AQA A-level PE · Paper 2 · Biomechanical movement · Linear motion: impulse

F × t = mΔv

Every foot contact gives a sprinter a push back (braking) then a push forward (propulsion). The difference between the two areas decides whether they speed up, hold speed or slow down.

+ impulse0Ns · propulsive
− impulse0Ns · braking
Net impulse0Ns
Δv per step0m/s · 70 kg sprinter

Model: horizontal ground reaction force during one foot contact. Each half-sine pulse has an area of (2/π) × peak force × duration.

Slow motion ×0.4 Accelerating Speed 0.0 m/s
Race phase:
The exam graph, drawn live

Area = impulse.

Force is on the y-axis, time on the x-axis. Braking force is drawn below the axis (negative impulse) and propulsive force above it (positive impulse). Compare the two shaded areas.

Move the sliders and watch the shaded areas and the runner change together.

01 · AQA definitions

Learn these word for word

Momentum
The quantity of motion possessed by a moving body. Momentum = mass × velocity. kg m/s
Impulse
Force × time. Impulse is equal to the change in momentum of a body. It is used to add speed to a body or object, and to reduce the force of an impact. Ns
N2L
Newton's second law (law of acceleration) The rate of change of momentum of a body is proportional to the force applied and takes place in the direction in which the force acts. F = ma.
Net impulse
Positive impulse minus negative impulse. It decides whether the performer accelerates, keeps a constant velocity or decelerates.

Where the formula comes from. F = ma and a = Δv ÷ t, so F = mΔv ÷ t. Multiply both sides by t: net F × t = mΔv. The net impulse (net force × time) equals the change in momentum. A bigger net impulse means a bigger change in velocity.

02 · Reading the graph

Start, middle, end

On a force–time graph the area under the curve is the impulse. For a sprinter, the foot lands slightly ahead of the centre of mass and the ground pushes back (negative impulse, below the axis). The foot then pushes off behind the centre of mass and the ground pushes forward (positive impulse, above the axis).

Phase of 100 mGraph shapeNet impulseEffect on velocity
StartLarge area above the axis, small area belowNet positiveAcceleration
MiddleArea above equals area belowZeroConstant velocity (no acceleration)
EndArea below the axis larger than area aboveNet negativeDeceleration
03 · Challenge

Catch the egg

The egg arrives with momentum. To stop it, your hand must give it an impulse equal to that momentum. You can't change the impulse, but you can change the time. Give with the hand to increase the time, and the force falls: a longer time means a smaller force. Too stiff and it cracks. Your arm can only give so far, so higher drops need your best technique.

Round 1/5Score 0Drop height –Egg speed –Momentum –

Same physics in sport: a cricketer draws the hands back when catching, a gymnast bends the knees on landing, and crash mats and cushioned trainers increase the time of impact.

04 · Quick quiz

Name that graph

Each graph shows one foot contact. Is the sprinter at the start, middle or end of the race? Compare the areas, not the heights.

This is the:
Correct 0/0Streak 0
05 · Applied

Using impulse in sport

To increase outgoing velocity, increase impulse. Apply a larger force, or apply the force for longer.

  • Follow-through in a tennis serve, golf drive or hockey hit keeps the force on the ball for longer, so impulse and outgoing velocity increase.
  • Rotation across the circle in the shot put and discus increases the time the force is applied before release.
  • Sprint start: a powerful drive from the blocks gives a large positive impulse, so the sprinter accelerates quickly.

To reduce the force of an impact, increase the time. The change in momentum is fixed, so a longer time means a smaller force.

  • Landing: a gymnast or long jumper bends at the hips, knees and ankles to extend the landing time, reducing the force and the risk of injury.
  • Catching: "giving" with the hands draws the ball in over a longer time, reducing the force so it is less likely to bounce out.
  • Equipment: crash mats, sand pits and cushioned soles increase the time of impact.
06 · Sketching

Getting the graph marks

  • Label the axes: force (N) on the y-axis, time (s) on the x-axis.
  • Draw the negative impulse below the time axis first (foot lands, braking), then the positive impulse above it (push-off).
  • Make the relative areas obvious. Start: positive area clearly bigger. Middle: areas equal. End: negative area clearly bigger.
  • Label each area "positive impulse" or "negative impulse" and state the net impulse and its effect (acceleration, constant velocity, deceleration).
  • Don't confuse it with a velocity–time graph. A force–time graph shows one foot contact, not the whole race.
07 · Exam practice

Try, then check

Question 1 · Define3 marks

Define impulse and momentum, and state the unit of impulse.

Show mark scheme
  • Impulse: force × time / the change in momentum
  • Momentum: the quantity of motion possessed by a moving body / mass × velocity
  • Newton seconds (Ns)
Question 2 · Calculate3 marks

An 80 kg sprinter's foot contact is shown on a force–time graph. The negative impulse is a triangle with a peak force of 300 N lasting 0.04 s. The positive impulse is a triangle with a peak force of 600 N lasting 0.10 s. Calculate the net impulse and the sprinter's change in velocity. Show your working.

Show mark scheme
  • Negative impulse = ½ × 300 × 0.04 = 6 Ns; positive impulse = ½ × 600 × 0.10 = 30 Ns
  • Net impulse = 30 − 6 = +24 Ns
  • Change in velocity = net impulse ÷ mass = 24 ÷ 80 = 0.3 m/s (the sprinter accelerates)
Question 3 · Sketch3 marks

Sketch and label force–time graphs to show the impulse acting on a sprinter at the start, middle and end of a 100 m race.

Show mark scheme
  • Start: positive impulse (above the axis) larger than negative impulse, so net positive impulse and acceleration
  • Middle: positive and negative impulse equal, so net impulse is zero and velocity is constant
  • End: negative impulse (below the axis) larger than positive impulse, so net negative impulse and deceleration
  • (Axes labelled force and time; area under the curve = impulse; negative impulse drawn first for the foot landing)
Question 4 · Explain3 marks

Using your knowledge of impulse, explain how a gymnast can reduce the risk of injury when landing from a vault.

Show mark scheme
  • The change in momentum on landing is fixed, so the impulse needed to stop is fixed
  • Flexing at the hips, knees and ankles increases the time of the impact / the time the force acts
  • Impulse = force × time, so a longer time means a smaller (peak) force on the body, reducing the risk of injury
Question 5 · Apply3 marks

Explain how a shot putter uses impulse to increase the distance of the throw.

Show mark scheme
  • Impulse = force × time; a larger impulse gives a larger change in momentum
  • The glide or rotation across the circle and a long pushing action increase the time the force is applied to the shot
  • A powerful leg and arm drive increases the force applied
  • So the shot leaves the hand with a greater release velocity and travels further (max 3)
Question 6 · Explain5 marks

Using Newton's second law and the concept of impulse, explain how a sprinter's velocity changes during a 100 m race. Refer to force–time graphs in your answer.

Show mark scheme
  • Newton's second law: the rate of change of momentum is proportional to the force applied and takes place in the direction the force acts (F = ma)
  • Impulse = force × time = change in momentum, so net impulse determines the change in velocity
  • Each foot contact has a negative impulse (foot lands, braking) and a positive impulse (push-off)
  • Start: positive impulse greater than negative, net positive impulse, so the sprinter accelerates
  • Middle: positive impulse equals negative impulse, net impulse zero, so constant velocity
  • End: negative impulse greater than positive, net negative impulse, so the sprinter decelerates
  • Sketch: force on the y-axis, time on the x-axis; area under the curve = impulse, with positive impulse above the axis and negative below (max 5)