LearnovaAll Physics notes
AQA PhysicsForces

Work, springs and momentum

Use work done, spring energy and momentum equations.

Start here

The key idea

Work is done when a force moves an object through a distance.

Equation to know

work done = force x distance

Work Springs And Momentum
spring extensionwork done = force x distance

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

Work done and energy transfer

Work is done when a force moves an object through a distance in the direction of the force.Work done = force x distance moved along the line of action of the force (W = F x s), measured in joules (J).One joule of work is done when a force of one newton moves an object one metre.Work done against friction is mainly transferred to thermal energy stores, causing heating.

2

Springs and Hooke's law

Hooke's law states that the extension of a spring is directly proportional to the force applied, provided the limit of proportionality is not exceeded: force = spring constant x extension (F = k x e).The spring constant k (in N/m) measures stiffness.Beyond the limit of proportionality the line on a force-extension graph curves and the spring may be permanently deformed.5 x k x e2.

3

Momentum

Momentum = mass x velocity (p = m x v), measured in kg m/s, and is a vector.In a closed system the total momentum before an event equals the total momentum after it (conservation of momentum).This applies to collisions and explosions.Because momentum has direction, opposite directions are given opposite signs in calculations.

4

Force, momentum and safety (higher)

Force equals the rate of change of momentum: F = (m x v - m x u) / t = change in momentum / time.For a given change in momentum, increasing the time of the collision reduces the force.Safety features such as crumple zones, airbags, seatbelts and cushioned playground surfaces all increase the contact time, reducing the force on the people involved.

Key terms

Definitions to learn

Work done

Energy transferred when a force moves an object: work = force x distance.

Spring constant

The force needed per metre of extension, in N/m; a measure of stiffness.

Limit of proportionality

The point beyond which extension is no longer proportional to force.

Momentum

The product of mass and velocity, p = m x v, in kg m/s.

Conservation of momentum

Total momentum before an event equals total momentum after, in a closed system.

Worked example

A force of 75 N moves a box 4.0 m. Calculate the work done.

1

Use W = Fs.

2

Substitute the values.

Final answer

300 J

Exam habit

For safety-device questions, always link longer stopping time → smaller force for the same momentum change.Show W = Fs clearly with perpendicular distance. Use the spring constant equation F = ke with consistent units.

Watch out

Do not use distance travelled if it is not in the force direction.

