Impulse and Force-Time Graphs

In the previous lesson, we used conservation of momentum to analyse collisions and explosions. But we often want to know more: how large were the forces involved? How long did the collision last? What is the relationship between force, time, and the change in momentum? These questions are answered by the concept of impulse.

Impulse

Impulse is defined as the product of force and the time for which it acts:

Impulse = F delta t

The unit of impulse is the newton-second (N s), which is equivalent to kg m/s (the same unit as momentum).

More importantly, impulse equals the change in momentum:

Impulse = F delta t = delta(mv) = mv - mu

This is actually Newton’s second law in its most general form. Newton originally stated his second law as “the rate of change of momentum is proportional to the applied force,” which gives:

F = delta(mv) / delta t

Rearranging: F delta t = delta(mv), which is the impulse-momentum theorem.

Key Relationships

Quantity	Formula	Units
Impulse	F delta t	N s
Change in momentum	mv - mu	kg m/s
Impulse = change in momentum	F delta t = mv - mu	N s = kg m/s
Average force	F = delta p / delta t	N

Worked Example 1

A 0.40 kg ball hits a wall at 10 m/s and bounces back at 8.0 m/s. What is the impulse on the ball?

Taking the initial direction as positive: u = +10 m/s, v = -8.0 m/s (reversed direction)

Impulse = mv - mu = (0.40)(-8.0) - (0.40)(10) = -3.2 - 4.0 = -7.2 N s

The magnitude is 7.2 N s, directed away from the wall. Note that the ball reverses direction, so we must account for the sign change — a common source of error.

Worked Example 2

A 60 kg person jumps from a wall and lands on the ground, coming to rest in 0.30 s. If they were falling at 5.0 m/s just before impact, what is the average force on their legs?

Impulse = change in momentum = mv - mu = (60)(0) - (60)(-5.0) = 0 + 300 = 300 N s (upward)

Average force = impulse / time = 300 / 0.30 = 1000 N

This is in addition to the person’s weight — the total force on the ground is greater.

Worked Example 3: Catching vs Deflecting

A 0.16 kg cricket ball arrives at 30 m/s.

(a) A fielder catches it (brings it to rest in 0.10 s): Impulse = mv - mu = 0 - (0.16)(30) = -4.8 N s Force = 4.8/0.10 = 48 N

(b) A batsman deflects it back at 30 m/s (contact time 0.005 s): Impulse = (0.16)(-30) - (0.16)(30) = -4.8 - 4.8 = -9.6 N s Force = 9.6/0.005 = 1920 N

The deflection requires double the impulse (the ball reverses direction) and the contact time is much shorter, resulting in a force 40 times larger.

Force-Time Graphs

A force-time graph plots force on the vertical axis against time on the horizontal axis.

The area under a force-time graph equals the impulse (and therefore the change in momentum).

This is directly analogous to how the area under a velocity-time graph equals displacement.

For a constant force, the area is a rectangle: impulse = F x delta t.

For a varying force, the area must be calculated by counting squares, using the trapezium rule, or integrating if the function is known.

Typical Force-Time Graph for a Collision

During a real collision (e.g. a ball hitting a bat), the force rises rapidly from zero to a peak and then falls back to zero. The shape is roughly a tall, narrow peak (like a triangle or bell curve). The area under this peak equals the impulse.

Summary: Graph Areas in Mechanics

Graph Type	Area Under Graph Represents
Velocity-time	Displacement
Force-time	Impulse (change in momentum)
Force-displacement	Work done
Power-time	Energy transferred

Reducing Force by Increasing Collision Time

The impulse-momentum theorem, F delta t = delta(mv), shows that for a given change in momentum, increasing the collision time reduces the average force:

F = delta(mv) / delta t

If delta(mv) is fixed (the object undergoes the same change in momentum), then a longer delta t means a smaller F.

This principle is the basis of many safety features:

flowchart TD
    A["Same momentum change Δp"]
    A --> B["Short collision time"]
    A --> C["Long collision time"]
    B --> D["LARGE force\nF = Δp / small Δt"]
    C --> E["SMALL force\nF = Δp / large Δt"]
    D --> F["More injury/damage"]
    E --> G["Less injury/damage"]

Safety Features Explained

Safety Feature	How It Works	Typical Time Increase
Crumple zones	Car body crushes progressively, extending deceleration time	0.05 s to 0.3 s
Airbags	Inflatable cushion extends head deceleration time	0.005 s to 0.05 s (factor of 10)
Seat belts	Stretch slightly under load, spreading force over time and area	0.01 s to 0.1 s
Helmets	Foam liner crushes, extending impact time on skull	0.001 s to 0.01 s
Landing mats (gymnastics)	Compress on impact, increasing deceleration time	0.01 s to 0.2 s
Bending knees on landing	Muscles absorb energy over longer time	0.01 s to 0.3 s

Worked Example 4: Crumple Zone Comparison

A 1200 kg car travelling at 15 m/s crashes and stops. Compare the average force when stopping in 0.10 s (rigid car) versus 0.50 s (with crumple zone).

