Perfectly inelastic collisions — maximum KE loss: Question bank

Definition Symbols and idealizations used on this page

Throughout we picture two point masses (all mass at a single point, no spinning, no size) colliding on a frictionless line with no external horizontal force during the brief impact — these are the standard idealizations. The symbols:

$m_{1}, m_{2}$ — the two masses.
$u_{1}, u_{2}$ — their velocities before impact (vectors: a sign carries the direction).
$v$ — the single common velocity after they stick.
$v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ — the centre-of-mass velocity: the momentum-weighted average velocity of the pair. It never changes in a collision, and after sticking $v = v_{c m}$ . See Centre of mass motion.
$u_{r e l} = u_{1} - u_{2}$ — the relative velocity: how fast body 1 approaches body 2 as seen sitting on body 2.
$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ — the reduced mass: a single effective mass that captures the two-body relative motion (always smaller than the smaller mass). See Reduced mass.
$e$ — the coefficient of restitution $= \frac{relative speed after}{relative speed before}$ . Perfectly inelastic $\Rightarrow e = 0$ . See Coefficient of restitution.

Formula Why

$Δ K E = \frac{1}{2} μ u_{r e l}^{2}$ — derived by hopping into the COM frame The diagram below is the map for this whole derivation. Top row shows the two bodies in the lab frame with velocities $u_{1}$ (blue) and $u_{2}$ (orange). Bottom-left green arrow is the COM drift $v_{c m}$ that we subtract off. Bottom-right shows what is left after subtracting it: two equal-and-opposite arrows — the relative wobble. The red label marks the piece sticking destroys. Keep looking at it as we go.

Step 1 — Split the motion and find the COM-frame speeds. To "hop into the COM frame" we subtract $v_{c m}$ from every velocity. Call the leftover velocities $v_{1}^{'} = u_{1} - v_{c m}$ and $v_{2}^{'} = u_{2} - v_{c m}$ (the primed = "as seen in the COM frame"). Plug in $v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ for body 1: $v_{1}^{'} = u_{1} - \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}} = \frac{( m _{1} + m _{2} ) u _{1} - m _{1} u _{1} - m _{2} u _{2}}{m _{1} + m _{2}} = \frac{m _{2} ( u _{1} - u _{2} )}{m _{1} + m _{2}} = \frac{m _{2}}{m _{1} + m _{2}} u_{r e l} .$ The $m_{1} u_{1}$ terms cancel, leaving only the difference $u_{1} - u_{2} = u_{r e l}$ — that is why the COM-frame speed is built purely from the relative velocity. The same subtraction for body 2 gives $v_{2}^{'} = \frac{m _{1} ( u _{2} - u _{1} )}{m _{1} + m _{2}} = - \frac{m _{1}}{m _{1} + m _{2}} u_{r e l}$ (the minus sign is why the two arrows point opposite in the diagram). Notice the momenta are equal and opposite: $m_{1} v_{1}^{'} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{r e l} = - m_{2} v_{2}^{'}$ , so total momentum in this frame is zero, as it must be.

Step 2 — Add up the KE and watch it collapse. Now add the two kinetic energies using those speeds: $K E_{rel} = \frac{1}{2} m_{1} v_{1}^{'2} + \frac{1}{2} m_{2} v_{2}^{'2} = \frac{1}{2} m_{1} \frac{m _{2}^{2}}{( m _{1} + m _{2} ) ^{2}} u_{r e l}^{2} + \frac{1}{2} m_{2} \frac{m _{1}^{2}}{( m _{1} + m _{2} ) ^{2}} u_{r e l}^{2} .$ Both terms share the factor $\frac{m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} u_{r e l}^{2}$ ; pulling it out leaves $m_{2} + m_{1}$ inside: $K E_{rel} = \frac{1}{2} \frac{m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} (m_{2} + m_{1}) u_{r e l}^{2} = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{r e l}^{2} = \frac{1}{2} μ u_{r e l}^{2} .$ The key cancellation is one factor of $(m_{1} + m_{2})$ in the numerator against one in the denominator — that is exactly what turns the awkward two-mass sum into the tidy reduced mass $μ$ . This is the kinetic energy of the relative motion — the only KE that isn't just the whole system drifting together.

Step 3 — Sticking extinguishes exactly this. After they stick, both are at rest in the COM frame (they move together = no relative motion), so all of $K E_{rel}$ is destroyed while the drifting-together KE $\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}$ is untouched. Hence $Δ K E = \frac{1}{2} μ u_{r e l}^{2}$ .

