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

1. 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.

2. Momentum is a vector; its unit is kg m s⁻¹ (the same as N s).

3. In one dimension, choose a positive direction: momentum in the other direction is negative.

4. 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.

5. Collision method: positive direction → total momentum before → total momentum after → equate → interpret the sign.

6. Explosion from rest: total momentum is zero before, so the fragments have equal and opposite momenta after.

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 N = 1 kg m s⁻².

A slow, massive body can have more momentum than a fast, light one — it is the product that matters.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 Choose east as positive, so the car's velocity is −25 m s⁻¹. Van: 2400 ×15 = +36 000 kg m s⁻¹; car: 1200 ×(-25) = -30 000 kg m s⁻¹.

Step 2 Add the momenta, keeping their signs: 36 000 + (-30 000) = +6000 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.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 Take towards the wall as positive: u = +22 m s⁻¹, v = -16 m s⁻¹. So mu = +1.28 kg m s⁻¹ and mv = -0.928 kg m s⁻¹.

Step 2 Final minus initial: Δp = -0.928 - 1.28 = -2.20 kg m s⁻¹, i.e. 2.2 kg m s⁻¹ away from the wall

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

Conservation of momentum is not a separate law to memorise; it follows from Newton's laws in four steps.

1. 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 = Δp/Δt. For constant mass, Δp/Δt = mΔv/Δt = ma.

2. Newton's third law: when body A pushes body B, B pushes A with a force of equal size in the opposite direction.

3. The two forces act for exactly the same time Δt — the time the bodies are in contact.

4. So Δp_B = FΔt and Δp_A = -FΔt: 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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Total momentum before (direction of A positive): 32 000 ×2.5 + 18 000 ×0 = 80 000 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 ×(1.6 - 2.5) = -28 800 kg m s⁻¹. B: 18 000 ×(1.6 - 0) = +28 800 kg m s⁻¹. Equal size, opposite sign — exactly what Newton's third law predicts.

A reliable method

1. Sketch the bodies before and after, with an arrow for each velocity.

2. Write the positive direction ("taking right as positive") and give every velocity its sign.

3. Write total momentum before = total momentum after as an equation.

4. Solve for the unknown. A negative answer means the body moves in the negative direction — say so in words.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 East positive. Total momentum before: 240 ×3.0 + 300 ×(-2.0) = 720 - 600 = +120 kg m s⁻¹

Step 2 Total momentum after, with A moving west: 240 ×(-1.5) + 300v_B = -360 + 300v_B

Step 3 Equate and solve: -360 + 300v_B = 120, so v_B = 480 ÷300 = 1.6 m s⁻¹ east

Both cars have reversed direction. The positive answer tells you B now moves east.

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 m₁v₁ = -m₂v₂ — equal magnitudes, opposite directions.

The recoil speed of the heavier body is smaller in the ratio of the masses: v₁/v₂ = -m₂/m₁.

From a moving start: the total before is (m₁ + m₂)u, not zero.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 Total momentum before is zero. Take the bag's direction as positive: 0 = 4.0 ×3.5 + 95v

Step 2 Solve: v = -14 ÷95 = -0.147 m s⁻¹, so 0.147 m s⁻¹ in the opposite direction to the bag

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Total momentum before (original direction positive; working in kg km s⁻¹): 1500 ×7.60 = 11 400 kg km s⁻¹. After: 300 ×7.80 + 1200v = 11 400, so v = (11 400 - 2340)/1200 = 7.55 km s⁻¹ in the original direction

(b) In kg m s⁻¹: satellite 300 ×(7800 - 7600) = +6.0 ×10⁴; stage 1200 ×(7550 - 7600) = -6.0 ×10⁴. 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.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Step 1 Velocities from the light gates: u = 0.100 ÷0.125 = 0.800 m s⁻¹; v = 0.100 ÷0.205 = 0.488 m s⁻¹

(a) Before: 0.250 ×0.800 = 0.200 kg m s⁻¹. After: 0.400 ×0.488 = 0.195 kg m s⁻¹

(b) (0.200 - 0.195)/0.200 ×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.

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".

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.

1. Show that the units kg m s⁻¹ and N s are equivalent.

2. Calculate the momentum of a 0.16 kg cricket ball moving at 30 m s⁻¹.

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.

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.

5. Explain why a small rowing boat moves backwards when a person jumps forwards from it onto a jetty.

1. 1 N = 1 kg m s⁻², so 1 N s = 1 kg m s⁻² ×s = 1 kg m s⁻¹.

2. p = 0.16 ×30 = 4.8 kg m s⁻¹, in the direction of the ball's velocity.

3. 1.2 ×0.50 = (1.2 + 0.80)v, so v = 0.60 ÷2.0 = 0.30 m s⁻¹ in the original direction.

4. 0 = 70 ×1.5 + 50v, so v = -105 ÷50 = -2.1 m s⁻¹: 2.1 m s⁻¹ in the opposite direction.

5. 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.