Elastic collisions — 1D - solve for final velocities: Worked examples

Here $m_{1}, m_{2}$ are the two masses, $u_{1}, u_{2}$ their velocities before, $v_{1}, v_{2}$ their velocities after. A subscript is just a name tag: "1" is the left-hand object, "2" the right-hand one. Rightward is positive; a negative number means "moving left". Keep that in your bones — half of all mistakes are dropped minus signs.

The scenario matrix

Every 1D elastic problem falls into one of these cells. The examples below are labelled by cell so you can see the whole space is covered.

Cell	What varies	Example
A. Both at rest / trivial	$u_{1} = u_{2} = 0$ — degenerate	Ex 1
B. Equal masses, target still	$m_{1} = m_{2}, u_{2} = 0$	Ex 2
C. Equal masses, both moving (opposite signs)	$m_{1} = m_{2}$ , one negative	Ex 3
D. Unequal masses, target at rest	general $m$ , $u_{2} = 0$	Ex 4
E. Both moving, both nonzero, unequal masses	full general signs	Ex 5
F. Heavy → light limit	$m_{1} ≫ m_{2}$	Ex 6
G. Light → heavy wall limit	$m_{1} ≪ m_{2}$	Ex 7
H. Real-world word problem	pool/skater story	Ex 8
I. Exam twist — reverse the arrow	given $v$ 's, find $u$ 's	Ex 9

Ex 1 — Cell A · nothing is moving (degenerate)

Ex 2 — Cell B · equal masses, target at rest (the swap)

Ex 3 — Cell C · equal masses, head-on (both moving, opposite signs)

Worked example

$m_{1} = m_{2} = 3$ kg, $u_{1} = + 4$ m/s (right), $u_{2} = - 2$ m/s (left). Forecast: They rush at each other. Do they just bounce back? Guess signs of $v_{1}, v_{2}$ .

Step 1 — relative rule. $u_{1} - u_{2} = 4 - (- 2) = 6 = v_{2} - v_{1}$ . Why this step? Note the double minus: subtracting a leftward velocity adds. Missing this is the #1 sign error.

Step 2 — momentum. $3 (4) + 3 (- 2) = 12 - 6 = 6 = 3 (v_{1} + v_{2}) \Rightarrow v_{1} + v_{2} = 2$ . Why this step? Second equation; divide out the common mass.

Step 3 — solve. Add: $2 v_{2} = 6 + 2 = 8 \Rightarrow v_{2} = 4$ ; then $v_{1} = 2 - 4 = - 2$ . Why this step? Equal masses always swap velocities — and indeed $v_{1} = u_{2}$ , $v_{2} = u_{1}$ .

Verify: KE before $= \frac{1}{2} (3) (16) + \frac{1}{2} (3) (4) = 24 + 6 = 30$ J. After $= \frac{1}{2} (3) (4) + \frac{1}{2} (3) (16) = 6 + 24 = 30$ J. ✓ Look at the figure: red (object 1) came in fast-right, leaves slow-left; violet (object 2) came in slow-left, leaves fast-right. They traded velocities, signs and all.

Ex 4 — Cell D · unequal masses, target at rest (use the full formula)

Worked example

$m_{1} = 3$ kg, $m_{2} = 1$ kg, $u_{1} = 8$ m/s, $u_{2} = 0$ . Forecast: Heavier ball hits a lighter one. Does the heavy one keep going forward? Does the light one overtake?

Step 1 — plug the $v_{1}$ formula. $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} = \frac{3 - 1}{3 + 1} (8) = \frac{2}{4} (8) = 4$ m/s. Why this step? With $u_{2} = 0$ the second term of each formula dies, so the formula shrinks to one term — fastest route.

Step 2 — plug the $v_{2}$ formula. $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} = \frac{6}{4} (8) = 12$ m/s. Why this step? Same shortcut for the target.

Step 3 — sanity via relative rule. Separation $v_{2} - v_{1} = 12 - 4 = 8 = u_{1} - u_{2}$ . ✓ Why this step? Cross-check that our two numbers satisfy the approach = separation law.

Verify: Momentum before $= 3 (8) = 24$ ; after $= 3 (4) + 1 (12) = 12 + 12 = 24$ . ✓ KE before $= \frac{1}{2} (3) (64) = 96$ J; after $= \frac{1}{2} (3) (16) + \frac{1}{2} (1) (144) = 24 + 72 = 96$ J. ✓ Heavy ball keeps moving forward (positive) but slower; light ball shoots ahead faster than the heavy one arrived.

Ex 5 — Cell E · both moving, unequal masses, full signs

Worked example

$m_{1} = 4$ kg at $u_{1} = + 5$ m/s, $m_{2} = 2$ kg at $u_{2} = + 2$ m/s (both rightward, fast catches slow). Forecast: A fast rear cart catches a slow front cart. Can the rear one end up moving backward? Guess.

