Topic 1.6 · Unit 1
Momentum
Momentum and impulse, how the resultant force on an object is linked to its change in momentum, and how conservation of momentum is used to solve collision and explosion problems in one dimension.
Key points
All of this topic is Supplement content.
- Momentum is mass × velocity: p = mv. Extended
- The unit of momentum is kg m/s. Extended
- Momentum is a vector. It has the same direction as the velocity. Extended
- In a problem, choose one direction as positive. A velocity or momentum in the opposite direction is then negative. Extended
- Impulse is force × the time for which the force acts: impulse = FΔt. Extended
- The unit of impulse is the newton second (N s). 1 N s = 1 kg m/s. Extended
- Impulse equals the change in momentum: FΔt = Δ(mv). Extended
- The resultant force on an object is its change in momentum per unit time: F = Δp / Δt. Extended
- This is the same idea as F = ma. If the mass stays constant, Δp / Δt = m × Δv / Δt = ma (see 1.5). Extended
- Why longer impact times are safer: to stop a moving person, the change in momentum is fixed. If the stopping time Δt is longer, then F = Δp / Δt is smaller. A smaller force causes less injury. Extended
- Examples that make the time longer: Extended
- Crumple zones at the front and back of a car squash slowly in a crash.
- Airbags and seat belts stop the passenger over a longer time.
- A helmet has soft padding that squashes on impact.
- You bend your knees when you land after a jump.
- Rubber fenders (or old tyres) on the side of a jetty. A ferry that touches the jetty stops more slowly, so the force on the ferry and the jetty is smaller.
- Conservation of momentum: when no resultant external force acts on a group of objects, their total momentum stays the same. Extended
- So in a collision or an explosion: total momentum before = total momentum after. Extended
- In a collision, the objects may stick together and move off with the same velocity. Or they may bounce apart with different velocities. Momentum is conserved in both cases. Extended
- Explosion from rest: an object at rest splits into two parts, and no resultant external force acts. The total momentum before is zero. Afterwards the two parts move in opposite directions. Their momenta are equal in size and opposite in direction, so the total is still zero. Extended
- A person jumping forwards from a small boat at rest works in the same way as an explosion. The person moves forwards and the boat moves backwards. Extended
- Extra detail: in most collisions, some kinetic energy is transferred to internal (thermal) energy and sound. The total kinetic energy goes down, but the total momentum stays the same.
Model
Momentum and collisions: look at the total momentum arrow before and after each collision. In the last stage, compare the forces when the same change in momentum takes a longer time.
Equations
MomentumExtended
p = mv
p = momentum (kg m/s); m = mass (kg); v = velocity (m/s)
ImpulseExtended
impulse = FΔt = Δ(mv)
F = force (N); Δt = time for which the force acts (s); Δ(mv) = change in momentum (kg m/s); impulse in N s
Resultant force and momentumExtended
F = Δp / Δt
F = resultant force (N); Δp = change in momentum (kg m/s); Δt = time taken for the change (s)
Conservation of momentumExtended
total momentum before = total momentum after
applies when no resultant external force acts on the objects; momentum in kg m/s, with a sign for direction
Momentum Extended
A motorbike and its riders have a total mass of 250 kg. They travel at 8.0 m/s. Find their momentum.
- Given: m = 250 kg, v = 8.0 m/s
- p = mv
- p = 250 × 8.0
- p = 2000 kg m/s in the direction of motion
Impulse Extended
A trolley of mass 5.0 kg is at rest. A force of 20 N pushes it for 1.5 s. Find the impulse and the speed of the trolley afterwards. Ignore friction.
- Given: F = 20 N, Δt = 1.5 s, m = 5.0 kg, starting velocity = 0
- impulse = FΔt = 20 × 1.5 = 30 N s
- impulse = Δ(mv), so the change in momentum = 30 kg m/s
- The trolley started at rest, so its final momentum = 30 kg m/s
- v = 30 ÷ 5.0
- v = 6.0 m/s
Resultant force and momentum Extended
A ball of mass 0.060 kg hits a wall at 15 m/s. It bounces straight back at 10 m/s. It is in contact with the wall for 0.050 s. Find the average force on the ball.
