1.3 Momentum
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AS content in Unit 1 (Mechanics and Materials): specification points 13–14. Impulse calculations, collisions in two dimensions and the classification of collisions by kinetic energy are Unit 4 topics.
What this page covers
7 points
- Momentum \(p = mv\): a vector quantity, its unit and its sign in one dimension
- Change of momentum, including a body that rebounds
- The principle of conservation of linear momentum and the condition for it to apply
- How Newton's second and third laws lead to conservation of momentum
- Collisions in one dimension where bodies join together or move apart
- Explosions and recoil from rest, and separation of bodies that are already moving
- Testing conservation of momentum with gliders, light gates and a data logger
In one minute
6 points
- Momentum is mass × velocity, \(p = mv\), and the total momentum of a system stays constant if no resultant external force acts on it — one rule that solves every one-dimensional collision, explosion and recoil problem.
- Momentum is a vector; its unit is kg m s⁻¹ (the same as N s).
- In one dimension, choose a positive direction: momentum in the other direction is negative.
- Newton's third law: two bodies push each other with equal and opposite forces for the same time, so their momentum changes are equal and opposite.
- Collision method: positive direction → total momentum before → total momentum after → equate → interpret the sign.
- Explosion from rest: total momentum is zero before, so the fragments have equal and opposite momenta after.
Contents
13 sections
- What is momentum?
- How do you find a change in momentum?
- What is the principle of conservation of linear momentum?
- How do Newton's laws lead to conservation of momentum?
- How do you solve a collision problem?
- How do you handle explosions and recoil?
- How can you test conservation of momentum in the laboratory?
- Exam tips for 1.3
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is momentum?
A lorry and a bicycle moving at the same speed are not equally hard to stop, and neither are a bullet and a thrown pebble of the same mass. Momentum is the quantity that combines both ingredients — how much mass is moving and how fast — into one number with a direction.
- Vector: momentum has magnitude and direction. In one dimension, the direction is carried by a sign once you have chosen which way is positive.
- Unit: kg m s⁻¹. This is the same as N s, because \(1\ \text{N} = 1\ \text{kg m s}^{-2}\).
- A slow, massive body can have more momentum than a fast, light one — it is the product that matters.
A delivery van of mass 2400 kg travels east at 15 m s⁻¹. A car of mass 1200 kg travels west at 25 m s⁻¹ on the other side of the road. Calculate the momentum of each vehicle and the total momentum of the two vehicles.
Step 1 Choose east as positive, so the car's velocity is −25 m s⁻¹. Van: \(2400 \times 15 = +36\,000\ \text{kg m s}^{-1}\); car: \(1200 \times (-25) = -30\,000\ \text{kg m s}^{-1}\).
Step 2 Add the momenta, keeping their signs: \(36\,000 + (-30\,000) = +6000\ \text{kg m s}^{-1}\), i.e. 6000 kg m s⁻¹ east
Adding the magnitudes (66 000 kg m s⁻¹) ignores the directions and is wrong.
How do you find a change in momentum?
Change in momentum is final momentum minus initial momentum:
Because momentum is a vector, a body that reverses direction has a change in momentum larger than either momentum on its own: it must first lose all its momentum and then gain momentum the other way. Always substitute the velocities with their signs.
A tennis ball of mass 0.058 kg travels horizontally at 22 m s⁻¹ towards a wall and rebounds along the same line at 16 m s⁻¹. Calculate the change in momentum of the ball.
Step 1 Take towards the wall as positive: \(u = +22\ \text{m s}^{-1}\), \(v = -16\ \text{m s}^{-1}\). So \(mu = +1.28\ \text{kg m s}^{-1}\) and \(mv = -0.928\ \text{kg m s}^{-1}\).
Step 2 Final minus initial: \(\Delta p = -0.928 - 1.28 = -2.20\ \text{kg m s}^{-1}\), i.e. 2.2 kg m s⁻¹ away from the wall
Subtracting the speeds: \(0.058 \times (22 - 16) = 0.35\ \text{kg m s}^{-1}\). That treats the rebound as if the ball were still moving towards the wall. A reversal of direction means the two momenta have opposite signs, so their difference is large.
What is the principle of conservation of linear momentum?
- A system is the group of bodies you choose to consider, for example two colliding trolleys.
- Internal forces — forces the bodies exert on each other — cannot change the total momentum.
- External forces come from outside the system (friction from the ground, air resistance, a push from a hand). If they do not balance, the total momentum changes.
