ForcesAQA GCSE Physics (8463)

Momentum

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples8 Key formulae3 Definitions5 Common mistakes8

Step-by-step worked examples

01.Momentum of a runner
Higher
Question
A 70 kg runner moves at 6 m/s. Find their momentum.
Solution
1. 1
  Formula.
  $p = mv$
2. 2
  Substitute.
  $p = 70 \times 6 = 420\,\text{kg·m/s}$
Answer
420 kg·m/s (in direction of motion).
02.Momentum is a vector
Higher
Question
A 0.5 kg ball moves at 4 m/s right. What is its momentum (a) in magnitude and (b) if we redefine 'right' as the negative direction?
Solution
1. 1
  Magnitude.
  $|p| = 0.5 \times 4 = 2\,\text{kg·m/s}$
2. 2
  Sign convention.
  $p = -2\,\text{kg·m/s} \text{ if right is negative}$
Answer
Magnitude 2 kg·m/s. Sign depends on chosen positive direction.
03.Two trolleys stick together
Higher
Question
A 2 kg trolley moves at 5 m/s and hits a stationary 3 kg trolley; the two stick together. Find the common velocity afterwards.
Solution
1. 1
  Apply conservation of momentum.
  $2 \times 5 + 3 \times 0 = (2 + 3) v$
2. 2
  Solve.
  $10 = 5v \Rightarrow v = 2 \text{ m/s}$
Answer
2 m/s in the original direction.
04.Rifle recoil
Higher
Question
A 4 kg rifle fires a 0.025 kg bullet at 320 m/s. Find the rifle's recoil speed.
Solution
1. 1
  Total p before = 0.
  $0 = 0.025 \times 320 + 4 \times v_r$
2. 2
  Solve.
  $v_r = -8/4 = -2 \text{ m/s}$
Answer
2 m/s in the opposite direction to the bullet (the minus sign indicates direction).
05.Spring-released trolleys
Higher
Question
Two trolleys are at rest with a compressed spring between them. Trolley X (1 kg) moves left at 3 m/s when released. Trolley Y has mass 0.5 kg. Find Y's velocity.
Solution
1. 1
  Total momentum before = 0.
  $0 = 1 \times (-3) + 0.5 \times v_Y$
2. 2
  Solve.
  $v_Y = 3 / 0.5 = +6 \text{ m/s (right)}$
Answer
Trolley Y moves at 6 m/s to the right (opposite direction).
06.Tennis ball rebound
Higher
Question
A 0.06 kg ball at +30 m/s rebounds at −25 m/s in 5 ms. Find the average force on it.
Solution
1. 1
  Find Δv.
  $\Delta v = -25 - 30 = -55 \text{ m/s}$
2. 2
  Find Δp.
  $\Delta p = 0.06 \times -55 = -3.3 \text{ kg·m/s}$
3. 3
  Apply F = Δp/Δt.
  $F = -3.3 / 0.005 = -660 \text{ N}$
Answer
660 N back toward the racket.
07.Why crumple zones help
Higher
Question
A 60 kg passenger in a car going at 15 m/s comes to rest in (a) 0.1 s without an airbag (b) 0.5 s with airbag and seatbelt. Find the force in each case.
Solution
1. 1
  Δp.
  $\Delta p = 60 \times 15 = 900 \text{ kg·m/s}$
2. 2
  Force without airbag.
  $F_a = 900 / 0.1 = 9000 \text{ N}$
3. 3
  Force with airbag.
  $F_b = 900 / 0.5 = 1800 \text{ N}$
Answer
(a) 9000 N (b) 1800 N — the airbag reduces the force fivefold by extending the time.
08.Helmet design
Higher
Question
A cyclist's head (mass 5 kg) hits the ground at 6 m/s. Without a helmet it stops in 2 ms; with a helmet it stops in 20 ms. Compare the forces.
Solution
1. 1
  Δp.
  $\Delta p = 5 \times 6 = 30 \text{ kg·m/s}$
2. 2
  Without helmet.
  $F = 30 / 0.002 = 15{,}000 \text{ N}$
3. 3
  With helmet.
  $F = 30 / 0.020 = 1500 \text{ N}$
Answer
Without helmet: 15 kN. With helmet: 1.5 kN — a 10× reduction in force.

