Perfectly inelastic collisions — maximum KE loss: Exercises

Everything on this page uses only two master results from the parent parent topic:

A reminder on signs, because half of these problems live or die by them:

Level 1 — Recognition

L1.1

Two lumps of putty stick together after colliding. Which quantity is conserved: momentum, kinetic energy, both, or neither?

Recall Solution

Momentum only. A perfectly inelastic collision has no external horizontal force, so total momentum is unchanged (this is Conservation of Linear Momentum). Kinetic energy is not conserved — some is spent squishing the putty and making heat/sound. Answer: momentum is conserved, KE is not.

L1.2

A $3 kg$ mass at $4 m/s$ strikes and sticks to a $1 kg$ mass at rest. Write the common velocity $v$ symbolically, then compute it.

Recall Solution

Pick the two given numbers and drop them into $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ . $v = \frac{( 3 ) ( 4 ) + ( 1 ) ( 0 )}{3 + 1} = \frac{12}{4} = 3 m/s$ Answer: $v = 3 m/s$ (same direction as the moving mass).

L1.3

For the same collision as L1.2, what is the reduced mass $μ$ ?

Recall Solution

$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{( 3 ) ( 1 )}{3 + 1} = \frac{3}{4} = 0.75 kg$ Answer: $μ = 0.75 kg$ . Notice $μ$ is always smaller than either individual mass — that's why it's called "reduced."

Level 2 — Application

L2.1

A $0.02 kg$ bullet travelling at $500 m/s$ embeds in a stationary $3.98 kg$ wooden block. Find (a) the common velocity, (b) the KE lost, (c) the percentage of initial KE lost.

Recall Solution

(a) Momentum first. Total mass $= 0.02 + 3.98 = 4.00 kg$ . $v = \frac{( 0.02 ) ( 500 ) + 0}{4.00} = \frac{10}{4.00} = 2.5 m/s$ (b) KE lost — use $\frac{1}{2} μ u_{r e l}^{2}$ . Relative speed $u_{r e l} = 500 - 0 = 500 m/s$ . $μ = \frac{( 0.02 ) ( 3.98 )}{4.00} = \frac{0.0796}{4.00} = 0.0199 kg$ This value is exact, not rounded: $3.98 \times 0.02 = 0.0796$ and $0.0796/4.00 = 0.0199$ terminates cleanly, so we carry all four significant figures with confidence. $Δ K E = \frac{1}{2} (0.0199) (500)^{2} = \frac{1}{2} (0.0199) (250000) = 2487.5 J$ (c) Percentage. $K E_{i} = \frac{1}{2} (0.02) (500)^{2} = 2500 J$ . $\frac{Δ K E}{K E _{i}} = \frac{2487.5}{2500} = 0.995 = 99.5%$ Answers: $v = 2.5 m/s$ , $Δ K E = 2487.5 J$ , $99.5%$ lost. A tiny mass forced to share momentum with a huge one ends up slow, so nearly all the energy is dumped into deformation — the Ballistic pendulum works on exactly this fact.

L2.2

Two identical railway cars, each $m = 8000 kg$ , approach on the same track: one at $+ 2 m/s$ , the other at $+ 0.5 m/s$ (the faster catches up and couples with the slower). Find $v$ and $Δ K E$ .

Recall Solution

Both move the same way, so both are $+$ . $v = \frac{8000 ( 2 ) + 8000 ( 0.5 )}{16000} = \frac{16000 + 4000}{16000} = \frac{20000}{16000} = 1.25 m/s$ Relative speed $u_{r e l} = 2 - 0.5 = 1.5 m/s$ , and $μ = \frac{8000 \cdot 8000}{16000} = 4000 kg$ . $Δ K E = \frac{1}{2} (4000) (1.5)^{2} = \frac{1}{2} (4000) (2.25) = 4500 J$ Answers: $v = 1.25 m/s$ , $Δ K E = 4500 J$ .

L2.3

A $2 kg$ ball moving right at $6 m/s$ collides head-on and sticks to a $4 kg$ ball moving left at $3 m/s$ . Find $v$ (state its direction) and $Δ K E$ .