- In many collisions the external forces are small compared with the large internal forces during the brief contact, so momentum is conserved to a good approximation.
For two bodies in one dimension the principle becomes
When asked to state the principle, give both parts: total momentum is constant (or total before = total after) and the condition — no resultant external force. "Momentum is always conserved" on its own is incomplete.
How do Newton's laws lead to conservation of momentum?
Conservation of momentum is not a separate law to memorise; it follows from Newton's laws in four steps.
- Newton's second law can be written in terms of momentum: the resultant force on a body equals its rate of change of momentum, \(F = \Delta p/\Delta t\). For constant mass, \(\Delta p/\Delta t = m\Delta v/\Delta t = ma\).
- Newton's third law: when body A pushes body B, B pushes A with a force of equal size in the opposite direction.
- The two forces act for exactly the same time \(\Delta t\) — the time the bodies are in contact.
- So \(\Delta p_{\text{B}} = F\Delta t\) and \(\Delta p_{\text{A}} = -F\Delta t\): the changes in momentum are equal and opposite. Adding them gives zero — the total momentum does not change. Momentum is transferred from one body to the other, not created or destroyed.
A railway wagon A of mass 32 000 kg rolls at 2.5 m s⁻¹ along a level track and hits a stationary wagon B of mass 18 000 kg. The wagons couple together. Friction is negligible. (a) Calculate the velocity of the wagons just after coupling. (b) Show that the change in momentum of A is equal and opposite to the change in momentum of B.
(a) Total momentum before (direction of A positive): \(32\,000 \times 2.5 + 18\,000 \times 0 = 80\,000\ \text{kg m s}^{-1}\). After coupling the total mass is 50 000 kg: \(80\,000 = 50\,000\,v\), so \(v = \) 1.6 m s⁻¹ in the original direction
(b) A: \(32\,000 \times (1.6 - 2.5) = -28\,800\ \text{kg m s}^{-1}\). B: \(18\,000 \times (1.6 - 0) = +28\,800\ \text{kg m s}^{-1}\). Equal size, opposite sign — exactly what Newton's third law predicts.
Thinking the heavier wagon exerts the bigger force. The forces are always equal and opposite. The lighter body has the larger change in velocity because the same change in momentum is shared over a smaller mass.
How do you solve a collision problem?
A reliable method
- Sketch the bodies before and after, with an arrow for each velocity.
- Write the positive direction ("taking right as positive") and give every velocity its sign.
- Write total momentum before = total momentum after as an equation.
- Solve for the unknown. A negative answer means the body moves in the negative direction — say so in words.
| Type of event | Before | After | Equation |
|---|---|---|---|
| Bodies join | two velocities | one common velocity | \(m_1u_1 + m_2u_2 = (m_1 + m_2)v\) |
| Bodies move apart after a collision | two velocities | two velocities | \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\) |
| Explosion from rest | total momentum zero | two velocities, opposite signs | \(0 = m_1v_1 + m_2v_2\) |
At a fairground, bumper car A (total mass 240 kg) moves east at 3.0 m s⁻¹ and collides head-on with bumper car B (total mass 300 kg) moving west at 2.0 m s⁻¹. After the collision A moves west at 1.5 m s⁻¹. Determine the velocity of B after the collision.
Step 1 East positive. Total momentum before: \(240 \times 3.0 + 300 \times (-2.0) = 720 - 600 = +120\ \text{kg m s}^{-1}\)
Step 2 Total momentum after, with A moving west: \(240 \times (-1.5) + 300v_{\text{B}} = -360 + 300v_{\text{B}}\)
Step 3 Equate and solve: \(-360 + 300v_{\text{B}} = 120\), so \(v_{\text{B}} = 480 \div 300 = \) 1.6 m s⁻¹ east
Both cars have reversed direction. The positive answer tells you B now moves east.
Momentum is conserved in this collision, but kinetic energy is not: it falls from 1680 J to 654 J, the rest being transferred to thermal energy and sound. Do not try to use kinetic energy to solve a collision unless the question tells you it is conserved.
How do you handle explosions and recoil?
In an explosion or a push-off, internal forces drive the bodies apart. Energy stored in a spring, a chemical or the muscles becomes kinetic energy, but the momentum of the system is unchanged.
- From rest: total momentum is zero before, so after the event \(m_1v_1 = -m_2v_2\) — equal magnitudes, opposite directions.
- The recoil speed of the heavier body is smaller in the ratio of the masses: \(v_1/v_2 = -m_2/m_1\).