Key formulae

Momentum
$p = m v$
- $p$
  — Momentum (kg·m/s)
- $m$
  — Mass (kg)
- $v$
  — Velocity (m/s)
When to use
Whenever momentum is needed. Recall — not on equation sheet.
Conservation of momentum (two bodies, 1D)
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
- $m$
  — Mass (kg)
- $u$
  — Initial velocity (m/s, with sign)
- $v$
  — Final velocity (m/s, with sign)
When to use
Any collision or explosion in a closed system. Recall — derivable from p = mv.
Newton's 2nd Law (momentum form)
$F = \frac{\Delta p}{\Delta t} = \frac{m \, \Delta v}{\Delta t}$
- $F$
  — Resultant force (N)
- $\Delta p$
  — Change in momentum (kg·m/s)
- $\Delta t$
  — Time taken (s)
- $m$
  — Mass (kg)
- $\Delta v$
  — Change in velocity (m/s)
When to use
Collisions, safety problems, anywhere you need to relate force to a change in motion over time. Recall — not on the equation sheet.

Practice with flashcards

Card 1 of 50 known · 0 to review

Key definitions and keywords

Momentum (p)Examiner keyword: The product of an object's mass and velocity. A vector quantity. p = mv (kg·m/s).
Conservation of momentumExaminer keyword: In a closed system, total momentum before an event equals total momentum after.
Closed systemExaminer keyword: A system in which no external (resultant) force acts. Required for momentum to be conserved.
Change in momentum (Δp)Examiner keyword: Final momentum minus initial momentum: Δp = mv − mu = m(v − u).
Newton's 2nd Law (momentum form)Examiner keyword: Force equals the rate of change of momentum: F = Δp/Δt.

Common mistakes and misconceptions

Mistake
Quoting momentum in kg·m/s² or N.
Why it happens
Confusing with force.
How to avoid it
Momentum is kg·m/s. Force is N = kg·m/s².
Mistake
Reporting momentum as a scalar number.
Why it happens
Just multiplying m × v.
How to avoid it
Always state direction (or sign) for momentum.
Mistake
Treating velocities as positive only.
Why it happens
Forgetting momentum is a vector.
How to avoid it
Use + and − consistently for the chosen positive direction.
Mistake
Calculating different final velocities for two stuck-together objects.
Why it happens
Treating as elastic collision.
How to avoid it
Stuck-together objects share the same final velocity. Use (m_A + m_B)v on the right.
Mistake
Claiming the lighter and heavier pieces in an explosion have equal speeds.
Why it happens
Confusing equal momenta with equal speeds.
How to avoid it
Equal magnitudes of MOMENTUM (mv), not speed. Heavier piece is slower.
Mistake
Calculating Δv as 'u − v' for a rebound (giving too small a value).
Why it happens
Forgetting sign change.
How to avoid it
Take outgoing velocity as negative if incoming is positive: Δv = v − u = (−25) − (+30) = −55 m/s.
Mistake
Saying 'the airbag absorbs the momentum / energy'.
Why it happens
Vague language.
How to avoid it
Say: 'the airbag increases the time over which the head decelerates, so by F = Δp/Δt the force is reduced'.
Mistake
Forgetting to convert ms to s before substituting.
Why it happens
Used to seeing ms quoted in collisions.
How to avoid it
Always convert: 5 ms = 0.005 s before dividing.

