Conservation of linear momentum — derivation from Newton's third law: Worked examples

The scenario matrix

Before working anything, let us list every distinct situation a momentum problem can be. Each later example is tagged with the cell it fills.

#	Case class	What is different about it	Example
A	Starts at rest, splits apart	Total $P = 0$ before → afterwards the two momenta are equal and opposite	Ex 1
B	Both moving, head-on, they bounce	Opposite signs before; both change	Ex 2
C	Both moving, same direction, catch-up	Same signs before; slower one sped up	Ex 3
D	They stick (perfectly inelastic)	One shared final velocity; KE drops	Ex 4
E1	Two dimensions — explosion	Conserve $x$ and $y$ separately, start from rest	Ex 5
E2	Two dimensions — scattering	Two moving-off pieces; solve both component equations	Ex 6
F	External force present (a wall)	Momentum is NOT conserved — the trap case	Ex 7
G	Limiting mass ( $m_{2} \to \infty$ or $m_{2} = 0$ )	Degenerate: ball off a wall / feather hit by truck	Ex 8
H	Exam twist — one unknown mass, everything else given	Rearrange for the hidden quantity	Ex 9

We hit A–H below. Notice the axis we choose ( $+ x$ = right) and the sign of every velocity — that discipline is what separates a right answer from a sign-flipped one.

Example 1 — Cell A: starts at rest, splits apart

Worked example Firecracker on ice

A $4 kg$ block sits still on frictionless ice. It bursts into two pieces: piece $P$ of $1 kg$ flies right at $6 m/s$ . Find piece $Q$ 's velocity ( $3 kg$ ).

Forecast: Before the burst nothing moves, so total momentum is zero. Something flew right, so — guess the direction of $Q$ before reading on. It must move left to keep the total at zero. Guess its speed too.

Step 1 — Choose axis and write "before". Let $+ x$ = right. $P_{i} = 0 (block at rest)$ Why this step? No external horizontal force (frictionless), so momentum is conserved. Starting at rest fixes the total at exactly zero, the cleanest possible constraint.

Step 2 — Write "after". $P_{f} = m_{P} v_{P} + m_{Q} v_{Q} = (1) (+ 6) + (3) v_{Q}$ Why this step? After the burst the only momentum is carried by the two flying pieces.

Step 3 — Set them equal and solve. $0 = 6 + 3 v_{Q} \Rightarrow v_{Q} = - 2 m/s$ Why this step? Conservation says $P_{f} = P_{i} = 0$ . The negative sign is the physics: $Q$ moves left at $2 m/s$ , exactly as forecast.

Verify: Total after $= (1) (6) + (3) (- 2) = 6 - 6 = 0$ ✓, matches "before". Units: $kg \cdot m/s$ throughout ✓.

This is the twin of the gun-recoil example — same skeleton as Rocket Equation, where the "piece flying out" is exhaust gas.

Example 2 — Cell B: head-on, both bounce

Worked example Two carts collide head-on

Cart 1 ( $2 kg$ ) moves right at $3 m/s$ ; cart 2 ( $1 kg$ ) moves left at $4 m/s$ . After an elastic collision cart 1 is measured moving left at $\frac{7}{3} m/s$ . Find cart 2's final velocity.

Forecast: They approach from opposite sides, so their momenta partly cancel. Cart 2 was heading left fast — after being struck by the heavier cart, guess: does it end up moving right or left?

Step 1 — Axis and signed "before". $+ x$ = right. $v_{1} = + 3, v_{2} = - 4$ Why this step? Head-on means opposite signs. Writing the leftward cart as $- 4$ is the entire point of a sign convention.

Step 2 — Total momentum before. $P_{i} = (2) (+ 3) + (1) (- 4) = 6 - 4 = + 2 kg\cdotm/s$ Why this step? Adding signed momenta lets the cancellation happen automatically; the net drift is gently rightward.

Step 3 — Conserve and solve for $v_{2}^{'}$ . Given $v_{1}^{'} = - \frac{7}{3}$ : $+ 2 = (2) (- \frac{7}{3}) + (1) v_{2}^{'} \Rightarrow v_{2}^{'} = 2 + \frac{14}{3} = \frac{20}{3} m/s$ Why this step? Same conservation equation, now the unknown is cart 2. Positive $\Rightarrow$ cart 2 ends up moving right at $\frac{20}{3} \approx 6.67 m/s$ — it got knocked back the way it came.

