Elastic collisions — 1D - solve for final velocities: Question bank

Where the symbols mean: $m$ = mass (how much "stuff", always positive), $u$ = velocity before with a sign (right = $+$ , left = $-$ ), $v$ = velocity after with a sign. See Elastic collisions — 1D - solve for final velocities (index 1.4.5) for the full derivation these all lean on.

True or false — justify

The first ball always slows down after an elastic collision.

False. If it hits something heavier it can bounce backward (velocity reverses, so it doesn't merely slow), and if the target is moving toward it fast enough the first ball can even speed up. "Slows down" only holds for special mass/velocity combos.

In an elastic collision the two speeds are simply swapped between the balls.

False in general. Velocities swap only when the masses are equal. With unequal masses you must use the full formulas; speed-swapping violates momentum conservation for

m_{1} \neq = m_{2}

Momentum is conserved but kinetic energy is not, in an elastic collision.

False. "Elastic" is defined by kinetic energy being conserved as well. Both quantities are conserved — that is exactly why we get two equations for two unknowns.

If both balls end up moving to the right, momentum was not conserved.

False. Momentum is about the signed sum

m_{1} v_{1} + m_{2} v_{2}

, not about directions being opposite. Both moving right is perfectly fine as long as that sum equals the initial total momentum.

The relative-velocity rule is a separate physical law added on top of momentum and energy.

False. It is algebraically equivalent to energy conservation (it's what you get after dividing the factored energy equation by the momentum equation). It replaces the messy squared equation with a clean linear one.

A stationary ball can never move faster than the incoming ball after being hit elastically.

False. When a heavy ball hits a light ball at rest, the light one leaves at up to

2 u_{1}

— nearly double the incoming speed. The cap

2 u_{1}

is approached as

m_{1} ≫ m_{2}

If the target is initially at rest, the incoming ball can never stop dead.

False. Equal masses with target at rest is the exact case where the incoming ball stops dead (

v_{1} = 0

) and hands all its velocity over. This is the Newton's-cradle result — see Newton's Cradle.

Total momentum being zero before the collision forces both final velocities to be zero.

False. Zero total momentum means the two contributions cancel (

m_{1} u_{1} = - m_{2} u_{2}

), not that motion stops. Elastically, each object simply reverses in the Center of Mass Frame; energy is still there, so they fly apart, not freeze.

Spot the error

"Speeds are conserved in an elastic collision, so I'll just keep each speed and swap directions."

Wrong: individual speeds are not conserved unless masses are equal. What's conserved is total momentum and total KE. Reversing directions alone generally breaks momentum.

u_{2} = - 1

m/s means the target's speed is

1

, so I plug in

+ 1

Sign dropped. The velocity is

- 1

(it's moving left). Plugging

+ 1

changes the approach speed

u_{1} - u_{2}

and flips the whole answer. Always use signed velocities.

"Heavy hits light, so

v_{1} = u_{1}

exactly — the heavy ball keeps its speed."

That's an approximation, valid only in the limit

m_{1} ≫ m_{2}

. For any finite mass ratio the heavy ball loses a little speed; conserve momentum properly when numbers are given.

"Energy conservation gives

u_{1} + u_{2} = v_{1} + v_{2}

No — energy is quadratic. The correct linear consequence is the relative-velocity rule

u_{1} - u_{2} = v_{2} - v_{1}

(note the minus and the swapped order on the right).

"I used only momentum, so

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

fully determines

v_{1}

and

v_{2}

One equation, two unknowns — infinitely many solutions. You need the second equation (energy / relative-velocity rule) to pin down the answer.

"The formula

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}

works even when the target is moving."

Only if

u_{2} = 0

. The full formula has a second term

\frac{2 m _{2}}{m _{1} + m _{2}} u_{2}

; dropping it when the target moves gives a wrong answer.

"In

u_{1} - u_{2} = - (v_{1} - v_{2})

the minus sign is a typo, it should be

+

The minus is essential. Approach (

u_{1} - u_{2}

) and separation (

v_{2} - v_{1}

) are the same magnitude but the pair reverses which one is catching up — the sign flip encodes that reversal.

Why questions

Why do we need exactly two conservation laws, no more and no less?

There are exactly two unknowns (

v_{1}, v_{2}

). Two independent equations give a unique solution — one leaves it undetermined, three would over-constrain (and would clash if the collision weren't truly elastic).