ForcesAQA GCSE Physics (8463)

Momentum

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

PreviousNext

Master the topic

Worked examples8 Key formulae3 Definitions5 Common mistakes8

Step-by-step worked examples

01.Momentum of a runner
Higher
Question
A 70 kg runner moves at 6 m/s. Find their momentum.
Solution
1. 1
  Formula.
  $p = mv$
2. 2
  Substitute.
  $p = 70 \times 6 = 420\,\text{kg·m/s}$
Answer
420 kg·m/s (in direction of motion).
02.Momentum is a vector
Higher
Question
A 0.5 kg ball moves at 4 m/s right. What is its momentum (a) in magnitude and (b) if we redefine 'right' as the negative direction?
Solution
1. 1
  Magnitude.
  $|p| = 0.5 \times 4 = 2\,\text{kg·m/s}$
2. 2
  Sign convention.
  $p = -2\,\text{kg·m/s} \text{ if right is negative}$
Answer
Magnitude 2 kg·m/s. Sign depends on chosen positive direction.
03.Two trolleys stick together
Higher
Question
A 2 kg trolley moves at 5 m/s and hits a stationary 3 kg trolley; the two stick together. Find the common velocity afterwards.
Solution
1. 1
  Apply conservation of momentum.
  $2 \times 5 + 3 \times 0 = (2 + 3) v$
2. 2
  Solve.
  $10 = 5v \Rightarrow v = 2 \text{ m/s}$
Answer
2 m/s in the original direction.
04.Rifle recoil
Higher
Question
A 4 kg rifle fires a 0.025 kg bullet at 320 m/s. Find the rifle's recoil speed.
Solution
1. 1
  Total p before = 0.
  $0 = 0.025 \times 320 + 4 \times v_r$
2. 2
  Solve.
  $v_r = -8/4 = -2 \text{ m/s}$
Answer
2 m/s in the opposite direction to the bullet (the minus sign indicates direction).
05.Spring-released trolleys
Higher
Question
Two trolleys are at rest with a compressed spring between them. Trolley X (1 kg) moves left at 3 m/s when released. Trolley Y has mass 0.5 kg. Find Y's velocity.
Solution
1. 1
  Total momentum before = 0.
  $0 = 1 \times (-3) + 0.5 \times v_Y$
2. 2
  Solve.
  $v_Y = 3 / 0.5 = +6 \text{ m/s (right)}$
Answer
Trolley Y moves at 6 m/s to the right (opposite direction).
06.Tennis ball rebound
Higher
Question
A 0.06 kg ball at +30 m/s rebounds at −25 m/s in 5 ms. Find the average force on it.
Solution
1. 1
  Find Δv.
  $\Delta v = -25 - 30 = -55 \text{ m/s}$
2. 2
  Find Δp.
  $\Delta p = 0.06 \times -55 = -3.3 \text{ kg·m/s}$
3. 3
  Apply F = Δp/Δt.
  $F = -3.3 / 0.005 = -660 \text{ N}$
Answer
660 N back toward the racket.
07.Why crumple zones help
Higher
Question
A 60 kg passenger in a car going at 15 m/s comes to rest in (a) 0.1 s without an airbag (b) 0.5 s with airbag and seatbelt. Find the force in each case.
Solution
1. 1
  Δp.
  $\Delta p = 60 \times 15 = 900 \text{ kg·m/s}$
2. 2
  Force without airbag.
  $F_a = 900 / 0.1 = 9000 \text{ N}$
3. 3
  Force with airbag.
  $F_b = 900 / 0.5 = 1800 \text{ N}$
Answer
(a) 9000 N (b) 1800 N — the airbag reduces the force fivefold by extending the time.
08.Helmet design
Higher
Question
A cyclist's head (mass 5 kg) hits the ground at 6 m/s. Without a helmet it stops in 2 ms; with a helmet it stops in 20 ms. Compare the forces.
Solution
1. 1
  Δp.
  $\Delta p = 5 \times 6 = 30 \text{ kg·m/s}$
2. 2
  Without helmet.
  $F = 30 / 0.002 = 15{,}000 \text{ N}$
3. 3
  With helmet.
  $F = 30 / 0.020 = 1500 \text{ N}$
Answer
Without helmet: 15 kN. With helmet: 1.5 kN — a 10× reduction in force.

