IGCSE Physics · Unit 1: Motion, Forces and Energy · Extended only

Momentum Extended only

This entire note is Extended-only content — there is no Core equivalent. Momentum measures how hard it is to stop a moving object, combining both how much mass it has and how fast it’s going. This note builds directly on your work with F = ma and resultant force from the Forces note, and ends with one of the most powerful ideas in physics: momentum is always conserved.

Why this matters A fast-moving tennis ball and a slow-moving truck can be compared using a single, simple quantity — momentum — and that quantity behaves in a beautifully predictable way whenever objects collide or push apart. This idea underpins everything from car safety design (crumple zones, airbags) to rocket propulsion, and it reappears throughout physics well beyond IGCSE.

1. What Is Momentum?

Momentum is defined as mass × velocity.

p = mv

Momentum is measured in kg m/s. Because velocity is a vector, momentum is also a vector — it has both a magnitude and a direction, and that direction matters enormously once objects start interacting with each other.

Worked example — calculating momentum A 60 kg cyclist travels at 8 m/s. Find her momentum.
p = mv = 60 × 8 = 480 kg m/s
Worked example — comparing momenta A 1200 kg car travels at 15 m/s. A 20000 kg truck travels at 2 m/s. Which has more momentum?
Car: p = 1200 × 15 = 18000 kg m/s
Truck: p = 20000 × 2 = 40000 kg m/s
The truck has more momentum, even though it is travelling much more slowly — because momentum depends on both mass and velocity together.
Try it — Calculating Momentum
  1. A 0.15 kg cricket ball travels at 30 m/s. Calculate its momentum.
  2. A 900 kg car has a momentum of 13500 kg m/s. Calculate its velocity.
  3. Two objects have the same momentum: a 2 kg object and a 500 kg object. Explain why the 2 kg object must be moving much faster than the 500 kg object.
Show answers
  1. p = mv = 0.15 × 30 = 4.5 kg m/s
  2. v = p/m = 13500/900 = 15 m/s
  3. Since momentum = mass × velocity and both objects have equal momentum, the object with the much smaller mass (2 kg) must have a much larger velocity to make up for it — momentum stays constant only if a decrease in mass is balanced by a proportional increase in velocity.

2. Momentum, Impulse and Force

Impulse is defined as force × the time for which the force acts, and it is equal to the change in momentum it produces.

impulse = FΔt = Δ(mv)

This is an extremely useful equation whenever a force acts for a known time to change an object’s velocity — for example, a bat hitting a ball, or a rocket engine firing for a set duration.

Worked example — impulse changing momentum A 0.5 kg ball is struck by a bat. Before being hit, it is travelling at 4 m/s; after being hit, it travels at 22 m/s in the same direction. The bat is in contact with the ball for 0.01 s. Find the average force exerted by the bat.
Change in momentum = m(v − u) = 0.5 × (22 − 4) = 0.5 × 18 = 9 kg m/s
Impulse = FΔt = Δ(mv), so F = Δ(mv) / Δt = 9 / 0.01 = 900 N
Force as the rate of change of momentum Resultant force can also be defined as the change in momentum per unit time:
F = Δp / Δt
This is really the same relationship as F = ma written in a different (more general) form — it explains, for instance, why a car crumple zone reduces the force on passengers: it increases the time (Δt) over which the momentum change happens, which reduces the force needed to produce that same change in momentum.
Worked example — F = Δp/Δt in safety design A 900 kg car travelling at 20 m/s crashes and comes to rest. Compare the force experienced if the collision takes (a) 0.1 s (rigid car, no crumple zone) and (b) 0.5 s (car with a crumple zone that extends the collision time).
Change in momentum = m(v − u) = 900 × (0 − 20) = −18000 kg m/s (momentum decreases by 18000 kg m/s)
(a) F = Δp / Δt = 18000 / 0.1 = 180000 N
(b) F = Δp / Δt = 18000 / 0.5 = 36000 N
Extending the collision time from 0.1 s to 0.5 s reduces the force on the car (and its occupants) by a factor of 5 — this is exactly why crumple zones make cars safer.
Try it — Impulse and Force
  1. A 0.2 kg hockey ball is hit, changing its velocity from 0 to 25 m/s. The stick is in contact with the ball for 0.008 s. Calculate the average force exerted by the stick.
  2. A resultant force of 60 N acts on an object for 0.5 s. Calculate the impulse (and hence the change in momentum) this produces.
  3. Explain, using F = Δp/Δt, why bending your knees when landing from a jump reduces the force on your legs compared to landing with straight, stiff legs.
Show answers
  1. Change in momentum = 0.2 × (25 − 0) = 5 kg m/s. F = Δp/Δt = 5/0.008 = 625 N
  2. Impulse = FΔt = 60 × 0.5 = 30 N s = 30 kg m/s change in momentum
  3. Bending your knees increases the time (Δt) taken for your momentum to reduce to zero on landing. Since F = Δp/Δt, and the change in momentum (Δp) is fixed by your mass and landing speed, increasing Δt reduces the force F needed to bring you to rest — reducing the force on your legs and joints.