- Take the direction towards the wall as positive.
- Momentum before = 0.060 × (+15) = +0.90 kg m/s
- Momentum after = 0.060 × (−10) = −0.60 kg m/s
- Δp = after − before = −0.60 − 0.90 = −1.5 kg m/s
- F = Δp / Δt = −1.5 ÷ 0.050
- F = −30 N, so the force is 30 N away from the wall
- Note: the ball changes direction, so the change in momentum is 0.90 + 0.60, not 0.90 − 0.60.
A car of mass 1000 kg moving at 15 m/s stops in a crash. Compare the force if it stops in 0.10 s with the force if a crumple zone makes it stop in 0.50 s.
- Δp = 1000 × 15 = 15 000 kg m/s (in size)
- Stopping in 0.10 s: F = 15 000 ÷ 0.10 = 150 000 N
- Stopping in 0.50 s: F = 15 000 ÷ 0.50 = 30 000 N
- The same change in momentum over five times the time gives one fifth of the force.
Conservation of momentum Extended
Trolley A (mass 200 kg) moves at 5.0 m/s. It hits trolley B (mass 300 kg), which is at rest. They stick together. Find their velocity after the collision.
- Total momentum before = (200 × 5.0) + (300 × 0) = 1000 kg m/s
- After, they move together as one mass: 200 + 300 = 500 kg
- Total momentum after = 500 × v
- Conservation of momentum: 500 × v = 1000
- v = 2.0 m/s in the direction trolley A was moving
A person of mass 60 kg jumps forwards from a boat of mass 120 kg. The person’s velocity is 2.0 m/s relative to the shore. The boat and the person start at rest. Find the velocity of the boat relative to the shore. Ignore the force of the water on the boat.
- Total momentum before = 0
- Take forwards as positive. Person’s momentum after = 60 × 2.0 = +120 kg m/s
- Total momentum after = 0, so the boat’s momentum = −120 kg m/s
- Boat’s velocity = −120 ÷ 120
- v = −1.0 m/s, so the boat moves at 1.0 m/s backwards relative to the shore
Common mistakes
- Students write the unit of momentum as kg/m s or N. / The mark scheme wants kg m/s for momentum and N s for impulse. Extended
- Students ignore direction when an object bounces back. / The mark scheme wants a sign for each direction, so the change in momentum is larger when the object reverses. Extended
- Students write that an airbag reduces the force because it reduces the change in momentum. / The mark scheme wants: the change in momentum is the same; the airbag makes the time longer, so the force (Δp / Δt) is smaller. Extended
- Students use only one mass after two objects stick together. / The mark scheme wants the combined mass × the common velocity. Extended
- Students state conservation of momentum with no condition. / The mark scheme wants the condition: the total momentum is conserved when no resultant external force acts. Extended
- Students define momentum as “mass × speed”. / The mark scheme wants mass × velocity, because momentum has a direction. Extended
Exam tips
- Define momentum and impulse in words, then give the equation. Extended
- Calculate in collision problems: write “total momentum before = total momentum after”, then put in every object, with signs. Extended
- Draw a quick sketch with arrows and the positive direction marked before you start. Extended
- Explain safety features with three linked steps: same change in momentum → longer time → smaller force. Extended
- Give the direction of a velocity or force in your final answer when the question involves objects moving in opposite directions. Extended
- A typical 1-mark answer: “Momentum is mass multiplied by velocity.” Extended
- A typical 2-mark answer to “Define impulse”: “Force multiplied by the time for which the force acts (1). It equals the change in momentum (1).” Extended
- A typical 3-mark answer to “Explain how a crumple zone protects the driver”: “The car’s change in momentum in the crash is fixed (1). The crumple zone squashes, so the time to stop is longer (1). Force = change in momentum ÷ time, so the force on the car and the driver is smaller (1).” Extended