Perfectly inelastic collisions — maximum KE loss: Worked examples

Before anything: a symbol dictionary so nothing enters unexplained.

Definition The symbols, in plain words

$m_{1}, m_{2}$ — the two masses (kilograms). Always positive numbers.
$u_{1}, u_{2}$ — velocities before the hit. These are signed: pick one direction as $+$ (say, rightward), then leftward is $-$ . A velocity is a speed with a direction attached.
$v$ — the single velocity both share after they stick.
$p$ — the total momentum, defined as $p = m_{1} u_{1} + m_{2} u_{2}$ (mass times velocity, summed, signs included). It is the same before and after — that is the one law we never break. See Conservation of Linear Momentum.
$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ — the reduced mass. Think of it as the "effective mass of the relative motion." See Reduced mass.
$u_{r e l} = u_{1} - u_{2}$ — the relative velocity: how fast object 1 approaches object 2 as seen from object 2.
$K E = \frac{1}{2} m v^{2}$ — kinetic energy, the "energy of motion." Because of the square, it never cares about sign — moving left or right, same KE.

The scenario matrix

Every perfectly inelastic collision is fixed by three yes/no questions: same direction or opposite?, is one at rest?, equal or unequal masses? — plus the degenerate edges. The total momentum column uses $p = m_{1} u_{1} + m_{2} u_{2}$ (defined above). The table below lists every cell; each example tag [C#] maps to a row.

Cell	Sign of $u_{1}, u_{2}$	Masses	Total momentum $p = m_{1} u_{1} + m_{2} u_{2}$	What's special	Example
C1	both $+$ , one is $0$	equal	$> 0$	textbook baseline	Ex 1
C2	both $+$ , one is $0$	very unequal	$> 0$	huge KE loss $%$	Ex 2
C3	$+$ and $-$	unequal	$= 0$	they stop dead, 100 $%$ loss	Ex 3
C4	$+$ and $-$	unequal	$\neq = 0$	survivors keep moving	Ex 4
C5	both $+$	any	$> 0$	both already moving, catch-up	Ex 5
C6 (degenerate)	$u_{1} = u_{2}$	any	$\neq = 0$	no relative motion ⇒ zero loss	Ex 6
C7 (degenerate)	any	$m_{2} \to 0$	$\approx m_{1} u_{1}$	limiting behaviour	Ex 7
C8 (word)	real-world	unequal	$> 0$	ballistic-pendulum height, Ballistic pendulum	Ex 8
C9 (exam twist)	fraction asked	equal	$> 0$	"what fraction of KE survives?"	Ex 9

Figure — Perfectly inelastic collisions — maximum KE loss

Worked Examples

Worked example Ex 8 — Word problem: ballistic pendulum height

[C8] A $0.05 kg$ dart at $u_{1} = + 120 m/s$ embeds in a $1.95 kg$ block hanging at rest. How high $h$ does the block-plus-dart swing? Use $g = 10 m/s^{2}$ . See Ballistic pendulum.

Forecast: two energy regimes — one where KE is lost, one where it's conserved. Which is which?

Step 1 — Collision (momentum only). $v = \frac{0.05 ( 120 ) + 1.95 ( 0 )}{2} = \frac{6}{2} = 3 m/s$ . Why this step? During embedding, KE is lost — so we may not use energy here, only momentum. See Conservation of Linear Momentum.

Step 2 — Swing (energy conserved). After sticking, no more squishing, so the blob's KE climbs into height: $\frac{1}{2} (2) v^{2} = (2) g h$ . See Work–Energy theorem. Why this step? Post-collision there's no non-conservative work, so mechanical energy is now conserved.

Step 3 — Solve for height. $h = \frac{v ^{2}}{2 g} = \frac{3 ^{2}}{2 ( 10 )} = \frac{9}{20} = 0.45 m$ .

Verify: $K E$ at bottom $= \frac{1}{2} (2) (3)^{2} = 9 J$ ; $P E$ at top $= 2 \cdot 10 \cdot 0.45 = 9 J$ ✓. Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ ✓.

Worked example Ex 9 — Exam twist: fraction of KE that

survives [C9] Mass $m$ at speed $u$ sticks to an identical mass $m$ at rest. What fraction of the initial KE remains? (Answer symbolically, then check numerically with $m = 1, u = 4$ .)

Forecast: we found "half is lost" in Ex 1 — so does half survive?

Step 1 — Common velocity from momentum. Conserve momentum: $m u + m (0) = (m + m) v$ , so $2 m v = m u$ , giving $v = \frac{u}{2}$ . Why this step? We can't guess $v$ ; the only law that survives a sticking collision is momentum conservation, and solving it for $v$ forces $v = u /2$ . Everything downstream depends on this.

Step 2 — Form the fraction. $K E_{f} = \frac{1}{2} (2 m) (u /2)^{2} = \frac{1}{4} m u^{2}$ , $K E_{i} = \frac{1}{2} m u^{2}$ , so $\frac{K E _{f}}{K E _{i}} = \frac{\frac{1}{4} m u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{1}{2}$ . Why this step? Keeping $m, u$ symbolic proves the fraction is universal — masses and speeds cancel to a pure number.

Step 3 — Interpret. Exactly $\frac{1}{2}$ survives and $\frac{1}{2}$ is lost, for any equal-mass one-at-rest sticking collision.

Verify: with $m = 1, u = 4$ : $K E_{i} = 8 J$ , $v = 2$ , $K E_{f} = \frac{1}{2} (2) (2)^{2} = 4 J$ , ratio $= 4/8 = 0.5$ ✓.

Recall Self-check: which cell is which?

A blob ends up at rest after the collision — what must have been true? ::: Total momentum was zero (equal and opposite), the C3 cell — 100 $%$ of KE lost. The collision loses zero KE — what does that tell you? ::: The two bodies had identical velocities ( $u_{r e l} = 0$ ), the degenerate C6 cell. Bullet-into-block loses ~99 $%$ of KE — why? ::: Wildly unequal masses (C2/C7 regime); the tiny mass shares momentum with a huge mass so $v$ and $K E_{f}$ are tiny. In a ballistic pendulum, why can't you use energy conservation through the whole problem? ::: The embedding phase (C8) loses KE; only the swing after sticking conserves energy.

Connections

Parent — full derivation
Conservation of Linear Momentum — the one law used in every cell above.
Reduced mass — the $μ$ that carries all the lost energy.
Ballistic pendulum — the two-phase logic of Ex 8.
Work–Energy theorem — turns the surviving KE into swing height.
Coefficient of restitution — all these cells sit at $e = 0$ .
Elastic collisions — the opposite extreme, $e = 1$ , zero loss.
Centre of mass motion — $v$ here is always the COM velocity.