Classical-Mechanics · Unit 19 · Video 2 · Interactive Practice

Elastic, Inelastic, Superelastic: Classifying Collisions and Solving the One-Dimensional Elastic Case

IKey Formulas

Formula	Name	What you need
$e = \dfrac{v_B}{v_A}$	Coefficient of restitution	The relative speed after and before; both are frame-independent, so every inertial observer measures the same $e$
$\Delta K = \tfrac{1}{2}\mu v_A^2\left(e^2 - 1\right)$	Kinetic energy change	The reduced mass $\dfrac{1}{\mu} = \dfrac{1}{m_1} + \dfrac{1}{m_2}$ and the approach speed $v_A$
$\vec{v}^{\,\text{rel}}_i = -\vec{v}^{\,\text{rel}}_f$	One-dimensional energy–momentum principle	A head-on elastic collision — and nothing else: no mass appears
$v_{1x,f} = \dfrac{m_1 - m_2}{m_1 + m_2}\,v_{1x,i} + \dfrac{2m_2}{m_1 + m_2}\,v_{2x,i}$	Final velocities, elastic collision	Both masses and both initial $x$ -components; swap the labels $1 \leftrightarrow 2$ to get $v_{2x,f}$

Key Insight: Factoring the energy equation as a difference of squares makes each side contain the corresponding side of the momentum equation, $m_1\!\left(v_{1x,i} - v_{1x,f}\right) = m_2\!\left(v_{2x,f} - v_{2x,i}\right)$ . Dividing one equation by the other cancels those factors — and with them both masses — leaving $v_{1x,i} + v_{1x,f} = v_{2x,i} + v_{2x,f}$ , which is the reversal of the relative velocity. The four labels classify $\Delta K$ alone; momentum is conserved in every one of them.

IIVisualization 1 — One Ratio Sorts Every Collision

The ratio of relative speeds after and before decides the sign of $\Delta K$ — and nothing else does.

Restitution e 1.00 Time t = −1.40 s

A superelastic collision creates nothing: a latch, a compressed spring or a chemical charge releases energy that was already stored inside the system.

IIIVisualization 2 — The Relative Velocity Turns Around

An elastic head-on collision returns the same approach speed, whatever the two initial velocities are.

Masses

The same two conservation equations also admit the root $v_{1x,f} = v_{1x,i}$ , $v_{2x,f} = v_{2x,i}$ — no collision at all — which the division by $m_1\!\left(v_{1x,i} - v_{1x,f}\right)$ quietly discards.

IVVisualization 3 — What the Mass Ratio Decides

The mass ratio moves both final velocities, but never the gap between them.

Mass ratio m₁/m₂ 100

Both first-order corrections are the same size: at $m_1/m_2 = 100$ the tempting shortcut $v_{1x,f} \approx v_{1x,i} + \tfrac{2m_2}{m_1}v_{2x,i}$ gives $9.90\ \text{m/s}$ — a change of only $-0.10\ \text{m/s}$ , one third of the true $-0.30\ \text{m/s}$ .

VQuiz Questions

Problem 1 · Sorting a Collision

Given: two carts approach with relative speed $v_A = 8.0\ \text{m/s}$ and separate with relative speed $v_B = 6.0\ \text{m/s}$ ; their reduced mass is $\mu = 1.2\ \text{kg}$ — find $\Delta K$ .

✅ Correct!

e = 6.0/8.0 = 0.75 < 1

, so the collision is inelastic and

e^2 - 1 = -0.4375

makes

\Delta K

negative.

❌ Not quite. That is the value for

e = 0

— the totally inelastic case, where the carts leave stuck together. Here they separate at

6.0\ \text{m/s}

❌ Not quite.

\Delta K

is not

\tfrac{1}{2}\mu(v_A - v_B)^2

: the two relative speeds enter squared and separately, as

\tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right)

❌ Check the sign. The magnitude is right, but

e < 1

forces

e^2 - 1 < 0

: an inelastic collision loses kinetic energy.

Show solution

Step 1 — the coefficient of restitution:

e = \frac{v_B}{v_A} = \frac{6.0}{8.0} = 0.75

Since $0 < e < 1$ , this is an inelastic collision.