Key formulae

Momentum
$p = m v$
- $p$
  — Momentum (kg·m/s)
- $m$
  — Mass (kg)
- $v$
  — Velocity (m/s)
When to use
Whenever momentum is needed. Recall — not on equation sheet.
Conservation of momentum (two bodies, 1D)
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
- $m$
  — Mass (kg)
- $u$
  — Initial velocity (m/s, with sign)
- $v$
  — Final velocity (m/s, with sign)
When to use
Any collision or explosion in a closed system. Recall — derivable from p = mv.
Newton's 2nd Law (momentum form)
$F = \frac{\Delta p}{\Delta t} = \frac{m \, \Delta v}{\Delta t}$
- $F$
  — Resultant force (N)
- $\Delta p$
  — Change in momentum (kg·m/s)
- $\Delta t$
  — Time taken (s)
- $m$
  — Mass (kg)
- $\Delta v$
  — Change in velocity (m/s)
When to use
Collisions, safety problems, anywhere you need to relate force to a change in motion over time. Recall — not on the equation sheet.

Practice with flashcards

Card 1 of 50 known · 0 to review

Key definitions and keywords

Momentum (p)Examiner keyword: The product of an object's mass and velocity. A vector quantity. p = mv (kg·m/s).
Conservation of momentumExaminer keyword: In a closed system, total momentum before an event equals total momentum after.
Closed systemExaminer keyword: A system in which no external (resultant) force acts. Required for momentum to be conserved.
Change in momentum (Δp)Examiner keyword: Final momentum minus initial momentum: Δp = mv − mu = m(v − u).
Newton's 2nd Law (momentum form)Examiner keyword: Force equals the rate of change of momentum: F = Δp/Δt.

Common mistakes and misconceptions

Mistake
Quoting momentum in kg·m/s² or N.
Why it happens
Confusing with force.
How to avoid it
Momentum is kg·m/s. Force is N = kg·m/s².
Mistake
Reporting momentum as a scalar number.
Why it happens
Just multiplying m × v.
How to avoid it
Always state direction (or sign) for momentum.
Mistake
Treating velocities as positive only.
Why it happens
Forgetting momentum is a vector.
How to avoid it
Use + and − consistently for the chosen positive direction.
Mistake
Calculating different final velocities for two stuck-together objects.
Why it happens
Treating as elastic collision.
How to avoid it
Stuck-together objects share the same final velocity. Use (m_A + m_B)v on the right.
Mistake
Claiming the lighter and heavier pieces in an explosion have equal speeds.
Why it happens
Confusing equal momenta with equal speeds.
How to avoid it
Equal magnitudes of MOMENTUM (mv), not speed. Heavier piece is slower.
Mistake
Calculating Δv as 'u − v' for a rebound (giving too small a value).
Why it happens
Forgetting sign change.
How to avoid it
Take outgoing velocity as negative if incoming is positive: Δv = v − u = (−25) − (+30) = −55 m/s.
Mistake
Saying 'the airbag absorbs the momentum / energy'.
Why it happens
Vague language.
How to avoid it
Say: 'the airbag increases the time over which the head decelerates, so by F = Δp/Δt the force is reduced'.
Mistake
Forgetting to convert ms to s before substituting.
Why it happens
Used to seeing ms quoted in collisions.
How to avoid it
Always convert: 5 ms = 0.005 s before dividing.

Momentum

Master the topic

Step-by-step worked examples

01.Momentum of a runner

02.Momentum is a vector

03.Two trolleys stick together

04.Rifle recoil

05.Spring-released trolleys

06.Tennis ball rebound

07.Why crumple zones help

08.Helmet design

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Momentum

Master the topic

Step-by-step worked examples

01.Momentum of a runner

02.Momentum is a vector

03.Two trolleys stick together

04.Rifle recoil

05.Spring-released trolleys

06.Tennis ball rebound

07.Why crumple zones help

08.Helmet design

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions