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How do Newton's laws, force, momentum and impulse describe linear motion in sport?

Newton's three laws of motion applied to sport, the quantities of linear motion (distance, displacement, speed, velocity, acceleration), and the calculation and use of force, momentum and impulse.

A focused answer to OCR A-Level PE on linear motion and Newton's laws: the three laws applied to sporting movements, the linear quantities (distance, displacement, speed, velocity, acceleration), and the calculation of force, momentum and impulse with the impulse-momentum relationship read from a force-time graph.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Newton's three laws describe linear motion: the first (inertia) says a body stays at rest or constant velocity unless acted on by a net force; the second (acceleration) says force equals mass times acceleration, $F = ma$ , so force is proportional to the rate of change of momentum; the third (action-reaction) says every force has an equal and opposite reaction. The linear quantities are distance and displacement (m), speed and velocity (m/s) and acceleration (m/s squared). Force is in newtons, momentum is $\text{mass} \times \text{velocity}$ (kg m/s), and impulse is $\text{force} \times \text{time}$ (Ns), which equals the change in momentum. On a force-time graph the area under the curve is the impulse; a net positive area means a positive change in momentum, so the performer accelerates.

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What this dot point is asking
Newton's three laws
The quantities of linear motion
Force and momentum
Impulse and the impulse-momentum relationship

What this dot point is asking

OCR wants you to state and apply Newton's three laws to sport, define the quantities of linear motion, and calculate force ( $F = ma$ ), momentum and impulse, using the impulse-momentum relationship and the area under a force-time graph.

Newton's three laws

The quantities of linear motion

For example, a sprinter whose velocity rises from $0$ to $8$ m/s in $2$ s has an acceleration of $a = \frac{8 - 0}{2} = 4$ m/s squared.

Force and momentum

A $70$ kg sprinter moving at $9$ m/s has a momentum of $70 \times 9 = 630$ kg m/s. A heavier or faster body has more momentum and is harder to stop, which matters in contact sports.

Impulse and the impulse-momentum relationship

For a sprinter, the net impulse over a foot contact is positive (the forward propulsion area exceeds the backward braking area), so there is a positive change in momentum and the runner accelerates; at top speed the areas are roughly equal, so the net impulse is near zero and velocity is maintained.

Calculating impulse and the change in velocity

Write the relationship

Use $\text{impulse} = F \times t = \Delta(mv)$ , the impulse-momentum relationship.

Substitute the impulse values

A $70$ kg sprinter applies an average net forward force of $1400$ N for $0.4$ s in driving from the blocks. The impulse is $1400 \times 0.4 = 560$ Ns.

Find the change in momentum

Because impulse equals the change in momentum, the change in momentum is $560$ kg m/s.

Find the change in velocity

The change in velocity is $\Delta v = \frac{\Delta(mv)}{m} = \frac{560}{70} = 8$ m/s. Starting from rest, the sprinter leaves the blocks at $8$ m/s. Always keep the units (Ns for impulse, kg m/s for momentum, m/s for velocity) and remember the area under a force-time graph gives the impulse.

Common traps

Naming the wrong law for $F = ma$: $F = ma$ is Newton's second law (acceleration). The first law is inertia; the third is action-reaction.
Confusing distance with displacement, or speed with velocity: Distance and speed are scalars (size only); displacement and velocity are vectors (size and direction). A 400 m runner finishing where they started has run a 400 m distance but zero displacement.
Forgetting impulse equals change in momentum: Impulse ( $F \times t$ ) is not just a force; it equals the change in momentum, which is why follow-through and longer ground contact matter. State the link.
Misreading the force-time graph: The impulse is the area under the curve, not the peak force. A net positive area means acceleration; equal positive and negative areas mean constant velocity.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20193 marksA 70 kg sprinter accelerates out of the blocks at 4.5 m/s squared. Calculate the horizontal force they produce, give the unit, and state which of Newton's laws this illustrates.

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A Component 01 Section C calculation. One mark for the value, one for the unit, one for the law.

Use Newton's second law $F = m \times a$ , so $F = 70 \times 4.5 = 315$ N. The unit is newtons (N). This illustrates Newton's second law (the law of acceleration): the force produced is proportional to the rate of change of momentum, so a greater force gives a greater acceleration for a given mass.

A common dropped mark is omitting the unit or naming the wrong law; $F = ma$ is the second law.

OCR 20226 marksFigure 3 shows a force-time graph for a sprinter's foot contact, with a small negative (braking) phase followed by a larger positive (propulsion) phase. Explain, using impulse, how the shape of the graph relates to the sprinter accelerating, and define impulse.

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A Component 01 Section C data-response question. Markers reward defining impulse and reading the graph correctly.

Award marks for: impulse is the product of force and the time it acts, $\text{impulse} = F \times t$ , measured in newton-seconds (Ns), and it equals the change in momentum. On the graph, the area under the curve is the impulse. The small negative phase is a braking impulse (backward) at footstrike that slows the sprinter slightly; the larger positive phase is a propulsive impulse (forward) that speeds them up. Because the positive area is greater than the negative area, the net impulse is positive (forward), so there is a positive change in momentum and the sprinter accelerates. As a sprinter reaches top speed the two areas become more equal, so the net impulse approaches zero and velocity is maintained.

A top answer links the net (positive) area to a positive change in momentum and therefore acceleration, not just describes the graph.

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Sources & how we know this

OCR A Level Physical Education (H555) specification — OCR (2016)