Conservation of Momentum
On 21 December 2015, SpaceX landed the first orbital-class Falcon 9 rocket booster at Cape Canaveral. The 22,200 kg booster decelerated using nine Merlin engines, with the impulse $J = 630{,}000\text{ N·s}$ delivered over 25 s — a controlled momentum change that relied on conservation of momentum applied to the rocket-exhaust system. No momentum was created or destroyed; it was transferred from booster to exhaust gases.
Practise this lesson
Three worksheets — foundations to mastery.
An astronaut floating in space (no gravity, no air resistance) throws a 2 kg wrench forward at 4 m/s. The astronaut has a mass of 80 kg. What happens to the astronaut — and how fast do they move?
In which type of collision is kinetic energy conserved?
Know
- Law of conservation of momentum: $\Sigma p_{before} = \Sigma p_{after}$
- Elastic collision: both momentum and KE conserved
- Inelastic collision: momentum conserved, KE not
- Perfectly inelastic: objects stick together
Understand
- Why total momentum is conserved (Newton's Third Law)
- Why the astronaut moves backward when the wrench goes forward
- Why elastic collisions conserve KE but inelastic ones don't
Can Do
- Apply $\Sigma p_{before} = \Sigma p_{after}$ to solve collision/explosion problems
- Classify a collision as elastic or inelastic by calculating KE before and after
- Find the velocity after a perfectly inelastic collision
Core Content
Two ice skaters face each other and push off simultaneously. Neither was moving before. After the push, both glide away in opposite directions. No external force acted. Where did their momentum come from? It did not come from anywhere — the system had zero total momentum before, and it still has zero total momentum after: equal magnitudes, opposite directions, cancelling to zero.
This law follows from Newton's Third Law: action and reaction forces are equal and opposite, and act for the same time. The impulses are equal and opposite, so the changes in momentum cancel — total momentum is unchanged.
Astronaut answer: Before: total momentum = 0 (at rest). After: $0 = 2 \times 4 + 80 \times v_{astronaut}$. $v_{astronaut} = -8/80 = -0.1\text{ m/s}$ — astronaut moves at 0.1 m/s in the opposite direction to the wrench.
A 1200 kg car moving at 15 m/s (east) collides and sticks to a stationary 800 kg car. Find the final velocity of the combined mass.
Law of conservation of momentum (closed system): $\Sigma p_{before} = \Sigma p_{after}$, i.e. $m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$; this follows from Newton's Third Law; for a perfectly inelastic collision: $v_f = \Sigma p_{before}/(m_1+m_2)$; for an explosion from rest: $0 = m_1v_1' + m_2v_2'$.
Pause — copy the highlighted law and both special cases into your book before moving on.
True or false: In a perfectly inelastic collision, kinetic energy is conserved.
We just saw that momentum is conserved in all closed-system collisions. That raises a question: is kinetic energy also conserved in every collision, or only in some? This card answers it → all collisions conserve $\Sigma p$; only elastic collisions also conserve $\Sigma KE$; test by calculating both sides.
All collisions conserve momentum. Only elastic collisions also conserve kinetic energy.
| Type | Momentum conserved? | KE conserved? | Example |
|---|---|---|---|
| Elastic | Yes | Yes | Ideal billiard balls, alpha particles on nuclei |
| Inelastic | Yes | No (some lost to heat/sound/deformation) | Car crash where cars separate |
| Perfectly inelastic | Yes | No (maximum KE lost) | Cars that stick together; clay catching a ball |
To verify if a collision is elastic: calculate total KE before and after. If $\Sigma KE_{before} = \Sigma KE_{after}$, it is elastic. Even if only 1% of KE is lost to heat, it is inelastic.
All collisions conserve $\Sigma p$; elastic collisions additionally conserve $\Sigma KE$; inelastic and perfectly inelastic collisions lose kinetic energy as heat/sound/deformation; to classify: calculate $\Sigma KE_{before}$ and $\Sigma KE_{after}$ — if equal, the collision is elastic.
Pause — copy the highlighted classification rule into your book before moving on.
A 3 kg ball at 4 m/s collides with a 1 kg stationary ball. After the collision, the 3 kg ball moves at 2 m/s and the 1 kg ball moves at 6 m/s (all in the original direction). Is this collision elastic?
An explosion from rest conserves ______ but the total kinetic energy ______ because ______ provides the energy.
Activities
Two trolleys on a frictionless track. Trolley A (2 kg, 3 m/s east) collides with Trolley B (2 kg, at rest). After: A is stationary, B moves at 3 m/s east. (a) Show momentum is conserved. (b) Show KE is conserved. (c) Classify the collision.
UnderstandBand 3(3 marks) 1. Explain why momentum is conserved when two ice skaters push off each other from rest, even though each skater's momentum changes.
ApplyBand 4(3 marks) 2. A 1000 kg car moving at 12 m/s north collides and joins with a 2000 kg stationary truck. Find the final velocity of the combined vehicles.
AnalyseBand 5(4 marks) 3. Ball A (2 kg, 6 m/s east) collides with Ball B (2 kg, at rest). After the collision, Ball A moves at 1 m/s east and Ball B moves at 5 m/s east. (a) Check if momentum is conserved. (b) Calculate KE before and after. (c) Classify the collision and account for any KE discrepancy.
Show all answers
Short Answer — Model Answers
Q1 (3 marks): Before they push, both skaters are at rest, so total momentum = 0. By Newton's Third Law, the forces they exert on each other are equal and opposite. These forces act for the same time, producing equal and opposite impulses. The changes in momentum are equal and opposite, so they cancel. Total momentum after = $m_1v_1' + m_2v_2' = 0$ — momentum is conserved. Each skater's individual momentum changed, but the total of the system did not.
Q2 (3 marks): $\Sigma p_{before} = 1000 \times 12 + 0 = 12\,000\text{ kg m/s}$ north. $\Sigma p_{after} = (1000 + 2000)v_f = 3000v_f$. $v_f = 12\,000/3000 = 4\text{ m/s}$ north.
Q3 (4 marks): (a) $\Sigma p_{before} = 2 \times 6 + 0 = 12\text{ kg m/s}$. $\Sigma p_{after} = 2 \times 1 + 2 \times 5 = 2 + 10 = 12\text{ kg m/s}$. Momentum is conserved ✓. (b) $KE_{before} = \frac{1}{2}(2)(6^2) = 36\text{ J}$. $KE_{after} = \frac{1}{2}(2)(1^2) + \frac{1}{2}(2)(5^2) = 1 + 25 = 26\text{ J}$. (c) $KE_{after} < KE_{before}$ by 10 J — this is an inelastic collision. The 10 J was converted to heat, sound, or deformation during the collision.
The astronaut moves at 0.1 m/s backward — before: zero total momentum; after: $+2 \times 4 + 80 \times v = 0$, so $v = -0.1\text{ m/s}$. The SpaceX Falcon 9 landing on 21 December 2015 uses the same law in reverse: the 22,200 kg booster expelled exhaust gas downward; by conservation of momentum, the booster gained upward momentum equal and opposite to the exhaust's downward momentum, decelerating from free-fall to a controlled landing.