Step 1 — relative rule. $u_{1} - u_{2} = 5 - 2 = 3 \Rightarrow v_{2} - v_{1} = 3 \Rightarrow v_{2} = v_{1} + 3$ . Why this step? Turns two unknowns into one substitution.

Step 2 — momentum. $4 (5) + 2 (2) = 20 + 4 = 24 = 4 v_{1} + 2 v_{2}$ . Why this step? Total $p$ is our anchor; it must equal 24 after too.

Step 3 — substitute & solve. $4 v_{1} + 2 (v_{1} + 3) = 24 \Rightarrow 6 v_{1} + 6 = 24 \Rightarrow v_{1} = 3$ ; then $v_{2} = 6$ . Why this step? One linear equation in $v_{1}$ once we substitute.

Verify: Momentum after $= 4 (3) + 2 (6) = 12 + 12 = 24$ . ✓ KE before $= \frac{1}{2} (4) (25) + \frac{1}{2} (2) (4) = 50 + 4 = 54$ J. After $= \frac{1}{2} (4) (9) + \frac{1}{2} (2) (36) = 18 + 36 = 54$ J. ✓ The rear cart slows from 5 to 3 (still forward, doesn't reverse — because it's heavier), the front cart speeds from 2 to 6.

Ex 6 — Cell F · heavy hits light (limiting behaviour)

Worked example

$m_{1} = 1000$ kg at $u_{1} = 3$ m/s hits $m_{2} = 1$ kg at rest ( $u_{2} = 0$ ). Forecast: The parent note says the light ball leaves at ~double. Guess $v_{1}$ and $v_{2}$ , then check how close to $2 u_{1}$ .

Step 1 — $v_{1}$ formula. $v_{1} = \frac{1000 - 1}{1000 + 1} (3) = \frac{999}{1001} (3) = \frac{2997}{1001} \approx 2.994$ m/s. Why this step? We want the exact value to see how tiny the slow-down really is — the parent warned $v_{1} \approx u_{1}$ is only an approximation.

Step 2 — $v_{2}$ formula. $v_{2} = \frac{2 ( 1000 )}{1001} (3) = \frac{6000}{1001} \approx 5.994$ m/s. Why this step? Compare to the "double" prediction $2 u_{1} = 6$ .

Step 3 — interpret. $v_{1} \approx 2.994$ (heavy barely slows), $v_{2} \approx 5.994 \approx 2 u_{1}$ . Why this step? Confirms the limit: as $m_{1} \to \infty$ , $v_{1} \to u_{1}$ and $v_{2} \to 2 u_{1}$ , but for finite mass it is slightly less.

Verify: Momentum before $= 1000 (3) = 3000$ ; after $= 1000 (2.994...) + 1 (5.994...) = 2994.006 + 5.994 = 3000$ . ✓ Exactly the double-speed limit, softened by the finite ratio — this is why the " $v_{1} = u_{1}$ exactly" mistake is wrong.

Ex 7 — Cell G · light ball off a heavy wall (limiting behaviour)

Worked example

$m_{1} = 0.5$ kg at $u_{1} = 7$ m/s hits $m_{2} = 500$ kg at rest. Forecast: A pebble hits a boulder. Does it bounce straight back? At what speed?

Step 1 — $v_{1}$ formula. $v_{1} = \frac{0.5 - 500}{0.5 + 500} (7) = \frac{- 499.5}{500.5} (7) \approx - 6.986$ m/s. Why this step? The numerator is negative because $m_{2} > m_{1}$ , so $v_{1}$ flips sign — the pebble reverses.

Step 2 — $v_{2}$ formula. $v_{2} = \frac{2 ( 0.5 )}{500.5} (7) = \frac{7}{500.5} \approx 0.0140$ m/s. Why this step? Shows the wall barely moves — tiny but nonzero (a real wall recoils a hair).

Step 3 — interpret. $v_{1} \approx - 7$ (bounces back at nearly the same speed), $v_{2} \approx 0$ . Why this step? Confirms the wall limit: $m_{2} \to \infty$ gives $v_{1} \to - u_{1}$ , $v_{2} \to 0$ .

Verify: Momentum before $= 0.5 (7) = 3.5$ ; after $= 0.5 (- 6.986...) + 500 (0.01398...) = - 3.493 + 6.993 = 3.5$ . ✓ The pebble carries away nearly all the energy, the wall almost none — the elastic "wall bounce".

Ex 8 — Cell H · real-world word problem

Worked example A

$0.17$ kg cue ball rolls at $2$ m/s and strikes a stationary $0.16$ kg billiard ball head-on. Assume a perfectly elastic hit. Find both speeds after. Forecast: Masses are almost equal. Guess: do they nearly swap?