Figure — Perfectly inelastic collisions — maximum KE loss

True or false — justify

Momentum is conserved in a perfectly inelastic collision.

True. With no external horizontal force, the impact forces are internal and equal-and-opposite (Newton's third law), so they sum to zero and cannot change the total momentum — sticking or not. Only kinetic energy is free to leak away.

"Maximum KE loss" means all kinetic energy is destroyed.

False. It means the maximum loss allowed by momentum conservation. If total momentum is nonzero the centre of mass must keep moving, so the drifting KE

K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2} > 0

always survives.

You can lose more KE than in a perfectly inelastic collision by choosing a cleverer collision.

False. Sticking already brings both bodies to rest in the COM frame (zero relative motion), the lowest KE momentum allows. Anything less would need a body with negative KE, which is impossible.

In a perfectly inelastic collision the two bodies must have zero velocity afterward.

False. They share one common velocity

v = v_{c m}

, which is zero only when total momentum is zero. Generally they move off together at a nonzero speed.

The coefficient of restitution for a perfectly inelastic collision is

e = 0

True. By definition

e = \frac{relative speed after}{relative speed before}

; sticking means the relative speed after is zero, so

e = \frac{0}{u _{r e l}} = 0

If two objects collide and stick, they cannot possibly have moved apart afterward.

True. "Stick" means zero relative velocity after impact — moving apart would require the numerator of

e

to be positive, i.e.

e > 0

, contradicting the perfectly inelastic (

e = 0

) case.

The KE lost equals the kinetic energy of the relative motion of the two bodies.

True. As derived above, the relative-motion KE is

\frac{1}{2} μ u_{r e l}^{2}

where

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

; sticking destroys exactly that motion and nothing else.

A perfectly inelastic collision always converts KE into heat only.

False. The lost KE becomes heat, sound, and permanent deformation of the bodies — not heat alone.

Doubling both initial velocity vectors quadruples the KE lost, in 1D and in 2D alike.

True.

Δ K E = \frac{1}{2} μ ∣ u_{1} - u_{2} ∣^{2}

; doubling both velocity vectors doubles the relative-velocity vector

u_{1} - u_{2}

, and its magnitude squared then grows fourfold. This holds generally — the only catch is you must combine velocities as vectors, not as bare numbers.

If one body is infinitely more massive than the other, essentially 100% of the light body's KE is lost.

True. The huge mass barely moves, so the common velocity is tiny; the light body's motion energy goes almost entirely into deformation/heat (the bullet-in-block limit).

Spot the error

"KE is conserved, so I'll set

K E_{i} = K E_{f}

to find

v

Error: KE is not conserved here — some leaks into deformation, so

K E_{i} > K E_{f}

. You must use momentum conservation,

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

, to find

v

, then compute the loss.

"They collided head-on with equal speeds, so all momentum is destroyed."

Error: Momentum is never destroyed. The impact forces are internal Newton's-third-law pairs, so they cancel; if the momenta (

m u

) are equal and opposite their sum is zero — but that zero total is conserved, not lost.

m_{1} = 2

kg at

3

m/s hits

m_{2} = 1

kg at

6

m/s (opposite), so

v = \frac{2 ( 3 ) + 1 ( 6 )}{3} = 4

Error: Velocity is a vector — the second body moves in the opposite direction, so

u_{2} = - 6

. Correct:

v = \frac{6 - 6}{3} = 0

. Forgetting the sign is the classic trap.

"Reduced mass

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

is bigger than either mass."

Error:

μ

is always smaller than the smaller mass. For equal masses

μ = m /2

; as one mass

\to \infty

μ \to

the smaller mass but never exceeds it — so it can never be bigger than either.

"After sticking, the two bodies are at rest in the lab frame because they lost all their relative KE."

Error: They lose relative-motion KE, but the centre-of-mass drift is untouched. In the lab frame they glide off at

v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

unless that COM velocity happens to be zero.

"Since the objects deform, momentum leaks into the deformation."

Error: Deformation absorbs energy, not momentum. Why? The deformation forces between the bodies are an internal equal-and-opposite pair (Newton's third law), so their impulses cancel and total momentum stays fixed; energy has no such cancellation and can pile up as heat.

Why questions

Why is momentum conserved but kinetic energy not?

The collision forces are internal and equal-and-opposite, so their net impulse is zero and total momentum can't change; but each force still does negative work on its body (they squeeze against the motion), and that work drains KE into heat and deformation. See Work–Energy theorem.

Why does sticking give the minimum possible final KE?