Recall Solution

Right $= +$ . So $u_{1} = + 6$ , $u_{2} = - 3$ . $v = \frac{2 ( 6 ) + 4 ( - 3 )}{6} = \frac{12 - 12}{6} = \frac{0}{6} = 0 m/s$ The blob is momentarily at rest! $K E_{i} = \frac{1}{2} (2) (6)^{2} + \frac{1}{2} (4) (3)^{2} = 36 + 18 = 54 J, K E_{f} = 0$ $Δ K E = 54 J = 100% lost$ Cross-check with $\frac{1}{2} μ u_{r e l}^{2}$ : $u_{r e l} = 6 - (- 3) = 9$ , $μ = \frac{2 \cdot 4}{6} = \frac{4}{3}$ , so $Δ K E = \frac{1}{2} \cdot \frac{4}{3} \cdot 81 = 54 J$ ✓. Answers: $v = 0$ , $Δ K E = 54 J$ (all of it).

Level 3 — Analysis

L3.1

Prove that when a moving mass $m_{1}$ (speed $u$ ) sticks to an equal stationary mass $m_{2} = m_{1}$ , exactly half the initial KE is lost — regardless of the value of $m_{1}$ or $u$ .

Recall Solution

Let each mass $= m$ , $u_{1} = u$ , $u_{2} = 0$ . $v = \frac{m u + 0}{2 m} = \frac{u}{2}$ $K E_{i} = \frac{1}{2} m u^{2}, K E_{f} = \frac{1}{2} (2 m) (\frac{u}{2})^{2} = \frac{1}{2} (2 m) \frac{u ^{2}}{4} = \frac{1}{4} m u^{2}$ $\frac{Δ K E}{K E _{i}} = \frac{\frac{1}{2} m u ^{2} - \frac{1}{4} m u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{\frac{1}{4} m u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{1}{2}$ The $m$ and $u$ cancel completely, so the fraction is always $1/2$ . Proven. The geometry: the moving mass hands half its speed to its partner, and halving the speed while doubling the mass quarters — not halves — the KE, leaving a clean one-half deficit.

L3.2

Two masses collide and stick. In terms of the Coefficient of restitution $e$ , a perfectly inelastic collision has $e = 0$ . Explain why $e = 0$ forces the objects to share one velocity, and connect this to "maximum KE loss."

Recall Solution

First pin down the labels. Put both bodies on a line moving in the $+$ direction. Call body 1 the one behind (velocity $u_{1}$ ) and body 2 the one ahead (velocity $u_{2}$ ), and choose them so that body 1 is catching up: $u_{1} > u_{2}$ , i.e. the approach difference $u_{1} - u_{2} > 0$ . Their final velocities are $v_{1}^{'}$ (body 1) and $v_{2}^{'}$ (body 2). The coefficient of restitution is the ratio of the two speeds — which are magnitudes, so we use absolute-value bars to keep them positive regardless of sign conventions: $e = \frac{speed of separation}{speed of approach} = \frac{∣ v _{2}^{'} - v _{1}^{'} ∣}{∣ u _{1} - u _{2} ∣}$ Setting $e = 0$ forces the numerator to vanish: $∣ v_{2}^{'} - v_{1}^{'} ∣ = 0$ , i.e. $v_{2}^{'} = v_{1}^{'}$ . That is precisely "they move together with one common velocity" — and this shared value is exactly the bare $v$ we compute everywhere else: $v_{1}^{'} = v_{2}^{'} = v$ . Because they separate at zero speed, all the relative-motion KE (the $\frac{1}{2} μ u_{r e l}^{2}$ chunk) is destroyed — the maximum a momentum-conserving collision can lose. Any $e > 0$ leaves some separation speed, hence some relative KE survives, hence less is lost. Answer: $e = 0 \Rightarrow v_{1}^{'} = v_{2}^{'} = v$ , killing the entire relative-motion KE — the maximum loss.

L3.3

A $5 kg$ mass at $u_{1}$ (unknown, moving right) sticks to a $5 kg$ mass at rest, and the collision loses $250 J$ . Find $u_{1}$ .

Recall Solution

Equal masses, one at rest. Reduced mass $μ = \frac{5 \cdot 5}{10} = 2.5 kg$ , relative speed $= u_{1}$ . $Δ K E = \frac{1}{2} μ u_{1}^{2} \Rightarrow 250 = \frac{1}{2} (2.5) u_{1}^{2} = 1.25 u_{1}^{2}$ $u_{1}^{2} = \frac{250}{1.25} = 200 \Rightarrow u_{1} = \pm 200 = \pm 102$ The algebra gives two roots. Our sign convention fixed "moving right" as $+$ , and the problem states the mass moves right, so we reject the negative root $- 102$ (that would describe a leftward-moving mass, contradicting the given). Keep the positive one: $u_{1} = 102 \approx 14.14 m/s$ Answer: $u_{1} = 102 \approx 14.14 m/s$ . We inverted the energy formula because it directly links the loss to the one unknown speed — no need to find $v$ at all.