Verify (elastic ⇒ KE also conserved): Before: $\frac{1}{2} (2) (3^{2}) + \frac{1}{2} (1) (4^{2}) = 9 + 8 = 17 J$ . After: $\frac{1}{2} (2) (\frac{7}{3})^{2} + \frac{1}{2} (1) (\frac{20}{3})^{2} = \frac{49}{9} + \frac{200}{9} = \frac{249}{9} \approx 27.7 J$ . ⚠️ These do not match — so the given "elastic" numbers are inconsistent. Momentum is still satisfied, but energy flags the error. Lesson: momentum alone can't tell you a collision is elastic; check KE separately (see Elastic vs Inelastic Collisions).

Example 3 — Cell C: same direction, catch-up

Worked example Fast cart rear-ends a slow one

Cart 1 ( $2 kg$ ) at $+ 5 m/s$ catches cart 2 ( $3 kg$ ) at $+ 1 m/s$ (both moving right). After they bounce, cart 1 slows to $+ 2 m/s$ . Find $v_{2}^{'}$ .

Forecast: Both move right, so both momenta are positive — nothing cancels. Cart 1 lost speed, so it handed momentum forward; cart 2 must speed up. Guess by how much.

Step 1 — Signed before (same sign this time). $P_{i} = (2) (+ 5) + (3) (+ 1) = 10 + 3 = + 13 kg\cdotm/s$ Why this step? Same-direction motion means like signs add, contrasting Example 2 where they subtracted.

Step 2 — Conserve. $13 = (2) (+ 2) + (3) v_{2}^{'} \Rightarrow 3 v_{2}^{'} = 13 - 4 = 9 \Rightarrow v_{2}^{'} = + 3 m/s$ Why this step? One conservation equation, one unknown. Cart 2 rose from $1$ to $3 m/s$ — it sped up, as forecast, and still moves right.

Verify: After $= (2) (2) + (3) (3) = 4 + 9 = 13$ ✓. Sanity: cart 1 ( $2 \to$ still ahead? $2 < 3$ ) is now slower than cart 2, so they've separated — physically consistent, no interpenetration.

Example 4 — Cell D: they stick (perfectly inelastic)

Worked example Clay ball into a block

A $0.5 kg$ clay ball flies right at $8 m/s$ into a stationary $1.5 kg$ block; they stick. Find the common speed $v^{'}$ and the kinetic energy lost.

Forecast: Only $0.5$ of the $2 kg$ total was moving, so the combined lump should crawl — well under $8 m/s$ . Guess the fraction.

Step 1 — One shared final velocity. $m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$ Why this step? "Stick together" means one object afterward, so a single $v^{'}$ — that is what makes this case algebraically simplest.

Step 2 — Plug and solve. $(0.5) (8) + (1.5) (0) = (2.0) v^{'} \Rightarrow 4 = 2 v^{'} \Rightarrow v^{'} = 2 m/s$ Why this step? Momentum conserved (no external horizontal force during the brief impact). $2 m/s$ is $\frac{1}{4}$ of $8$ — the ball shares its motion with $4 \times$ its mass.

Step 3 — Energy audit. $K E_{i} = \frac{1}{2} (0.5) (8^{2}) = 16 J, K E_{f} = \frac{1}{2} (2.0) (2^{2}) = 4 J$ Lost $= 12 J$ to heat/sound/deformation. Why this step? Inelastic collisions conserve momentum but destroy KE — the parent's Example 2 point, quantified here.

Verify: $P$ : before $= 4$ , after $= (2) (2) = 4$ ✓. Fraction of KE kept $= 4/16 = 25% = \frac{m _{1}}{m _{1} + m _{2}}$ — the exact rule for one body hitting a stationary one. ✓

Decomposing a vector into components (build the tool for 2D)

Before the two-dimensional examples, we must earn the notation $P_{i x}$ and $P_{i y}$ .

Example 5 — Cell E1: two-dimensional explosion

Worked example L-shaped explosion

A $5 kg$ object at rest bursts into three: A ( $1 kg$ , $+ x$ at $9 m/s$ ), B ( $2 kg$ , $+ y$ at $3 m/s$ ), and C ( $2 kg$ , unknown $v_{C}$ ). Find $v_{C}$ .

Figure 1 (Cell E1). Momentum-space diagram. Orange arrow: A's momentum,

+ 9

along the horizontal (x-momentum) axis. Teal arrow: B's momentum,

+ 6

along the vertical (y-momentum) axis. Dashed ink arrow: their combined pull

(9, 6)

. Plum arrow: C's momentum

(- 9, - 6)

, pointing into the third quadrant to cancel that pull so the three arrows sum back to the origin (rest).

Forecast: Two pieces fly out along $+ x$ and $+ y$ ; C must fly into the third quadrant (both components negative) to cancel them. In the figure the orange and teal arrows point right and up, so the plum arrow must point down-left.

Step 1 — Momentum is a vector: split into axes. $P_{i x} = 0, P_{i y} = 0$ Why this step? $P$ conserves independently in $x$ and $y$ (just defined above). Starting at rest sets both totals to zero.