- From a moving start: the total before is \((m_1 + m_2)u\), not zero.
An astronaut of mass 95 kg (including her suit) floats at rest outside a space station. She throws a 4.0 kg tool bag directly away from her at 3.5 m s⁻¹. Calculate her velocity after the throw.
Step 1 Total momentum before is zero. Take the bag's direction as positive: \(0 = 4.0 \times 3.5 + 95v\)
Step 2 Solve: \(v = -14 \div 95 = -0.147\ \text{m s}^{-1}\), so 0.147 m s⁻¹ in the opposite direction to the bag
A rocket stage of mass 1200 kg and a satellite of mass 300 kg move together along a straight line at 7.60 km s⁻¹. Springs push them apart. Afterwards the satellite moves at 7.80 km s⁻¹ in the original direction. Treat the motion as one-dimensional and ignore external forces during the separation. (a) Calculate the velocity of the rocket stage after the separation. (b) Show that the changes in momentum of the two bodies are equal and opposite.
(a) Total momentum before (original direction positive; working in kg km s⁻¹): \(1500 \times 7.60 = 11\,400\ \text{kg km s}^{-1}\). After: \(300 \times 7.80 + 1200v = 11\,400\), so \(v = \dfrac{11\,400 - 2340}{1200} = \) 7.55 km s⁻¹ in the original direction
(b) In kg m s⁻¹: satellite \(300 \times (7800 - 7600) = +6.0 \times 10^4\); stage \(1200 \times (7550 - 7600) = -6.0 \times 10^4\). Equal and opposite.
The stage is not pushed backwards — it simply moves forwards a little more slowly. An explosion from a moving start need not reverse either body.
Setting the total momentum before to zero when the bodies were already moving. Zero is correct only for an explosion from rest.
How can you test conservation of momentum in the laboratory?
- Apparatus: two gliders on a level air track (or two trolleys on a runway tilted just enough to compensate for friction). Each carries a card of measured length; light gates connected to a data logger time the cards.
- Velocity from a light gate: \(v = \) card length ÷ time for the card to pass through the beam.
- Masses from a top-pan balance, including the card and any pads.
- Collisions to try: Velcro or a pin and plasticine makes the gliders stick; magnets or a spring makes them move apart; a spring-loaded plunger released between two gliders at rest gives an explosion.
- Checks: a single glider given a gentle push should pass both gates in the same time — this shows the track is level and friction is negligible.
- Repeat each collision several times and compare total momentum before and after.
Glider A (mass 0.250 kg) carries a card of length 0.100 m. It passes light gate 1 in 0.125 s, then collides with and sticks to glider B (mass 0.150 kg), which was at rest. The joined gliders pass light gate 2 in 0.205 s. (a) Calculate the total momentum before and after the collision. (b) Calculate the percentage difference and suggest a reason why the momentum after is smaller.
Step 1 Velocities from the light gates: \(u = 0.100 \div 0.125 = 0.800\ \text{m s}^{-1}\); \(v = 0.100 \div 0.205 = 0.488\ \text{m s}^{-1}\)
(a) Before: \(0.250 \times 0.800 = \) 0.200 kg m s⁻¹. After: \(0.400 \times 0.488 = \) 0.195 kg m s⁻¹
(b) \(\dfrac{0.200 - 0.195}{0.200} \times 100 = \) 2.4%. A small external force — air resistance on the cards, or a track that slopes slightly upwards — acts between the gates and removes a little momentum from the system. A difference this small is consistent with conservation within experimental uncertainty.
If asked how to improve the experiment, link each change to a source of error: a longer card reduces the percentage uncertainty in its length; placing the gates close to the collision point reduces the time over which friction can act; repeating and averaging reduces the effect of random errors in timing.
Exam tips for 1.3
Write "taking right as positive" (or east, or the original direction) before any momentum equation. It costs one line and makes every sign that follows easy to check.
Write total momentum before = total momentum after, with every term substituted, before you rearrange. A bare final answer gives nothing to follow if the arithmetic slips.
A velocity of −0.15 m s⁻¹ is not a finished answer. Translate it: "0.15 m s⁻¹ in the opposite direction to the bag".
For an "explain" question about why momentum is conserved, use three linked steps: the forces on the two bodies are equal and opposite (third law); they act for the same time; so the changes in momentum are equal and opposite and the total is unchanged.