3. The Principle of Conservation of Momentum

In any collision or interaction between objects, provided no external force acts on the system, total momentum before is equal to total momentum after. This is the principle of conservation of momentum, and it applies to collisions in one dimension (objects moving along a single straight line).

total momentum before = total momentum after

Sign convention: since momentum is a vector, you must choose a positive direction before you begin, and treat any velocity in the opposite direction as negative.

Worked example — two objects colliding and sticking together A 3 kg trolley moving at 4 m/s collides with a stationary 1 kg trolley, and they stick together after the collision. Find their combined velocity afterward.
Total momentum before = (3 × 4) + (1 × 0) = 12 + 0 = 12 kg m/s
Total momentum after = total momentum before (conservation of momentum)
Combined mass = 3 + 1 = 4 kg
12 = 4 × v → v = 3 m/s, in the same direction as the original moving trolley
Worked example — objects moving in opposite directions A 2 kg ball moving at 5 m/s to the right collides head-on with a 3 kg ball moving at 4 m/s to the left. After the collision, they stick together. Find their combined velocity.
Taking “right” as positive:
Total momentum before = (2 × 5) + (3 × −4) = 10 − 12 = −2 kg m/s
Combined mass = 2 + 3 = 5 kg
−2 = 5 × v → v = −0.4 m/s
Answer: 0.4 m/s to the left (the negative sign shows the combined objects move in the direction originally taken as negative)
Worked example — objects separating (like an explosion or a gun firing) A 50 kg cannon, initially at rest, fires a 2 kg cannonball at 60 m/s to the right. Find the recoil velocity of the cannon.
Total momentum before = 0 (everything is at rest)
Total momentum after = total momentum before = 0
(2 × 60) + (50 × v) = 0
120 + 50v = 0 → v = −2.4 m/s
Answer: the cannon recoils at 2.4 m/s to the left (opposite to the direction of the cannonball, so the total momentum remains zero)
Try it — Conservation of Momentum
  1. A 4 kg trolley moving at 6 m/s collides with a stationary 2 kg trolley and they stick together. Find their combined velocity after the collision.
  2. A 1500 kg car moving at 10 m/s to the right collides head-on with a 1000 kg car moving at 8 m/s to the left. They lock together on impact. Find their combined velocity, stating its direction.
  3. A 0.02 kg bullet is fired at 300 m/s from a 4 kg rifle that is initially at rest. Find the recoil velocity of the rifle.
Show answers
  1. Momentum before = (4×6)+(2×0) = 24 kg m/s. Combined mass = 6 kg. v = 24/6 = 4 m/s (same direction as the moving trolley).
  2. Taking right as positive: momentum before = (1500×10)+(1000×−8) = 15000 − 8000 = 7000 kg m/s. Combined mass = 2500 kg. v = 7000/2500 = 2.8 m/s to the right.
  3. Momentum before = 0. (0.02×300)+(4×v)=0 → 6 + 4v = 0 → v = −1.5 m/s, i.e. 1.5 m/s in the opposite direction to the bullet (recoil).
Common mistakes to watch for
  • Forgetting that momentum is a vector — objects moving in opposite directions must have opposite signs in your calculation, not just their speeds added together.
  • Not choosing a positive direction before starting a conservation-of-momentum problem, leading to sign errors.
  • Assuming “more momentum” always means “moving faster” — a slow, massive object can have more momentum than a fast, light one.
  • Mixing up impulse (FΔt, measured in N s, equal to a change in momentum) with momentum itself (mv, measured in kg m/s) — note these units are actually equivalent, but the two quantities describe different things (a total amount vs. a change).
  • Forgetting that “total momentum before = total momentum after” applies to the whole system, not to each object individually.
  • In recoil/explosion problems, forgetting that if the objects start at rest, total momentum is zero before AND after — so the two resulting momenta must be equal in size and opposite in direction.