Examiner tips

How to score full marks

  • 1Hooke's law uses extension (change in length), not the total length of the spring.
  • 2In momentum problems set one direction as positive and be careful with signs after a collision.
  • 3For safety questions, always link a longer time to a smaller rate of change of momentum, hence smaller force.
Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A spring has spring constant 250 N/m and extends by 0.040 m. Find the force.
Mark scheme
  1. 1.Use F = ke.
10 N
2A 0.20 kg ball travels at 15 m/s. Calculate its momentum.
Mark scheme
  1. 1.Use p = mv.
3.0 kg m/s
3Explain why seat belts reduce injury in a collision.
Mark scheme
  1. 1.Use momentum and stopping time.
A seat belt increases the stopping time, reducing the force needed for the same change in momentum.
4A force of 50 N moves a box 3 m in the direction of the force. Calculate the work done.[2 marks]
Mark scheme
  1. 1.Use W = F x s.
  2. 2.Substitute.
W = F x s = 50 × 3 (1) = 150 J (1).
5A spring extends by 0.04 m when a force of 6 N is applied. Calculate the spring constant.[2 marks]
Mark scheme
  1. 1.Use F = k x e, rearrange to k = F / e.
  2. 2.Substitute.
k = F / e = 6 / 0.04 (1) = 150 N/m (1).
6Calculate the momentum of a 0.5 kg ball moving at 12 m/s.[2 marks]
Mark scheme
  1. 1.Use p = m x v.
  2. 2.Substitute.
p = m x v = 0.5 × 12 (1) = 6 kg m/s (1).
7A spring has a spring constant of 200 N/m and is stretched by 0.1 m. Calculate the elastic potential energy stored.[3 marks]
Mark scheme
  1. 1.Use Ee = 0.5 x k x e2.
  2. 2.Substitute.
  3. 3.Evaluate.
Ee = 0.5 x k x e2 = 0.5 × 200 × 0.12 (1) = 0.5 × 200 × 0.01 (1) = 1 J (1).
8A 1000 kg car moving at 20 m/s collides with a stationary 1500 kg van and they move off together. Calculate their common velocity after the collision, and explain how a crumple zone reduces the force on the passengers.[4 marks]
Mark scheme
  1. 1.Momentum before = momentum after.
  2. 2.Total mass after = 2500 kg.
  3. 3.Solve for v.
  4. 4.Crumple zone increases time.
  5. 5.Reduced rate of change of momentum.
Momentum before = 1000 × 20 = 20000 kg m/s (1). After: 20000 = 2500 x v so v = 20000 / 2500 = 8 m/s (1). A crumple zone increases the time over which the momentum changes (1), and since F = change in momentum / time, a longer time gives a smaller force on the passengers (1).
9A force of 80 N is applied to a 20 kg trolley for 5.0 s from rest. Calculate the impulse, the change in momentum, and the final velocity of the trolley.[3 marks]
Mark scheme
  1. 1.Impulse = F x t.
  2. 2.Impulse = change in momentum.
  3. 3.Change in momentum = m x (v - u); u = 0 so v = change in momentum / m.
Impulse = F x t = 80 × 5.0 = 400 N s (1); change in momentum = 400 kg m/s (1); v = 400 / 20 = 20 m/s (1)
10A spring is stretched from its natural length of 0.25 m to a total length of 0.37 m by a force. The spring constant is 500 N/m. Calculate the force applied and the elastic potential energy stored.[3 marks]
Mark scheme
  1. 1.Extension = total length - natural length = 0.37 - 0.25 = 0.12 m.
  2. 2.Force F = k x e.
  3. 3.Ee = 0.5 x k x e2.
Extension = 0.37 - 0.25 = 0.12 m (1); F = 500 × 0.12 = 60 N (1); Ee = 0.5 × 500 × 0.122 = 0.5 × 500 × 0.0144 = 3.6 J (1)
11Two ice skaters, A (mass 60 kg) and B (mass 80 kg), are stationary and then push off each other. Skater A moves to the right at 4.0 m/s. Calculate the velocity of skater B, stating its direction.[3 marks]
Mark scheme
  1. 1.Total momentum before = 0 (both stationary).
  2. 2.Conservation of momentum: p_A + p_B = 0.
  3. 3.p_A = 60 × 4.0; p_B = -p_A.
  4. 4.v_B = p_B / m_B.
Total initial momentum = 0 (1); p_A after = 60 × 4.0 = 240 kg m/s; p_B = -240 kg m/s (1); v_B = 240 / 80 = 3.0 m/s to the left (1)
12A student stretches a spring and plots a force-extension graph. The graph is a straight line through the origin up to 8 N, then curves. Explain what the straight-line section tells us and what happens beyond the curved point. Calculate the spring constant from the straight section if 8 N gives an extension of 0.040 m.[3 marks]
Mark scheme
  1. 1.Straight line through origin means extension proportional to force — Hooke's law obeyed.
  2. 2.Gradient = k = F / e.
  3. 3.Beyond the curve: limit of proportionality exceeded, permanent deformation possible.
  4. 4.State Hooke's law is no longer obeyed.
The straight-line section shows extension is directly proportional to force — Hooke's law is obeyed in this region (1); k = F / e = 8 / 0.040 = 200 N/m (1); beyond the curved point, the limit of proportionality has been exceeded so extension is no longer proportional to force; the spring may be permanently deformed and will not return to its original length (1)
13A 0.080 kg cricket ball travelling at 35 m/s is caught by a fielder who brings it to rest in 0.20 s. Calculate the force exerted on the fielder's hands, and explain why a fielder moves their hands backwards when catching to reduce injury.[4 marks]
Mark scheme
  1. 1.Change in momentum = m x (v - u) = 0.080 × (0 - 35).
  2. 2.Force = change in momentum / time.
  3. 3.Moving hands back increases time of contact.
  4. 4.Longer time reduces force by F = delta p / t.
Change in momentum = 0.080 × (0 - 35) = -2.8 kg m/s; magnitude = 2.8 N s (1); F = 2.8 / 0.20 = 14 N (1); by moving hands backwards the fielder increases the time over which the ball decelerates (1); since F = change in momentum / time, a longer time produces a smaller force on the hands, reducing the risk of injury (1)
14A bullet of mass 0.010 kg is fired from a rifle of mass 3.0 kg. The bullet leaves at 600 m/s. (a) Calculate the recoil speed of the rifle. (b) Explain why the rifle recoils more slowly than the bullet even though the forces on each are equal and opposite. (c) Calculate the kinetic energy of both the bullet and the rifle immediately after firing, and comment on which carries more energy.[4 marks]
Mark scheme
  1. 1.(a) Conservation of momentum: total before = 0; m_bullet x v_bullet = m_rifle x v_rifle.
  2. 2.(b) Newton's 2nd law: same force, rifle has much larger mass so smaller acceleration.
  3. 3.(c) Ek_bullet = 0.5 × 0.010 × 6002; Ek_rifle = 0.5 × 3.0 x v_rifle2.
(a) 0 = 0.010 × 600 + 3.0 x v_rifle; v_rifle = -6 / 3.0 = 2.0 m/s (in opposite direction to bullet) (1); (b) the impulse (force x time) on both is equal and opposite, but the rifle has 300 times more mass than the bullet, so by a = F/m it accelerates 300 times less, giving a much lower recoil speed (1); (c) Ek_bullet = 0.5 × 0.010 × 6002 = 1800 J; Ek_rifle = 0.5 × 3.0 × 2.02 = 6.0 J (1); the bullet carries vastly more kinetic energy (1800 J vs 6 J) because kinetic energy depends on v2, and the bullet's much higher speed outweighs its lower mass (1)
Official exam-board sources
Specification Past papers
Browse all Physics topics