Level 4 — Synthesis

L4.1 — Ballistic pendulum

A bullet of mass $m = 0.05 kg$ fired horizontally embeds in a block of mass $M = 4.95 kg$ hanging from strings. The block+bullet then swings up to a height $h = 0.20 m$ . Take $g = 10 m/s^{2}$ . Find the bullet's original speed $u$ .

The figure below shows the two-phase story. Left (Phase 1): the magenta arrow is the bullet of mass $m$ flying in at speed $u$ toward the violet block $M$ that hangs at rest from two vertical strings; during this fast impact momentum is conserved (labelled in magenta). Right (Phase 2): the stuck lump $M + m$ (orange) swings along the dashed orange arc, rising by the vertical height $h$ (the navy double-headed arrow); the small magenta arrow $v$ at the lowest point is the common velocity just after sticking, and during this slow rise energy is conserved (labelled in orange). Notice the two phases obey two different laws.

Figure — Perfectly inelastic collisions — maximum KE loss

Recall Solution

Combine the masses first (and mind the precision). $M + m = 4.95 + 0.05 = 5.00 kg$ exactly — the two decimals add to a clean $5.00$ , so no rounding is involved and we keep three significant figures $5.00 kg$ throughout. Phase 2 first (the swing) — use energy. After sticking, the combined mass $M + m$ moves at $v$ , then rises to height $h$ , converting all KE to potential energy (this is the Work–Energy theorem with gravity): $\frac{1}{2} (M + m) v^{2} = (M + m) g h \Rightarrow v = 2 g h = 2 (10) (0.20) = 4 = 2 m/s$ Phase 1 (the collision) — use momentum. The bullet+block stick, so: $m u = (M + m) v \Rightarrow u = \frac{( M + m ) v}{m} = \frac{( 5.00 ) ( 2 )}{0.05} = \frac{10}{0.05} = 200 m/s$ Answer: $u = 200 m/s$ . Why two laws? During the collision, energy leaks into deformation — so KE conservation is forbidden there; only momentum survives. During the swing, nothing sticks or squishes, so energy conservation is valid there. Using the right law in the right phase is the whole trick.

L4.2 — Fraction lost depends only on the mass ratio

Show that when mass $m_{1}$ hits a stationary mass $m_{2}$ , the fraction of KE lost is $\frac{m _{2}}{m _{1} + m _{2}}$ . Then evaluate it for a bullet ( $m_{1} = 0.01$ ) and block ( $m_{2} = 1.99$ ).

Recall Solution

With $u_{2} = 0$ : $u_{r e l} = u_{1}$ , and $K E_{i} = \frac{1}{2} m_{1} u_{1}^{2}$ . $\frac{Δ K E}{K E _{i}} = \frac{\frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} u _{1}^{2}}{\frac{1}{2} m _{1} u _{1}^{2}} = \frac{\frac{m _{1} m _{2}}{m _{1} + m _{2}}}{m _{1}} = \frac{m _{2}}{m _{1} + m _{2}}$ The $\frac{1}{2}$ , $m_{1}$ , and $u_{1}^{2}$ all cancel — the fraction is pure geometry of the masses. For the bullet/block: $\frac{1.99}{0.01 + 1.99} = \frac{1.99}{2.00} = 0.995 = 99.5%$ . Answer: fraction $= \frac{m _{2}}{m _{1} + m _{2}}$ ; numerically $99.5%$ . This is why a light projectile hitting a heavy target loses almost everything — the surviving KE fraction is $\frac{m _{1}}{m _{1} + m _{2}}$ , tiny when $m_{1} ≪ m_{2}$ .

Level 5 — Mastery

L5.1 — Minimising the loss with a movable second mass

A block of mass $m_{1} = 2 kg$ moving at $u_{1} = 9 m/s$ will stick to a second block $m_{2}$ at rest. You may choose $m_{2}$ . (a) As $m_{2} \to 0$ , what fraction of KE is lost? (b) As $m_{2} \to \infty$ ? (c) For $m_{2} = 2 kg$ , compute $Δ K E$ . Interpret the trend.