4. Quick-Fire Challenge Round

True or False?
  1. Momentum is a scalar quantity.
  2. Impulse is equal to the change in momentum produced by a force acting over a period of time.
  3. Extending the time over which a collision happens reduces the force involved, for the same change in momentum.
  4. In a collision with no external forces, total momentum before must equal total momentum after.
Show answers
  1. False — momentum is a vector, since it depends on velocity, which is itself a vector. Direction matters.
  2. True — impulse = FΔt = Δ(mv), the change in momentum.
  3. True — since F = Δp/Δt, a larger Δt (for the same Δp) gives a smaller force.
  4. True — this is the principle of conservation of momentum.
Spot the Error
  1. “A 2 kg ball moving at 3 m/s right and a 2 kg ball moving at 3 m/s left have the same momentum, since both have a mass of 2 kg and a speed of 3 m/s.”
  2. “A 5000 kg truck moving at 2 m/s definitely has less momentum than a 1000 kg car moving at 20 m/s, since the car is moving much faster.”
  3. “In a collision between two trolleys where they stick together, we can find their final velocity using: total momentum before = (mass of trolley 1 × velocity of trolley 1) + (mass of trolley 2 × velocity of trolley 2), added as simple positive numbers, regardless of direction.”
Show answers
  1. Momentum is a vector, so direction matters. The two balls have equal magnitude of momentum but in opposite directions, meaning their momenta are actually opposite (one is +6 kg m/s, the other is −6 kg m/s), not “the same.”
  2. Truck’s momentum = 5000 × 2 = 10000 kg m/s. Car’s momentum = 1000 × 20 = 20000 kg m/s. In this case the car does have more momentum, but the reasoning given (“moving faster, so more momentum”) isn’t reliable in general — momentum depends on mass and velocity together, and it should always be calculated, not assumed from speed alone.
  3. If the trolleys move in opposite directions, one velocity must be taken as negative when calculating total momentum, since momentum is a vector — simply adding magnitudes ignores direction and will give the wrong answer whenever the objects aren’t moving the same way.

5. Mixed Practice — Bringing It All Together

  1. A 70 kg sprinter runs at 9 m/s. Calculate her momentum.
  2. A resultant force acts on a 0.3 kg ball for 0.05 s, changing its velocity from 2 m/s to 14 m/s. Calculate the average force.
  3. A 5 kg trolley moving at 3 m/s collides with a stationary 5 kg trolley and they stick together. Find their combined velocity.
  4. A 60 kg astronaut at rest in space pushes off a 500 kg satellite, causing the astronaut to move away at 2 m/s. Find the recoil velocity of the satellite.
  5. Explain why airbags reduce the risk of injury in a car crash, using the relationship between force, momentum change, and time.
Show answers
  1. p = mv = 70 × 9 = 630 kg m/s
  2. Δp = m(v−u) = 0.3 × (14−2) = 3.6 kg m/s. F = Δp/Δt = 3.6/0.05 = 72 N
  3. Momentum before = (5×3)+(5×0) = 15 kg m/s. Combined mass = 10 kg. v = 15/10 = 1.5 m/s
  4. Momentum before = 0. (60×2)+(500×v)=0 → 120+500v=0 → v = −0.24 m/s, i.e. 0.24 m/s in the opposite direction to the astronaut.
  5. An airbag increases the time (Δt) over which a passenger’s momentum changes to zero during a crash, compared to hitting the dashboard or steering wheel directly. Since F = Δp/Δt and the change in momentum (Δp) is fixed by the passenger’s mass and speed, increasing Δt reduces the force F experienced by the passenger, lowering the risk of injury.

Practice at Home

This whole topic is Extended content, so every part of this set applies to Extended candidates.

Part A — Momentum
  1. A 0.058 kg tennis ball travels at 40 m/s. Calculate its momentum.
  2. A 1500 kg car has a momentum of 27000 kg m/s. Calculate its velocity.
  3. Explain why momentum is described as a vector quantity rather than a scalar.
Part B — Impulse and force
  1. A golf club is in contact with a 0.045 kg golf ball for 0.0005 s, changing its velocity from 0 to 60 m/s. Calculate the average force exerted by the club.
  2. A resultant force of 200 N acts on an object for 0.25 s. Calculate the impulse this produces.
  3. Explain, using F = Δp/Δt, why a boxer “rolling with the punch” (moving their head backward as they are hit) reduces the force of the impact.
Part C — Conservation of momentum
  1. A 6 kg trolley moving at 5 m/s collides with a stationary 3 kg trolley and they stick together. Find their combined velocity.
  2. A 1200 kg car moving at 12 m/s to the right collides head-on with a 900 kg car moving at 10 m/s to the left, and they lock together. Find their combined velocity, stating its direction.
  3. A 0.015 kg bullet is fired at 400 m/s from a 3 kg rifle initially at rest. Calculate the recoil velocity of the rifle.

Key Vocabulary Recap

Momentum — mass × velocity (p = mv); a vector quantity, measured in kg m/s
Impulse — force × time (FΔt); equal to the change in momentum it produces
Conservation of momentum — total momentum before an interaction equals total momentum after, provided no external force acts
Recoil — the backward movement of an object (e.g. a gun or cannon) as a result of firing or ejecting another object forward
F = Δp/Δt — resultant force defined as the rate of change of momentum
Sign convention — the choice of a positive direction, needed because momentum and velocity are vectors

Syllabus Reference

This note covers Cambridge IGCSE Physics (0625) section 1.6, Momentum — entirely Extended (Supplement) content: defining and calculating momentum (p = mv), defining and calculating impulse (impulse = FΔt = Δ(mv)), applying the principle of conservation of momentum to simple one-dimensional problems, and defining resultant force as the rate of change of momentum (F = Δp/Δt). There is no Core equivalent for this section.

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