Step 2 — $x$ -equation. $0 = (1) (9) + (2) (0) + (2) v_{C x} \Rightarrow v_{C x} = - 4.5 m/s$ Why this step? Only A has $x$ -momentum; C must cancel it → negative (leftward), matching the figure.

Step 3 — $y$ -equation. $0 = (1) (0) + (2) (3) + (2) v_{C y} \Rightarrow v_{C y} = - 3 m/s$ Why this step? Only B carries $y$ -momentum; C cancels it → downward.

Step 4 — Assemble the vector and its speed. $v_{C} = (- 4.5, - 3) m/s, ∣ v_{C} ∣ = 4. 5^{2} + 3^{2} = 29.25 \approx 5.41 m/s$ Why this step? The magnitude uses the Pythagorean theorem on the two components — the arrow's length in the figure.

Verify: $x$ -total after $= 9 + 2 (- 4.5) = 0$ ✓; $y$ -total $= 6 + 2 (- 3) = 0$ ✓. Third-quadrant direction confirmed. See Centre of Mass Motion: the centre of mass never moved, exactly why all momenta sum to zero.

Example 6 — Cell E2: two-dimensional scattering

Worked example Puck strikes a stationary puck and both scatter

A $2 kg$ puck moves right ( $+ x$ ) at $5 m/s$ and glances off a stationary $2 kg$ puck. After the hit, the incoming puck moves at $3 m/s$ along a direction $+ 3 0^{\circ}$ above the $x$ -axis. Find the struck puck's velocity components.

Forecast: The system was moving purely along $+ x$ , with zero $y$ -momentum. The first puck now drifts upward, so the struck puck must drift downward to keep $y$ -momentum at zero. Along $x$ , the struck puck should carry whatever the first one gave up.

Step 1 — Signed "before" in components. Incoming puck 1, target puck 2 (at rest). $P_{i x} = (2) (5) + (2) (0) = 10, P_{i y} = 0$ Why this step? The whole initial momentum lies along $x$ ; the $y$ -total starts at exactly zero, our tightest constraint.

Step 2 — Resolve puck 1's final velocity into components. Speed $3$ at $3 0^{\circ}$ : $v_{1 x}^{'} = 3 cos 3 0^{\circ} = 3 \cdot \frac{3}{2} = \frac{3 3}{2} \approx 2.598, v_{1 y}^{'} = 3 sin 3 0^{\circ} = 3 \cdot \frac{1}{2} = 1.5$ Why this step? We can only add momenta axis-by-axis, so an arrow given as "speed and angle" must first be broken into its $x$ and $y$ shadows (the tool defined above).

Step 3 — $x$ -equation for puck 2. $10 = (2) (\frac{3 3}{2}) + (2) v_{2 x}^{'} \Rightarrow 10 = 33 + 2 v_{2 x}^{'} \Rightarrow v_{2 x}^{'} = 5 - \frac{3 3}{2} \approx 2.402 m/s$ Why this step? Conserve $x$ -momentum; the struck puck carries the remainder — still positive, so it moves right.

Step 4 — $y$ -equation for puck 2. $0 = (2) (1.5) + (2) v_{2 y}^{'} \Rightarrow v_{2 y}^{'} = - 1.5 m/s$ Why this step? Total $y$ -momentum must stay zero; puck 1 went up ( $+ 1.5$ each unit mass), so puck 2 must go down ( $- 1.5$ ), as forecast.

Verify: $x$ -total after $= 2 (2.598) + 2 (2.402) = 5.196 + 4.804 = 10$ ✓; $y$ -total after $= 2 (1.5) + 2 (- 1.5) = 0$ ✓. Both axes balance independently — that is the essence of 2D conservation.

Example 7 — Cell F: external force, NOT conserved (the trap)

Worked example Ball bounces off a wall

A $0.2 kg$ ball hits a wall moving right at $10 m/s$ and rebounds left at $8 m/s$ . Is the ball's momentum conserved? Find the impulse the wall delivered.

Forecast: The wall clearly shoves the ball — an external force on the ball-alone system. So the ball's momentum should change. Guess the sign of that change.

Step 1 — Ball momentum before and after. $p_{i} = (0.2) (+ 10) = + 2, p_{f} = (0.2) (- 8) = - 1.6 kg\cdotm/s$ Why this step? Direction reversed → sign flipped. This alone shows momentum did not stay constant.

Step 2 — Impulse = change in momentum. $J = p_{f} - p_{i} = - 1.6 - 2 = - 3.6 kg\cdotm/s$ Why this step? The Impulse–Momentum Theorem says $J = Δ p$ . The wall pushed leftward ( $-$ ), which is why the ball's momentum dropped by $3.6$ units.

