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

The one tool for every problem. For an isolated system (net external force zero) the total linear momentum $p = m v$ obeys $P_{before} = P_{after}, P = \sum_{i} m_{i} v_{i} .$ Because $P$ is a vector, in two dimensions we conserve the $x$ -part and the $y$ -part separately — each is its own equation. Keep signs: a velocity to the left is a negative number, not just "2 m/s".

Level 1 — Recognition

Goal: decide whether momentum is conserved, and read off the trivial answer.

Exercise 1.1

A skater of mass $50 kg$ stands still on frictionless ice holding a $2 kg$ ball. She throws the ball east at $6 m/s$ . Is total momentum conserved, and what is it before the throw?

Recall Solution 1.1

Is it conserved? The only outside forces are gravity (down) and the ice's normal force (up); horizontally there is no external force (frictionless). So the horizontal system is isolated → momentum conserved. Before: everything at rest, so $P_{before} = 0.$ That single fact ("starts at rest → total stays zero") is the whole trick — the skater must recoil west so the two momenta still add to zero.

Exercise 1.2

A $1200 kg$ car drives at a steady $20 m/s$ on a straight road with its engine on and rolling friction present. Is the car's momentum conserved during this second of driving?

Recall Solution 1.2

No. Steady speed does not mean isolated. The engine pushes the car forward through the road (external), friction and drag push back (external). They happen to balance, so $P$ stays constant — but not because the system is isolated. It's a coincidence of cancellation, and momentum is only guaranteed constant when the net external force is genuinely zero. Here it is zero by balance, so $P = (1200) (20) = 24000 kg\cdotm/s$ and stays there only as long as that balance holds.

Level 2 — Application

Goal: plug numbers into $P_{before} = P_{after}$ in one dimension.

Exercise 2.1

A $0.02 kg$ bullet is fired at $500 m/s$ from a $3 kg$ rifle initially at rest. Find the rifle's recoil velocity.

Recall Solution 2.1

Take east ( $+$ ) as the bullet's direction. Before: at rest, $P_{i} = 0$ . $m_{b} v_{b} + m_{r} v_{r} = 0$ $(0.02) (500) + (3) v_{r} = 0 \Rightarrow v_{r} = - \frac{10}{3} = - 3.33 m/s .$ The minus means westward — the recoil. Magnitude $3.33 m/s$ .

Exercise 2.2

A $3 kg$ trolley moving right at $4 m/s$ collides and sticks to a $2 kg$ trolley moving right at $1 m/s$ . Find their common velocity.

Recall Solution 2.2

They stick → one final velocity $v^{'}$ . Right is $+$ . $m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$ $(3) (4) + (2) (1) = (5) v^{'} \Rightarrow 14 = 5 v^{'} \Rightarrow v^{'} = 2.8 m/s (right) .$

Exercise 2.3

Same two trolleys, but now the $2 kg$ one moves left at $1 m/s$ before sticking. Find the common velocity.

Recall Solution 2.3

Only the sign of $v_{2}$ changes: leftward means $v_{2} = - 1 m/s$ . $(3) (4) + (2) (- 1) = (5) v^{'} \Rightarrow 12 - 2 = 5 v^{'} \Rightarrow v^{'} = 2.0 m/s (right) .$ Notice how one flipped sign swings the answer from $2.8$ to $2.0$ — signs are the physics.

Level 3 — Analysis

Goal: two dimensions, or reasoning about what conservation forbids.

Exercise 3.1

A $4 kg$ shell at rest explodes into three pieces. Piece A ( $1 kg$ ) flies east at $9 m/s$ ; piece B ( $1 kg$ ) flies north at $9 m/s$ . Find the velocity (magnitude and direction) of piece C ( $2 kg$ ).

Recall Solution 3.1

Before: at rest → $P_{i} = 0$ , so both components must sum to zero after. x-axis (east +): $(1) (9) + (1) (0) + (2) v_{C x} = 0 \Rightarrow v_{C x} = - 4.5 m/s$ . y-axis (north +): $(1) (0) + (1) (9) + (2) v_{C y} = 0 \Rightarrow v_{C y} = - 4.5 m/s$ . So C moves south-west. Magnitude: $∣ v_{C} ∣ = (- 4.5)^{2} + (- 4.5)^{2} = 40.5 = 6.36 m/s .$ Direction: both components negative → third quadrant (down-left), $4 5^{\circ}$ below the west axis, i.e. exactly toward the south-west — as the red arrow in the figure shows, balancing the yellow (A) and green (B) arrows.

Exercise 3.2

A $1000 kg$ car heading east at $20 m/s$ collides with a $1500 kg$ van heading north at $10 m/s$ ; they lock together. Find their common velocity just after impact.

Recall Solution 3.2

Locked → one velocity $(v_{x}, v_{y})$ , total mass $2500 kg$ . Conserve each axis. x (east +): $(1000) (20) = 2500 v_{x} \Rightarrow v_{x} = 8 m/s$ . y (north +): $(1500) (10) = 2500 v_{y} \Rightarrow v_{y} = 6 m/s$ . $∣ v ∣ = 8^{2} + 6^{2} = 100 = 10 m/s .$ Direction: $θ = arctan (\frac{v _{y}}{v _{x}}) = arctan (\frac{6}{8}) = 36. 9^{\circ}$ north of east. Both components positive → first quadrant, so no sign correction needed on the arctan.