Why is the relative-velocity rule preferred over squaring for energy?

Squaring gives a quadratic equation that's painful to solve. The relative-velocity rule is linear, so paired with momentum you solve two easy linear equations — same physics, far less algebra.

Why does a light ball bounce straight back off a heavy wall at (nearly) the same speed?

The heavy wall barely moves, so in its frame the ball approaches at

u_{1}

and must separate at

u_{1}

(elastic) — that means reversing to

- u_{1}

in the lab. See the wall limit

m_{1} ≪ m_{2}

Why does a light ball leave at double speed when a heavy ball hits it?

In the heavy ball's frame the light ball is a wall approaching at

u_{1}

; it bounces back at

u_{1}

relative to the heavy ball, and adding the heavy ball's own

u_{1}

gives

u_{1} + u_{1} = 2 u_{1}

in the lab frame.

Why does swapping the labels

1 \leftrightarrow 2

turn the

v_{1}

formula into the

v_{2}

formula?

Physics doesn't care which ball you call "1". The equations are symmetric under relabelling, so the solution must be too — this symmetry is a free correctness check on your formulas.

Why can't a real steel-ball collision be perfectly elastic?

Some KE always leaks into sound, heat, and tiny deformation. Perfect elasticity is an idealisation (

e = 1

in Coefficient of Restitution); real collisions have

e < 1

and lose a sliver of energy.

Why is the Center of Mass Frame a slick way to think about elastic collisions?

In that frame total momentum is zero, and an elastic collision simply reverses each object's velocity. Transform to that frame, flip the signs, transform back — no quadratic ever appears.

Edge cases

What happens if

m_{2} \to 0

(target has essentially no mass)?

The incoming ball is unaffected (

v_{1} \to u_{1}

) because nothing pushes back, and the tiny target is flung to

v_{2} \to 2 u_{1} - u_{2}

— it can't carry away momentum or energy, so it just gets swept along at up to double the closing speed.

What happens if

m_{2} \to \infty

(immovable wall)?

The wall stays put (

v_{2} \to u_{2}

, usually

0

) and the incoming ball reflects:

v_{1} \to 2 u_{2} - u_{1}

, i.e.

- u_{1}

for a wall at rest. Energy and momentum still balance because the "infinite" mass absorbs vanishing velocity.

What if the two balls start with the same velocity (

u_{1} = u_{2}

Approach speed is

u_{1} - u_{2} = 0

, so there is no collision at all — they never close in. The formulas correctly return

v_{1} = u_{1}

v_{2} = u_{2}

: nothing changes.

What if

m_{1} = m_{2}

but both balls are moving?

They still simply exchange velocities:

v_{1} = u_{2}

v_{2} = u_{1}

. The equal-mass swap doesn't require the target to be at rest — that's just the most-quoted special case.

Can both objects be at rest after an elastic collision (with nonzero motion before)?

No. That would mean zero final KE, but KE is conserved and was positive before. At most one object can be at rest afterward (e.g. the incoming ball in the equal-mass swap).

What if

u_{2}

is chosen so total momentum is zero — do they stop?

They don't stop. Zero total momentum means the final momenta also cancel, but energy is conserved, so both reverse direction (in the CM frame each speed just flips) and fly apart. Zero momentum ≠ zero energy.

Is a head-on collision where the incoming ball speeds up possible?

Yes — if the target is moving toward it, the target can transfer momentum into the incoming ball. This differs from an Inelastic Collisions — 1D intuition; here energy is preserved so a fast approach can leave the first ball faster than it started.

Recall One-line self-test before you move on

Cover this and recite: the two conserved quantities, the exact form of the relative-velocity rule (with its minus sign), and the only condition under which velocities simply swap. Answer ::: Momentum and kinetic energy are conserved; $u_{1} - u_{2} = v_{2} - v_{1}$ ; velocities swap only when $m_{1} = m_{2}$ .

Connections

Elastic collisions — 1D - solve for final velocities (index 1.4.5) — the parent derivation every trap here tests.
Conservation of Momentum — the always-true law; distinguish it from energy.
Kinetic Energy — the extra elastic condition that makes speeds pin down.
Coefficient of Restitution — elastic is the $e = 1$ boundary; the edge cases blur toward $e < 1$ .
Center of Mass Frame — the "reverse each speed" viewpoint behind several answers.
Newton's Cradle — the equal-mass swap made physical.