Give momentum in kg m s⁻¹ (N s is equivalent). Keep unrounded intermediate values in your calculator, and give the final answer to the number of significant figures in the least precise data.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Each body keeps its own momentum in a collision. | The bodies push on each other, so each one's momentum changes. | Only the total momentum of the system is conserved. |
| Momentum can be added like a scalar. | Momentum has a direction; opposite momenta partly cancel. | Add momenta with signs (in one dimension). |
| The heavier body exerts the larger force in a collision. | Newton's third law makes the forces equal and opposite. | Equal forces, equal and opposite \(\Delta p\); the lighter body has the larger \(\Delta v\). |
| If momentum is conserved, kinetic energy must be too. | Kinetic energy is often transferred to thermal energy, sound or deformation. | Momentum is conserved whenever there is no resultant external force; kinetic energy usually is not. |
| An explosion creates momentum. | The fragments move in opposite directions. | From rest, the momenta are equal and opposite and the total stays zero. |
| A ball bouncing off a wall breaks the conservation law. | The wall and the Earth are part of the system and receive an equal and opposite change of momentum. | The Earth's mass is so large that its change in velocity is far too small to notice. |
How this topic connects
1.2 Forces and Newton's laws
Conservation of momentum is Newton's third law applied over the time of an interaction; \(F = ma\) is the constant-mass form of "force = rate of change of momentum".
1.5 Work and energy
Momentum and kinetic energy are different quantities: momentum is conserved in every collision with no resultant external force, while kinetic energy is usually partly transferred to other stores.
Further momentum and nuclear physics (Units 4 and 5)
Impulse, collisions in two dimensions and the difference between elastic and inelastic collisions build directly on the one-dimensional method on this page. A nucleus that emits an alpha particle recoils exactly like the astronaut in worked example 5, and conservation of momentum is used to analyse particle collisions and decays.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(p = mv\) | momentum of a body | Yes |
| \(\Delta p = mv - mu\) | change of momentum (signs!) | No — derive it from \(p = mv\) |
| \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\) | collision in one dimension, no resultant external force | No — learn the principle |
| \(m_1u_1 + m_2u_2 = (m_1 + m_2)v\) | bodies stick together | No — derive it |
| \(0 = m_1v_1 + m_2v_2\) | explosion or push-off from rest | No — derive it |
| \(F = \Delta p/\Delta t\) | link to Newton's second law (idea only on this page) | No — calculations with it come in Unit 4 |
| \(v = \) card length ÷ time | velocity from a light gate | No — learn it |
Check your understanding
1. Show that the units kg m s⁻¹ and N s are equivalent.
\(1\ \text{N} = 1\ \text{kg m s}^{-2}\), so \(1\ \text{N s} = 1\ \text{kg m s}^{-2} \times \text{s} = \) 1 kg m s⁻¹.
2. Calculate the momentum of a 0.16 kg cricket ball moving at 30 m s⁻¹.
\(p = 0.16 \times 30 = \) 4.8 kg m s⁻¹, in the direction of the ball's velocity.
3. A 1.2 kg trolley moving at 0.50 m s⁻¹ hits a stationary 0.80 kg trolley and they stick together. Find their common velocity.
\(1.2 \times 0.50 = (1.2 + 0.80)v\), so \(v = 0.60 \div 2.0 = \) 0.30 m s⁻¹ in the original direction.
4. Two ice skaters stand at rest facing each other. The 50 kg skater pushes the 70 kg skater, who moves off at 1.5 m s⁻¹. Find the velocity of the 50 kg skater.
\(0 = 70 \times 1.5 + 50v\), so \(v = -105 \div 50 = -2.1\ \text{m s}^{-1}\): 2.1 m s⁻¹ in the opposite direction.
5. Explain why a small rowing boat moves backwards when a person jumps forwards from it onto a jetty.
The person pushes the boat backwards and the boat pushes the person forwards with an equal and opposite force for the same time (Newton's third law). The total momentum was zero before, so the boat gains backward momentum equal in size to the person's forward momentum.
Specification checklist
- (13) I can calculate momentum with \(p = mv\), and I know it is a vector measured in kg m s⁻¹.
- (13) I can find a change in momentum, using signs when a body reverses direction.
- (14) I can state the principle of conservation of linear momentum, including the condition of no resultant external force.
- (14) I can explain, using Newton's second and third laws, why the momentum one body gains in an interaction equals the momentum the other loses.
- (14) I can solve one-dimensional problems for collisions in which bodies join or move apart, and for explosions and recoil, using signs for direction.
- (14) I can describe how light gates or a data logger can be used to test conservation of momentum, and comment on the results.