Recall Solution

Use the L4.2 result: fraction lost $= \frac{m _{2}}{m _{1} + m _{2}}$ . (a) $m_{2} \to 0$ : fraction $\to \frac{0}{2 + 0} = 0$ . Almost no mass to share with, so almost no KE lost — the collision barely happens. (b) $m_{2} \to \infty$ : fraction $\to \frac{m _{2}}{2 + m _{2}} \to 1$ , i.e. $100%$ . Hitting an immovable wall and sticking wastes all your KE. (c) $m_{2} = 2 kg$ : equal masses, so fraction $= \frac{1}{2}$ . Compute the actual loss: $μ = \frac{2 \cdot 2}{4} = 1 kg, Δ K E = \frac{1}{2} (1) (9)^{2} = 40.5 J$ (Check: $K E_{i} = \frac{1}{2} (2) (9^{2}) = 81 J$ ; half is $40.5 J$ ✓.) Answers: (a) $0%$ , (b) $100%$ , (c) $Δ K E = 40.5 J$ . The loss fraction climbs monotonically from $0$ to $1$ as the target grows.

L5.2 — Three-body sequential sticking

On a frictionless line, $m_{A} = 1 kg$ moves right at $12 m/s$ and sticks to $m_{B} = 2 kg$ at rest, forming a lump. That lump then continues and sticks to $m_{C} = 3 kg$ at rest. Find the final velocity of all three and the total KE lost across both collisions.

Recall Solution

Collision 1 ( $A$ into $B$ ): $v_{1} = \frac{( 1 ) ( 12 ) + ( 2 ) ( 0 )}{1 + 2} = \frac{12}{3} = 4 m/s (lump mass 3 kg)$ Collision 2 (lump into $C$ ): $v_{2} = \frac{( 3 ) ( 4 ) + ( 3 ) ( 0 )}{3 + 3} = \frac{12}{6} = 2 m/s (final, mass 6 kg)$ Total KE lost. Initial KE was all in $A$ : $K E_{i} = \frac{1}{2} (1) (12)^{2} = 72 J$ Final KE: $K E_{f} = \frac{1}{2} (6) (2)^{2} = 12 J$ $Δ K E_{total} = 72 - 12 = 60 J$ Answers: final velocity $= 2 m/s$ ; total KE lost $= 60 J$ . Sanity check by summing the two stage-losses: stage 1 loses $\frac{1}{2} μ_{1} u_{r e l 1}^{2} = \frac{1}{2} \cdot \frac{1 \cdot 2}{3} \cdot 1 2^{2} = 48 J$ ; stage 2 loses $\frac{1}{2} \cdot \frac{3 \cdot 3}{6} \cdot 4^{2} = \frac{1}{2} \cdot 1.5 \cdot 16 = 12 J$ ; total $48 + 12 = 60 J$ ✓. Notice: momentum was conserved throughout ( $1 \times 12 = 6 \times 2 = 12 kg\cdotm/s$ ), but KE steadily bled away at each stick.

L5.3 — When does a sticking collision lose exactly half its KE?

Prove: a moving mass $m_{1}$ sticking to a stationary mass $m_{2}$ loses exactly half the initial KE iff $m_{1} = m_{2}$ .

Recall Solution

From L4.2, fraction lost $= \frac{m _{2}}{m _{1} + m _{2}}$ . Set it equal to $\frac{1}{2}$ : $\frac{m _{2}}{m _{1} + m _{2}} = \frac{1}{2} \Rightarrow 2 m_{2} = m_{1} + m_{2} \Rightarrow m_{1} = m_{2}$ The step is reversible (each implication is an equivalence), so it holds if and only if the masses are equal. Proven. This is the deep reason L3.1's answer was independent of $m$ and $u$ : equal masses is the only condition that pins the fraction at one-half.

Recall One-line summary of the whole ladder

Momentum decides the speed ( $v = \frac{\sum m u}{\sum m}$ ); reduced mass and relative speed decide the loss ( $\frac{1}{2} μ u_{r e l}^{2}$ ); the fraction lost when a target is at rest is pure mass geometry ( $\frac{m _{2}}{m _{1} + m _{2}}$ ), reaching $100%$ only when total momentum is zero.

Connections

Parent topic — the theory these exercises drill.
Conservation of Linear Momentum — used in every single problem.
Ballistic pendulum — L4.1 is the classic application.
Work–Energy theorem — the swing phase of L4.1.
Coefficient of restitution — L3.2 links $e = 0$ to sticking.
Reduced mass — the $μ$ in the loss formula.
Elastic collisions — the $e = 1$ opposite extreme for contrast.