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Lesson plan · AS 9702 · 3.1 · Dynamics

Momentum and impulse, force over time

Linear momentum p = m v, the second law as the rate of change of momentum, and impulse as the area under a force-time graph. Built around the idea that a longer impact time means a smaller force, the physics behind every safety feature.

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At a glance

The shape of the lesson

Topic
Momentum and impulse (subtopic 3.1, momentum and Newton's laws)
Syllabus reference
Cambridge International AS & A Level Physics 9702, 3.1 (Topic 3: Dynamics)
Level
AS (first year)
Duration
60 minutes (single period)
Prior knowledge
Newton's laws and F = m a (the previous Dynamics lesson)
Central visual model
The same change in momentum over a longer time gives a smaller force
Simulation
Conservation of Momentum, momentum before and after a collision
Cooperative structure
Round Table (one sheet passes round; teacher example bank in the activity materials)
21st century skills
Collaboration, Communication
Assessment
An exit ticket, plus the group's Round Table sheet of named contributions
Learning objectives

By the end of the lesson, learners can

AS (all learners)
  • define linear momentum as p = m v and state its unit
  • state Newton's second law as the resultant force equals the rate of change of momentum, F = Δp / Δt
  • show that F = m a is a special case of this for constant mass
  • define the impulse of a force and relate it to the change in momentum, impulse = F Δt = Δp
  • recall that the area under a force-time graph is the impulse
  • explain how increasing the time of an impact reduces the force, and apply this to safety features

Key vocabulary

linear momentum, vector, rate of change of momentum, impulse, force-time graph, contact time, safety feature. Each term is introduced as it is first needed.

The core ideas

Momentum, force and impulse

Linear momentum is p = m v, a vector in the direction of the velocity, with unit kg m s−1, so a sign or direction is always needed. Newton's second law in its general form is F = Δp / Δt: the resultant force equals the rate of change of momentum. For constant mass this becomes F = m (v − u) / t = m a, the familiar special case.

A mass m moving with velocity v, with its momentum vector p equals m v in the same direction.
Momentum is a vector: p = m v
A ball rebounding from a wall with before and after velocity vectors in opposite directions, so the change in momentum is large.
In a rebound the velocity reverses, so Δp is large

The impulse of a force is F Δt and equals the change in momentum, with unit N s (the same as kg m s−1). On a force-time graph, the impulse is the area under the graph. For a fixed change in momentum, a longer contact time gives a smaller force, which is exactly why crumple zones, airbags and bending the knees reduce the force felt.

A force-time graph for an impact with the area under the graph shaded as the impulse, equal to the change in momentum.
Impulse = area under the force-time graph
Two force-time graphs with the same area; a short contact time gives a large force and a long contact time gives a small force.
Same Δp, longer time, smaller force
Lesson sequence

Sixty minutes, phase by phase

TimePhaseWhat happens in the roomResources
0 to 5 minStarterAsk why a high jumper lands on a thick crash mat rather than the floor; introduce impact time.Slide 1
5 to 20 minTeach: momentum and forceDefine momentum; state the second law in momentum form; show F = m a as a special case.Slides 2 to 6, fig-momentum, fig-force-momentum
20 to 28 minTeach: impulseDefine impulse; show it as the area under a force-time graph.Slides 7 to 9, fig-impulse-graph
28 to 50 minActivityRun Round Table: groups generate situations where impact time changes the force, each adding one example in turn.Round Table activity sheet
50 to 60 minPlenarySort the examples into reduce-the-force and increase-the-force; exit ticket.Slide 10, fig-airbag
Worked examples for the board

Momentum, a rebound, and a safety feature

Example 1: momentum

A car of mass 1500 kg travels at 20 m s−1. Find its momentum.

p = m v: p = 1500 × 20 = 3.0 × 104 kg m s−1, in the direction of travel

Example 2: force from a rebound

A 0.16 kg ball hits a wall at 8.0 m s−1 and rebounds at 6.0 m s−1. The contact lasts 0.020 s. Find the average force on the ball.

Sign convention (toward wall positive): u = +8.0, v = −6.0 m s−1
Change in momentum: Δp = m(v − u) = 0.16 × (−6.0 − 8.0) = −2.24 kg m s−1
Force: F = Δp / Δt = −2.24 / 0.020 = −112 N, that is 112 N directed away from the wall

Example 3: impulse and safety

A 70 kg passenger moving at 14 m s−1 is brought to rest. Compare the force when stopped in 0.10 s with the force when an airbag extends this to 0.70 s.

Change in momentum (both cases): Δp = 70 × 14 = 980 kg m s−1
Without the airbag: F = 980 / 0.10 = 9800 N
With the airbag: F = 980 / 0.70 = 1400 N (the same Δp over a longer time gives a much smaller force)
Running the cooperative task

Round Table on impact time

Each group is given one sheet and one pen, and the prompt: a situation where the time of a collision or push changes the force, with a one-line reason. In turn, each learner writes one example and a reason, then passes the sheet to the left. After several rounds, the group sorts the examples into those that lengthen the time to reduce the force and those that shorten it to increase the force. A full step-by-step facilitation guide, with the recording sheet and a teacher example bank, is provided as the activity in this bundle, so it can be run faithfully.

The Round Table flow: one recording sheet in the centre passes from learner to learner around the group.
One sheet passes round; each adds one example in turn

Why it suits this lesson. The idea learners most often miss is that the same change in momentum can give very different forces depending on the contact time. Round Table aims for breadth, surfacing many everyday examples, and each pass is one named contribution, which gives equal participation.

Examiner traps to pre-empt

What to head off, and how

Trap learners fall intoTeaching move that pre-empts it
Forgetting that momentum is a vector.In a rebound the velocity reverses, so the change in momentum is larger than it first appears; use a sign convention.
Using speeds without signs in a collision.Choose a positive direction first; a rebound speed takes the opposite sign.
Believing a longer impact time means a larger force.For the same change in momentum, a longer time means a smaller force, since F = Δp / Δt.
Confusing impulse with momentum itself.Impulse is a change in momentum (F Δt = Δp), not the momentum a body has.
Differentiation and assessment

Support, challenge and the checks

Assessment is formative. Exit ticket question 1: a 0.45 kg ball is kicked from rest to 24 m s−1 in 0.010 s, find the average force on the ball. Exit ticket question 2: explain, in terms of impulse, why bending the knees on landing reduces the force on the legs. The group's Round Table sheet shows that every learner has added an example and a reason.

Equipment and resources

Original work by the TheLucidSTEM team. Items are written in the style of the papers; no past paper question is reproduced. Supplied in editable formats so you can adapt them freely.
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