Change in momentum = mv - mu = 0 - (1200)(15) = -18 000 kg m/s. Magnitude: 18 000 N s.

Rigid car: F = 18 000 / 0.10 = 180 000 N With crumple zone: F = 18 000 / 0.50 = 36 000 N

The crumple zone reduces the force by a factor of 5.

Worked Example 5: Airbag

A person’s head (mass 5.0 kg) decelerates from 15 m/s to rest. Without airbag, the collision time is 0.003 s. With airbag, it is 0.060 s.

Change in momentum = 5.0 x 15 = 75 kg m/s

Without airbag: F = 75/0.003 = 25 000 N (extremely dangerous) With airbag: F = 75/0.060 = 1250 N (survivable)

Force reduction factor = 25000/1250 = 20

Worked Example 6: Reading a Force-Time Graph

A ball experiences a force described by a triangular force-time graph: the force rises linearly from 0 to a peak of 600 N over 0.010 s, then falls linearly back to 0 over another 0.010 s.

The area under the triangle = (1/2) x base x height = (1/2)(0.020)(600) = 6.0 N s

If the ball has a mass of 0.15 kg and was initially at rest, its velocity after the impact is:

Impulse = delta(mv) 6.0 = (0.15)(v) - 0 v = 6.0 / 0.15 = 40 m/s

Impulse in Different Directions

When calculating impulse for objects that bounce (change direction), remember that velocity is a vector. If an object approaches at +v and bounces back at -v, the change in momentum is:

delta p = m(-v) - m(+v) = -2mv

The magnitude of the impulse is 2mv, not zero. This is double the impulse for an object that simply stops (delta p = -mv).

Comparison of Impulse for Different Outcomes

Scenario	Impulse Magnitude
Object stops (v = 0)	mu
Object bounces back at same speed	2mu
Object bounces back at lower speed v	m(u + v)
Object speeds up in same direction to v	m(v - u)

Worked Example 7

A 0.060 kg tennis ball hits a racket at 20 m/s and is returned at 25 m/s. What is the impulse on the ball? If the contact time is 0.004 s, what is the average force?

Taking the initial direction as positive: Impulse = m(v) - m(u) = (0.060)(-25) - (0.060)(20) = -1.5 - 1.2 = -2.7 N s

Magnitude: 2.7 N s.

Average force = 2.7 / 0.004 = 675 N

Common Mistakes and Misconceptions

Forgetting the sign change when objects bounce: If a ball bounces, v and u have opposite signs. The change in momentum is m(v - u), which gives a larger impulse than if the ball simply stopped. Many students write impulse = mv instead of m(v - u).
Confusing impulse with force: Impulse is force x time, not just force. Saying "the impulse is 50 N" is dimensionally incorrect.
Thinking the area under a force-time graph is force: The area is impulse (N s), not force (N). The peak of the graph gives the maximum force.
Ignoring the weight when calculating total force: When a person lands, the impulse calculation gives the net force. The actual force on their feet includes this net force plus their weight.
Using average velocity instead of change in velocity: Impulse = m x (change in velocity), not m x (average velocity).

Exam Tips for Edexcel Physics

When reading a force-time graph, identify the shape (rectangle, triangle, trapezium) to calculate the area correctly.
For "explain how a safety feature reduces force," always state: (1) the momentum change is the same, (2) the time is increased, (3) therefore the force is reduced. Use F = delta p / delta t explicitly.
In calculations involving bouncing objects, draw a diagram showing the direction of velocity before and after, and assign signs consistently.
Edexcel questions often ask you to explain the physics of a safety feature using the concept of impulse. Structure your answer clearly with the impulse-momentum theorem.
If a graph is non-standard (curved, stepped), you may need to estimate the area by counting squares. Each square has an area of (force value) x (time value) in N s.
State the direction of the impulse as well as its magnitude.

Key Points

Impulse = F delta t = change in momentum (delta p = mv - mu). Units: N s or kg m/s.
The area under a force-time graph equals the impulse.
For a given momentum change, increasing the collision time reduces the force: F = delta p / delta t.
Safety features (crumple zones, airbags, helmets, seat belts) all work by increasing collision time to reduce impact force.
When objects bounce, the impulse is larger than if they simply stop — remember velocity is a vector.
Always define a positive direction and apply signs consistently when dealing with changes of direction.