Verify: Magnitude $3.6 kg\cdotm/s$ leftward on the ball; by Newton's Third Law the ball pushes the wall (and Earth) $+ 3.6$ rightward — so the ball + Earth system does conserve momentum. Momentum is only "lost" because we cut the system too small. ✓

Example 8 — Cell G: limiting masses (degenerate cases)

Worked example Two limits at once

(a) A $0.5 kg$ ball hits an immovable wall ( $m_{2} \to \infty$ ) elastically at $6 m/s$ . Find its rebound speed. (b) A $2000 kg$ truck at $20 m/s$ swallows a stationary $0.01 kg$ feather ( $m_{2} \to 0$ , they stick). Find the truck's new speed.

Forecast: (a) A wall is infinitely heavy — the ball should bounce back at the same speed. (b) A feather is basically massless — the truck shouldn't feel a thing.

Part (a) — infinite mass limit. $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1} m_{2} \to \infty \frac{- m _{2}}{m _{2}} v_{1} = - v_{1}$ Why this step? Dividing top and bottom by $m_{2}$ and letting it grow makes every finite term vanish, leaving $- v_{1}$ . So $v_{1}^{'} = - 6 m/s$ : same speed, reversed. Exactly the wall bounce.

Part (b) — zero mass limit (stick case). $v^{'} = \frac{m _{1} v _{1}}{m _{1} + m _{2}} = \frac{( 2000 ) ( 20 )}{2000 + 0.01} = \frac{40000}{2000.01} \approx 19.9999 m/s$ Why this step? As $m_{2} \to 0$ the denominator $\to m_{1}$ and $v^{'} \to v_{1}$ — the truck barely changes speed, as forecast.

Verify: (a) $∣ v_{1}^{'} ∣ = ∣ v_{1} ∣ = 6$ ✓, and KE $\frac{1}{2} (0.5) (6^{2}) = 9 J$ before and after ✓ (elastic). (b) $v^{'} < 20$ but by only $\sim 5 \times 1 0^{- 5} m/s$ ✓ — negligible, matching intuition.

Example 9 — Cell H: exam twist, hidden mass

Worked example Find the unknown mass

A ball of unknown mass $m$ moving right at $6 m/s$ strikes a stationary $4 kg$ ball. They stick and move off together at $2 m/s$ . Find $m$ .

Forecast: The pair ends at $2 m/s$ , which is $\frac{1}{3}$ of $6$ . So the moving mass must be roughly a third of the total — smaller than $4 kg$ . Guess.

Step 1 — Write conservation with $m$ as the unknown. $m (6) + (4) (0) = (m + 4) (2)$ Why this step? Same stick-together equation as Example 4, but now the final velocity is given and the mass is hidden — algebra runs backward.

Step 2 — Expand, collect the $m$ terms, and solve. $6 m = 2 m + 8 \Rightarrow 6 m - 2 m = 8 \Rightarrow 4 m = 8 \Rightarrow m = 2 kg$ Why this step? Bringing every term containing $m$ to one side isolates the unknown. Answer: $m = 2 kg$ — total mass $6 kg$ , so the moving ball is $\frac{2}{6} = \frac{1}{3}$ of it, exactly matching the $\frac{1}{3}$ speed ratio we forecast.

Verify: Before $= (2) (6) = 12$ ; after $= (2 + 4) (2) = (6) (2) = 12$ ✓. Units $kg\cdotm/s$ ✓.

Active recall

Recall In Cell A (splits from rest), what's the relation between the two pieces' momenta?

Equal in magnitude, opposite in direction — they sum to zero.

Recall Why do we conserve

$x$ and $y$ separately in 2D (Cells E1, E2)? Momentum is a vector; the $x$ and $y$ components are independent equations that don't mix, because perpendicular pushes don't affect each other.

Recall In Cell F, is the ball's momentum conserved? What restores conservation?

No — the wall is an external force. Including Earth in the system restores conservation.

Recall What is the same-speed-reversal result in the infinite-mass limit (Cell G)?

$v_{1}^{'} \to - v_{1}$ : the ball bounces back at the same speed.

Recall What single check distinguishes an elastic from an inelastic outcome once momentum balances?

The kinetic-energy check — momentum balances for both; only elastic conserves KE.

Connections

Conservation of Linear Momentum — Derivation from Newton's Third Law — the parent proof these examples exercise
Elastic vs Inelastic Collisions — Cells B, E2, G; the KE check and the derivation of $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1}$
Impulse–Momentum Theorem — Cell F, $J = Δ p$
Centre of Mass Motion — Cells A & E1, why $P = 0$ means the CM stays put
Newton's Third Law — Cell F, wall reaction restores conservation
Rocket Equation — Cell A's cousin with continuous mass ejection