Step 2 — the energy change:

\Delta K = \tfrac{1}{2}\mu v_A^2\left(e^2 - 1\right) = \tfrac{1}{2}(1.2)(8.0)^2\left(0.5625 - 1\right) = 38.4\,(-0.4375) = -16.8\ \text{J}

Check with the equivalent form $\Delta K = \tfrac{1}{2}\mu\left(v_B^2 - v_A^2\right) = 0.6\,(36 - 64) = -16.8\ \text{J}$ ✓

Problem 2 · A Latch Lets Go

Given: two carts collide with a compressed spring latched between them, and the latch releases during contact. They approach at $v_A = 3.0\ \text{m/s}$ and separate at $v_B = 4.5\ \text{m/s}$ , with no external force along the track — which statement is correct?

✅ Correct!

e = 4.5/3.0 = 1.5

gives

e^2 - 1 = 1.25 > 0

. The spring's stored elastic energy became kinetic energy, and with no external force the momentum is untouched.

❌ Not quite.

e = 1.5

is right, but nothing is created: the energy was already stored in the compressed spring. Energy conservation holds — only the kinetic share grew.

❌ Not quite. Momentum conservation follows from the absence of an external force, not from the energy budget. It holds in all four categories — elastic, inelastic, totally inelastic and superelastic.

❌ The ratio is upside down.

e = v_B/v_A

is after over before:

4.5/3.0 = 1.5

, not

3.0/4.5

Show solution

Step 1 — classify by $e$ :

e = \frac{v_B}{v_A} = \frac{4.5}{3.0} = 1.5 > 1 \quad\Rightarrow\quad \text{superelastic}

Step 2 — the sign of the energy change:

\Delta K = \tfrac{1}{2}\mu v_A^2\left(e^2 - 1\right) = \tfrac{1}{2}\mu (3.0)^2 (2.25 - 1) = +\tfrac{1}{2}\mu v_A^2 (1.25) > 0

Step 3 — the two cautions:

The kinetic energy gained came from the spring's stored elastic potential energy. A superelastic collision releases internal energy; it never creates energy.
Momentum is conserved whenever no external force acts, in all four categories. The label $e$ assigns describes $\Delta K$ only.

Problem 3 · Both Final Velocities

Given: $m_1 = 3.0\ \text{kg}$ with $v_{1x,i} = +6.0\ \text{m/s}$ collides head-on and elastically with $m_2 = 1.0\ \text{kg}$ with $v_{2x,i} = -2.0\ \text{m/s}$ — find both final $x$ -components.

What is $v_{1x,f}$ ?

What is $v_{2x,f}$ ?

✅ Correct! Momentum:

18 - 2 = 6 + 10

✓. Energy:

54 + 2 = 6 + 50

✓. And the relative velocity flips from

+8.0

-8.0\ \text{m/s}

✓.

❌ Check $v_{1x,f}$ . Both coefficients here equal

\tfrac{1}{2}

\dfrac{m_1 - m_2}{m_1 + m_2} = \dfrac{2}{4}

and

\dfrac{2m_2}{m_1 + m_2} = \dfrac{2}{4}

. Keep both terms, and keep the sign of

v_{2x,i}

❌ Check $v_{2x,f}$ . Swapping the labels gives

\dfrac{m_2 - m_1}{m_1 + m_2} = -\tfrac{1}{2}

and

\dfrac{2m_1}{m_1 + m_2} = \tfrac{3}{2}

. Velocities are exchanged outright only when

m_1 = m_2

Show solution

With $m_1 + m_2 = 4.0\ \text{kg}$ :

v_{1x,f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1x,i} + \frac{2m_2}{m_1 + m_2}\,v_{2x,i} = \tfrac{1}{2}(6.0) + \tfrac{1}{2}(-2.0) = 3.0 - 1.0 = +2.0\ \text{m/s}

v_{2x,f} = \frac{m_2 - m_1}{m_1 + m_2}\,v_{2x,i} + \frac{2m_1}{m_1 + m_2}\,v_{1x,i} = -\tfrac{1}{2}(-2.0) + \tfrac{3}{2}(6.0) = 1.0 + 9.0 = +10.0\ \text{m/s}

Three checks:

