqrt{2(9.8)(0.80)} = 3.96\ \text{m/s}

Step 2 — ride up with ball 2. Ball 2 rebounds first and moves up at $v_a$ ; ball 1 is still moving down at $v_a$ . In the frame rising at $u = v_a$ , ball 2 is at rest and ball 1 arrives at

v_1' = (-v_a) - (v_a) = -2v_a

Because $m_2 \gg m_1$ , ball 2 is an immovable wall and the elastic bounce simply reverses that velocity: $v_1' \to +2v_a$ .

Step 3 — return to the ground. Add the frame velocity back:

v_b = 2v_a + v_a = 3v_a = 3(3.96) = 11.9\ \text{m/s}

Step 4 — rise. With ball 1 and the Earth as the system, $\Delta K + \Delta U = 0$ :

h_f = \frac{v_b^2}{2g} = \frac{(11.88)^2}{2(9.8)} = \mathbf{7.2\ \text{m}} = 9h_i

Check: $9h_i = 9(0.80) = 7.2\ \text{m}$ ✓ — the factor of nine never depends on $h_i$ .

Problem 2 · The Elastic-Bounce Objection

Given: a student argues that because the collision is elastic and ball 1 arrives at $v_a$ , it must leave at $v_a$ and climb back to exactly $h_i$ . Find the statement that identifies the error.

✅ Correct! A rising wall returns a ball faster than it arrived: the ball's own

v_a

, plus the wall's

v_a

on the way in, plus the wall's

v_a

again on the way out —

3v_a

in all.

❌ Every collision here is elastic. And the energy runs the other way: ball 2 loses

4m_1v_a^2

, exactly what ball 1 gains — a fraction

8m_1/m_2

of its own kinetic energy, which vanishes as

m_2

grows.

❌ That is the definition of elastic, reversed. The total kinetic energy of the pair is conserved; what is not conserved is either ball's individual share, which is exactly why ball 1 can leave faster.

❌ The collision is far too brief. Over a contact time of milliseconds gravity supplies

g\Delta t \ll v_a

. The extra speed comes from ball 2's upward motion, not from gravity.

Show solution

"Elastic" fixes the relative speed, not either ball's own speed. For a one-dimensional elastic collision,

v_{1f} - v_{2f} = -(v_{1i} - v_{2i})

Here $v_{1i} = -v_a$ and $v_{2i} = +v_a$ , so the approach velocity is $v_{1i} - v_{2i} = -2v_a$ and the separation velocity must be $+2v_a$ . With $m_2 \gg m_1$ ball 2 keeps $v_{2f} \approx v_a$ , so

v_{1f} = v_{2f} + 2v_a = v_a + 2v_a = 3v_a

The student's argument silently assumes $v_{2i} = v_{2f} = 0$ — a floor that does not move. Count the three contributions to the outgoing speed: the ball's own $v_a$ , plus $v_a$ because the wall was closing, plus $v_a$ because the wall is still receding as the ball leaves.

Problem 3 · Who Pays the Bill

Given: $m_1 = 0.020\ \text{kg}$ rides on $m_2 = 1.00\ \text{kg}$ , dropped from $h_i = 1.00\ \text{m}$ so $v_a = 4.43\ \text{m/s}$ ; the collision takes ball 1 from $-v_a$ to $+3v_a$ . Use the heavy-object limit $v_{2f} \approx v_{2i} + \dfrac{2m_1}{m_2}\left(v_{1i} - v_{2i}\right)$ .

How much kinetic energy does ball 1 gain?

By how much does ball 2's speed drop?

✅ Correct! Ball 1 gains

8m_1gh_i = 1.57\ \text{J}

and ball 2 slows by

4m_1v_a/m_2 = 0.354\ \text{m/s}

— and

m_2v_a\,\Delta v_2 = -1.57\ \text{J}

, the same number. Ball 2 pays the whole bill.

❌ That is ball 1's final kinetic energy.

\tfrac{1}{2}m_1(3v_a)^2 = 1.76\ \text{J}

; the gain is the change, so subtract the

0.196\ \text{J}

it already had.

❌ That is $\tfrac{1}{2}m_1(\Delta v)^2$ . Kinetic energy is not a function of the velocity change: compute

\tfrac{1}{2}m_1v_f^2 - \tfrac{1}{2}m_1v_i^2

, not

\tfrac{1}{2}m_1(v_f - v_i)^2

❌ Check the change, not one endpoint.