Step 1 — name the symbols. $m_{1} = 0.17, u_{1} = 2, m_{2} = 0.16, u_{2} = 0$ . Why this step? Translating a story into $m, u$ tags is the real exam skill — do it first, every time.

Step 2 — $v_{1}$ formula. $v_{1} = \frac{0.17 - 0.16}{0.33} (2) = \frac{0.01}{0.33} (2) = \frac{0.02}{0.33} \approx 0.0606$ m/s. Why this step? Nearly-equal masses ⇒ nearly-zero for the cue ball, matching the "almost swap" forecast.

Step 3 — $v_{2}$ formula. $v_{2} = \frac{2 ( 0.17 )}{0.33} (2) = \frac{0.68}{0.33} \approx 2.0606$ m/s. Why this step? The target takes almost the full speed — just over $2$ .

Verify: Momentum before $= 0.17 (2) = 0.34$ ; after $= 0.17 (0.0606) + 0.16 (2.0606) = 0.0103 + 0.3297 = 0.34$ . ✓ Units: (kg)(m/s) throughout — momentum in kg·m/s, velocities in m/s. The cue ball almost stops, the target rolls off just above $2$ m/s: real pool behaviour.

Ex 9 — Cell I · exam twist, run it backwards

Worked example After an elastic collision two carts (

$m_{1} = 1$ kg, $m_{2} = 3$ kg) are seen moving at $v_{1} = - 3$ m/s and $v_{2} = 1$ m/s. Given that cart 2 started at rest ( $u_{2} = 0$ ), find the initial velocity $u_{1}$ . Forecast: Can we un-collide? The laws run both directions in time — guess whether one equation is enough.

Step 1 — momentum, solved for the unknown. $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ , i.e. $1 \cdot u_{1} + 0 = 1 (- 3) + 3 (1) = 0$ . Why this step? We know everything except $u_{1}$ ; momentum is one equation in one unknown.

Step 2 — read off. $u_{1} = 0$ ?? That can't be — a still cart can't cause a collision. Something is inconsistent, so cross-check with the relative-velocity rule. Why this step? If the given "after" state is genuinely elastic, both laws must agree. Test it.

Step 3 — relative-rule test. With $u_{2} = 0$ : approach $= u_{1} - 0 = u_{1}$ , separation $= v_{2} - v_{1} = 1 - (- 3) = 4$ . Elastic demands $u_{1} = 4$ . But momentum demanded $u_{1} = 0$ . Contradiction ⇒ the stated after-state is impossible for an elastic collision with these masses. Why this step? This is the exam trap: it hands you numbers that look fine but violate one law. Always test both.

Verify: The consistent scenario is $u_{1} = 4$ . Then momentum says $m_{1} (4) = 1 \cdot v_{1} + 3 \cdot v_{2}$ , i.e. $v_{1} + 3 v_{2} = 4$ , and the rule says $v_{2} - v_{1} = 4$ . Solve: add $3 \times$ (rule) to momentum... use formulas instead: $v_{1} = \frac{1 - 3}{4} (4) = - 2$ , $v_{2} = \frac{2 ( 1 )}{4} (4) = 2$ . Check momentum $1 (- 2) + 3 (2) = 4$ ✓ and KE $\frac{1}{2} (1) (16) = 8$ before, $\frac{1}{2} (1) (4) + \frac{1}{2} (3) (4) = 2 + 6 = 8$ after ✓. Lesson: a valid elastic outcome must satisfy BOTH laws; the printed $(- 3, 1)$ did not.

Recall Which cell was which?

Ex1 — both at rest (degenerate) ::: nothing happens, all zeros. Ex2 — equal masses target still ::: velocities swap, cue ball stops. Ex3 — equal masses head-on ::: swap including signs, $v_{1} = u_{2}, v_{2} = u_{1}$ . Ex4 — unequal, target at rest ::: use the $\frac{m _{1} - m _{2}}{m _{1} + m _{2}}$ form. Ex5 — both moving, unequal ::: full signed momentum + relative rule. Ex6 — heavy hits light ::: light ball $\to 2 u_{1}$ , heavy barely slows. Ex7 — light hits heavy ::: pebble reverses at $\approx - u_{1}$ , wall $\approx 0$ . Ex8 — pool balls ::: near-equal masses almost swap. Ex9 — reverse twist ::: a valid state must obey BOTH laws or it's impossible.

Connections

Parent topic — the derivation of the two tools used on every example here.
Conservation of Momentum — the equation used in every step 1 or 2 above.
Kinetic Energy — the quantity we verify at the end of each example.
Coefficient of Restitution — Ex 9's "does it obey both laws?" is the $e = 1$ test.
Center of Mass Frame — a slick alternate route for Ex 6 and Ex 7 limits.
Newton's Cradle — Ex 2 and Ex 3 are exactly this.
Inelastic Collisions — 1D — contrast: there KE is not conserved, so Ex 9-style checks fail on purpose.