Level 4 — Synthesis

Goal: combine momentum conservation with a second idea (energy, or a two-stage process).

Exercise 4.1 — Ballistic pendulum

A $0.05 kg$ bullet strikes and embeds in a $1.95 kg$ block hanging at rest. The block+bullet swing up to height $h = 0.20 m$ . Find the bullet's original speed. (Use $g = 10 m/s^{2}$ .)

Recall Solution 4.1

Two stages, two laws. Stage 1 (collision, momentum conserved — very fast, gravity does nothing yet): $m_{b} v_{b} = (m_{b} + m_{b l k}) V \Rightarrow (0.05) v_{b} = (2.0) V .$ Stage 2 (swing up, energy conserved — no more collision, only gravity): $\frac{1}{2} (2.0) V^{2} = (2.0) g h \Rightarrow V = 2 g h = 2 (10) (0.20) = 2 m/s .$ Back-substitute: $v_{b} = \frac{2.0 \times 2}{0.05} = 80 m/s .$ Why two different laws? During the embed, KE is lost to heat, so we may not use energy there — but momentum survives. After the embed, nothing is lost, so energy is the clean tool for the rise. Using the right law in each stage is the whole art.

Exercise 4.2 — Elastic 1D check

A $2 kg$ ball at $3 m/s$ hits a stationary $1 kg$ ball elastically (KE conserved). Find both final velocities.

Recall Solution 4.2

Two conserved quantities → two equations. Right is $+$ . Momentum: $(2) (3) = 2 v_{1}^{'} + 1 v_{2}^{'} \Rightarrow 6 = 2 v_{1}^{'} + v_{2}^{'}$ . KE: $\frac{1}{2} (2) (3^{2}) = \frac{1}{2} (2) v_{1}^{'2} + \frac{1}{2} (1) v_{2}^{'2} \Rightarrow 18 = 2 v_{1}^{'2} + v_{2}^{'2}$ . Use the elastic shortcut (relative speed reverses): $v_{2}^{'} - v_{1}^{'} = 3$ . Substitute $v_{2}^{'} = v_{1}^{'} + 3$ into momentum: $6 = 2 v_{1}^{'} + v_{1}^{'} + 3 \Rightarrow v_{1}^{'} = 1 m/s$ , and $v_{2}^{'} = 4 m/s$ . Check KE: $2 (1)^{2} + (4)^{2} = 2 + 16 = 18$ ✓. Momentum: $2 (1) + 4 = 6$ ✓.

Level 5 — Mastery

Goal: variable-mass / centre-of-mass reasoning, or a multi-step vector problem.

Exercise 5.1 — Rocket kick

A rocket of total mass $1000 kg$ (including $200 kg$ of fuel) drifts in deep space at $100 m/s$ . It ejects all $200 kg$ of fuel backward at $500 m/s$ relative to the rocket in one quick burst. Find the rocket's final speed. (See Rocket Equation for the continuous version.)

Recall Solution 5.1

Isolated system → momentum conserved. Let final rocket speed be $v_{r}$ (forward $+$ ), body mass after burn $= 800 kg$ . The fuel leaves at $500 m/s$ behind the rocket, so its ground velocity is $v_{fuel} = v_{r} - 500$ . $before (1000) (100) = rocket (800) v_{r} + fuel (200) (v_{r} - 500)$ $100000 = 800 v_{r} + 200 v_{r} - 100000$ $200000 = 1000 v_{r} \Rightarrow v_{r} = 200 m/s .$ The rocket speeds up from $100$ to $200 m/s$ — a $100 m/s$ gain from throwing mass backward.

Exercise 5.2 — Centre of mass never lies

On frictionless ice, a $60 kg$ person stands at the left end of a $40 kg$ , $4 m$ -long boat, all at rest. The person walks to the right end. How far does the boat move (relative to the ice), and in which direction? (Uses Centre of Mass Motion.)

Recall Solution 5.2

Key idea: the system starts at rest and is isolated, so $P = 0$ throughout → the centre of mass does not move relative to the ice. Let the boat slide left by distance $d$ . The person walks $4 m$ right relative to the boat, so relative to the ice the person moves $(4 - d)$ right. The centre of mass stays fixed: $m_{person} (4 - d) = m_{boat} d$ $(60) (4 - d) = (40) d \Rightarrow 240 - 60 d = 40 d \Rightarrow 240 = 100 d \Rightarrow d = 2.4 m .$ The boat slides $2.4 m$ left (opposite the walk), and the person actually moves $4 - 2.4 = 1.6 m$ right over the ice.

Quick self-check ladder

Rate yourself

L1 done = you know when momentum is conserved; L5 done = you can combine it with energy, frames, and centre of mass.

Which law during an inelastic crash?

Momentum only (KE is lost).

Which law for a frictionless swing/rise afterward?

Energy conservation.

2D collisions: how many equations?

One per axis — conserve

x

and

y

separately.

Connections

Newton's Third Law — why internal forces cancel in every problem here
Newton's Second Law (momentum form) — $F = d p / d t$ , the derivation's root
Elastic vs Inelastic Collisions — Exercises 2.2, 4.1 (inelastic) vs 4.2 (elastic)
Impulse–Momentum Theorem — the integral view of the recoil in 2.1
Centre of Mass Motion — the engine behind Exercise 5.2
Rocket Equation — continuous version of Exercise 5.1