\Delta K_1 = \tfrac{1}{2}m_1(3v_a)^2 - \tfrac{1}{2}m_1v_a^2 = 4m_1v_a^2

, and

v_a^2 = 2gh_i

❌ Check the velocity difference.

v_{1i} - v_{2i} = -v_a - (+v_a) = -2v_a

, not

-v_a

— ball 2 is moving up while ball 1 is still coming down, so their closing speed is doubled.

❌ Not quite.

\Delta v_2 = \dfrac{2m_1}{m_2}(v_{1i} - v_{2i}) = \dfrac{2m_1}{m_2}(-2v_a) = -\dfrac{4m_1v_a}{m_2}

Show solution

(a) Ball 1's gain. Its speed goes from $v_a$ to $3v_a$ , so

\Delta K_1 = \tfrac{1}{2}m_1(3v_a)^2 - \tfrac{1}{2}m_1v_a^2 = 4m_1v_a^2 = 8m_1gh_i = 8(0.020)(9.8)(1.00) = \mathbf{1.57\ \text{J}}

(b) Ball 2's loss of speed. With $v_{1i} = -v_a$ and $v_{2i} = +v_a$ , the heavy-object limit gives

\Delta v_2 = \frac{2m_1}{m_2}\left(-2v_a\right) = -\frac{4m_1v_a}{m_2} = -\frac{4(0.020)(4.43)}{1.00} = \mathbf{-0.354\ \text{m/s}}

Momentum check. $\Delta p_1 = m_1(3v_a + v_a) = 4m_1v_a = 0.354\ \text{kg}\cdot\text{m/s}$ and $\Delta p_2 = m_2\Delta v_2 = -0.354\ \text{kg}\cdot\text{m/s}$ ✓

Energy check. To first order in $m_1/m_2$ ,

\Delta K_2 \approx m_2v_a\,\Delta v_2 = -m_2v_a\cdot\frac{4m_1v_a}{m_2} = -4m_1v_a^2 = -1.57\ \text{J} = -\Delta K_1

As a fraction of ball 2's own kinetic energy the loss is $\dfrac{|\Delta K_2|}{\tfrac{1}{2}m_2v_a^2} = \dfrac{8m_1}{m_2} = 0.16$ — and it goes to zero as $m_2$ grows. Ball 2 pays for everything and barely notices.

Problem 4 · A Third Ball on Top

Given: a third ball of mass $m_0 \ll m_1$ rides on top of the stack, all three released together from $h_i$ ; each ball is far lighter than the one below it and every collision is elastic. Find the speed the top ball leaves with and the height it reaches.

✅ Correct! Ride with ball 1, which is already moving up at

3v_a

: the top ball closes at

4v_a

, reverses to

4v_a

, and the ground observer adds

3v_a

back for

7v_a

— and

7^2 = 49

❌ The bonus is not a fixed $2v_a$ . Each bounce adds twice the lower ball's velocity: the first stage adds

2v_a

, but the second adds

2(3v_a) = 6v_a

to the arriving

v_a

❌ The factor $3$ is not a rule that repeats. It came from a wall rising at

v_a

; here ball 1 is rising at

3v_a

, so the closing speed is

3v_a + v_a = 4v_a

, not

3v_a

❌ That is the two-ball answer. The top ball bounces off ball 1 after ball 1 has already been launched upward at

3v_a

, so it does far better than

3v_a

Show solution

Apply the same move twice. Every ball arrives at the floor moving down at $v_a$ .

Collision 1 (ball 2 with ball 1). Ball 2 rebounds at $+v_a$ ; ride with it. Ball 1 closes at $2v_a$ , reverses, and back in the ground frame

v_1 = 2v_a + v_a = 3v_a

Collision 2 (ball 1 with the top ball). Ball 1 is now rising at $3v_a$ while the top ball is still falling at $v_a$ . Ride with ball 1: the top ball closes at

v_0' = (-v_a) - (3v_a) = -4v_a

Ball 1 is the heavy one in this pair, so the bounce reverses that to $+4v_a$ . Adding the frame velocity back,

v_0 = 4v_a + 3v_a = \mathbf{7v_a}

Height. $h = \dfrac{v_0^2}{2g} = \dfrac{49v_a^2}{2g} = \mathbf{49h_i}$ .

The general pattern is $v_a,\, 3v_a,\, 7v_a,\, 15v_a,\dots = (2^n - 1)v_a$ for the $n$ -th ball up. Dropped from $1\ \text{m}$ , an ideal three-ball stack would send the top ball $49\ \text{m}$ up; real balls lose enough energy at each contact that the trick stops paying long before this.

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