In the COM frame total momentum is zero; the lowest KE state consistent with "momentum zero" is both bodies at rest — which is exactly "moving together." Any other outcome leaves some relative motion, and relative motion always carries positive KE

\frac{1}{2} μ u_{r e l}^{2}

Why is

Δ K E

independent of the frame's overall velocity?

It depends only on

u_{r e l} = u_{1} - u_{2}

, and adding a common velocity to both bodies changes each individual velocity (and

v_{c m}

) but leaves their difference unchanged — so the relative motion, and its KE, is the same in every inertial frame.

Why does the ballistic pendulum lose almost all the bullet's KE?

A tiny mass (

m_{1}

, the bullet) shares its momentum with a huge one (

m_{2}

, the block), so the common speed

v = \frac{m _{1} u _{1}}{m _{1} + m _{2}}

is minuscule; nearly all the bullet's kinetic energy converts to heat and deformation. See Ballistic pendulum.

Why can't you find the post-collision height of a ballistic pendulum using energy conservation through the impact?

The impact itself is inelastic and destroys KE, so energy isn't conserved there — use momentum for the impact to get

v

, then switch to energy conservation for the swing afterward.

Why does the reduced mass

μ

naturally appear in the KE-loss formula?

When you rewrite the two-body relative motion as one effective particle, that particle's mass turns out to be

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

, so its kinetic energy is

\frac{1}{2} μ u_{r e l}^{2}

— and that is exactly the energy sticking destroys. See Reduced mass.

Why is a perfectly inelastic collision the opposite extreme from an elastic one?

Elastic collisions conserve all KE (

e = 1

, bodies bounce apart at the same relative speed); perfectly inelastic conserve no relative KE (

e = 0

, bodies stick). Every real collision sits between them at

0 < e < 1

. See Elastic collisions.

Why does an oblique (2D) perfectly inelastic collision still work the same way?

Momentum is a vector, so it's conserved component by component (

x

and

y

separately); the common final velocity is the vector

v_{c m}

, and the KE lost is still

\frac{1}{2} μ ∣ u_{1} - u_{2} ∣^{2}

using the relative-velocity vector's magnitude.

Edge cases

Total momentum is exactly zero before impact — how much KE is lost?

100%. With zero total momentum the COM is at rest (

v_{c m} = 0

), so sticking brings everything to rest, destroying all kinetic energy.

Both bodies already move at the same velocity before touching — what happens if they stick?

Nothing changes and zero KE is lost. Here

u_{1} = u_{2}

, so

u_{r e l} = 0

and

Δ K E = \frac{1}{2} μ \cdot 0 = 0

; there was no relative motion to destroy.

One body has zero mass — is the formula still meaningful?

Degenerate. With

m_{2} = 0

μ = \frac{m _{1} \cdot 0}{m _{1}} = 0

Δ K E = 0

, and

v = u_{1}

; a massless body carries no momentum or KE, so the collision does nothing to the real mass.

One mass is enormous compared with the other and initially at rest — what is the common velocity?

Essentially zero. As

m_{2} \to \infty

v = \frac{m _{1} u _{1}}{m _{1} + m _{2}} \to 0

; the wall-like body absorbs the momentum without visibly moving, and the light body's KE is nearly all lost.

The two bodies have equal masses and one is at rest — what fraction of KE survives?

Exactly half survives, half is lost.

v = u /2

gives

K E_{f} = \frac{1}{4} m u^{2}

against

K E_{i} = \frac{1}{2} m u^{2}

, so precisely

50%

is destroyed.

What is the largest possible fraction of KE that can be lost, and when?

Up to 100%, reached only when total momentum is zero. Any nonzero total momentum forces a moving COM, capping the loss below 100% because the drift KE

\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}

must survive.

Two bodies meet at right angles (oblique impact) and stick — is any KE lost?

Yes. Even at

9 0^{\circ}

there is a nonzero relative-velocity vector

u_{1} - u_{2}

, so

Δ K E = \frac{1}{2} μ ∣ u_{1} - u_{2} ∣^{2} > 0

; only when the two velocity vectors are identical does the loss vanish.

Connections

Conservation of Linear Momentum — the law behind every "true" answer here.
Coefficient of restitution — the $e = 0$ edge case these traps orbit.
Reduced mass — why $μ$ and $u_{r e l}$ govern the loss.
Centre of mass motion — the surviving KE lives here.
Elastic collisions — the opposite extreme for contrast.
Ballistic pendulum — the "almost all KE lost" showcase.
Work–Energy theorem — where the lost KE actually goes.