Momentum: $3.0(6.0) + 1.0(-2.0) = 16$ and $3.0(2.0) + 1.0(10.0) = 16$ ✓
Kinetic energy: $54 + 2 = 56\ \text{J}$ and $6 + 50 = 56\ \text{J}$ ✓
Energy–momentum principle: $v_{1x,i} - v_{2x,i} = 8.0$ and $v_{1x,f} - v_{2x,f} = -8.0$ ✓

Problem 4 · Paddle and Ping-Pong Ball

Given: a paddle of mass $m_1 = 200\ \text{g}$ moving at $v_{1x,i} = +6.0\ \text{m/s}$ meets a ball of mass $m_2 = 2.0\ \text{g}$ moving at $v_{2x,i} = -4.0\ \text{m/s}$ , head-on and elastically — use the limit $m_1 \gg m_2$ , to first order in $m_2/m_1 = 0.010$ .

The ball's final velocity $v_{2x,f}$ ?

The paddle's velocity change $\Delta v_{1x} = v_{1x,f} - v_{1x,i}$ ?

✅ Correct! The ball leaves the paddle at the approach speed

10\ \text{m/s}

, so

v_{2x,f} \to v_{1x,i} + 10 = 16\ \text{m/s}

; the paddle pays for it with

\Delta v_{1x} = \tfrac{2m_2}{m_1}\left(v_{2x,i} - v_{1x,i}\right) = 0.020(-10) = -0.20\ \text{m/s}

❌ Not quite. The ball would simply reverse only off a paddle at rest. Here the paddle is advancing, and it is the relative velocity that reverses.

❌ Not quite. Leaving at the paddle's own speed is the totally inelastic outcome (

e = 0

), where the two move off together. An elastic bounce sends the ball away faster than the paddle.

❌ Close — that is the approach speed.

v_{1x,i} - v_{2x,i} = 10\ \text{m/s}

is the speed of the ball relative to the paddle. Add the paddle's own

+6.0\ \text{m/s}

to get its velocity in the lab frame.

❌ Check the paddle. Both first-order pieces,

\tfrac{2m_2}{m_1}v_{2x,i}

and

-\tfrac{2m_2}{m_1}v_{1x,i}

, are the same order in

m_2/m_1

— keeping only one of them is wrong.

Show solution

The ball, in the limit $m_1 \gg m_2$ . The coefficients of $v_{2x,f}$ tend to $\dfrac{m_2 - m_1}{m_1 + m_2} \to -1$ and $\dfrac{2m_1}{m_1 + m_2} \to 2$ , so

v_{2x,f} \to -v_{2x,i} + 2v_{1x,i} = v_{1x,i} + \left(v_{1x,i} - v_{2x,i}\right) = 6.0 + 10.0 = +16\ \text{m/s}

Equivalently $v_{2x,f} - v_{1x,i} = v^{\text{rel}}_{x,i}$ : the ball leaves the paddle at exactly the speed it approached it with.

The paddle, to first order. Expanding both coefficients of $v_{1x,f}$ in $m_2/m_1$ gives $\dfrac{m_1 - m_2}{m_1 + m_2} \approx 1 - \dfrac{2m_2}{m_1}$ and $\dfrac{2m_2}{m_1 + m_2} \approx \dfrac{2m_2}{m_1}$ , so

v_{1x,f} \approx v_{1x,i} + \frac{2m_2}{m_1}\left(v_{2x,i} - v_{1x,i}\right) = 6.0 + 0.020\,(-10.0) = 5.80\ \text{m/s}

\Delta v_{1x} \approx -0.20\ \text{m/s}

Compare with the exact formulas ( $m_1 + m_2 = 0.202\ \text{kg}$ ):

v_{1x,f} = \frac{0.198}{0.202}(6.0) + \frac{0.004}{0.202}(-4.0) = 5.802\ \text{m/s}, \qquad v_{2x,f} = \frac{-0.198}{0.202}(-4.0) + \frac{0.400}{0.202}(6.0) = 15.802\ \text{m/s}

Their difference is $5.802 - 15.802 = -10.0\ \text{m/s}$ , exactly the reverse of the initial $+10.0\ \text{m/s}$ ✓

The tempting shortcut. Keeping only $\tfrac{2m_2}{m_1}v_{2x,i} = 0.020(-4.0) = -0.08\ \text{m/s}$ discards the $-\tfrac{2m_2}{m_1}v_{1x,i}$ term and understates the paddle's change by more than half.

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