A-Level Deep Dive: Impulse and Force-Time Graphs

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.6 introduces impulse as the product of force and the time over which it acts, requires candidates to recognise that the area under a force-time graph represents the change in momentum, and demands the use of $F\Delta t = \Delta(mv)$ in collision and safety-feature contexts (refer to the official Pearson Edexcel specification document for exact wording). Although impulse is first formalised in Topic 2.6, the technique reappears throughout the qualification: Topic 4 (Materials) uses force-time analysis when evaluating brittle and ductile failure under shock loading, Topic 6 (Further mechanics) treats two-dimensional momentum exchange and circular-motion impulses, Topic 7 (Electric and magnetic fields) invokes $F = dp/dt$ when charged particles change velocity in field regions, and Topic 8 (Nuclear and particle physics) uses impulse implicitly when discussing scattering events and decay-product momenta. The Edexcel data, formulae and relationships booklet supplies $F = \Delta(mv)/\Delta t$ in symbolic form but does not state the area-under-graph rule in words — that interpretation must come from the candidate.

Worked example with full mark scheme

Question (8 marks):

A 1500 kg car travelling at 18.0 m s⁻¹ collides head-on with a rigid concrete barrier and comes to rest. The deceleration occurs over 0.120 s in a vehicle without crumple zones, and over 0.480 s in an equivalent vehicle fitted with crumple zones.

(a) Calculate the change in momentum of the car. (2)

(b) Calculate the average force exerted on the car (i) without crumple zones, and (ii) with crumple zones. (3)

Solution with mark scheme:

(a) Step 1 — apply $\Delta p = mv - mu$ with consistent signs. Take the initial direction of motion as positive.

$\Delta p = (1500)(0) - (1500)(18.0) = -27\,000 \text{ kg m s}^{-1}$

M1 — correct substitution into $\Delta p = mv - mu$ with consistent sign convention.

A1 — magnitude $27\,000$ kg m s⁻¹ (or equivalently $27\,000$ N s) with direction opposing initial motion.

(b) (i) Step 1 — apply $F = \Delta p / \Delta t$ with the short collision time.

$F = \dfrac{-27\,000}{0.120} = -225\,000 \text{ N} = -2.25 \times 10^5 \text{ N}$

M1 — correct use of $F = \Delta p / \Delta t$ .

A1 — magnitude $2.25 \times 10^5$ N (3 sf), direction stated.

(ii) Substituting the longer collision time:

$F = \dfrac{-27\,000}{0.480} = -56\,250 \text{ N} = -5.63 \times 10^4 \text{ N}$

A1 — magnitude $5.63 \times 10^4$ N (3 sf). The mark scheme typically allows ECF (error carried forward) from (a).

(c) B1 — state that the change in momentum is fixed by the initial mass and velocity, so $\Delta p$ is identical in both cases.

B1 — state that crumple zones extend the collision time $\Delta t$ by deforming progressively rather than rigidly.

B1 — conclude that since $F = \Delta p / \Delta t$ and $\Delta p$ is constant, increasing $\Delta t$ decreases the average force on the occupants, reducing peak deceleration and therefore the risk of internal injury.

Total: 8 marks (M2 A3 B3). A common error in (c) is to claim crumple zones "absorb" the momentum — they do not; momentum is transferred to the barrier in both cases. They reshape the force-time profile, not the integral under it.

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A tennis ball of mass 0.058 kg approaches a racket horizontally at 22.0 m s⁻¹ and is returned along the same line at 28.0 m s⁻¹. The variation of the contact force with time during the impact is approximately triangular: the force rises linearly from 0 to a peak $F_{\text{peak}}$ over 4.0 ms and then falls linearly back to 0 over a further 4.0 ms.

(a) Determine the impulse on the ball. (2)

(b) Hence calculate the peak force $F_{\text{peak}}$ exerted by the racket on the ball. (3)

Mark scheme decomposition by AO:

Part	AO weighting	Earned by
(a) M1	AO1.2	Substituting $v = -28.0$ m s⁻¹, $u = +22.0$ m s⁻¹ into $J = m(v - u)$ .
(a) A1	AO1.1b	$J = 0.058 \times (-50.0) = -2.90$ N s; magnitude 2.90 N s with direction opposite to the incoming ball.
(b) M1	AO2.1	Setting impulse equal to the area under the force-time graph: $J = \tfrac{1}{2} \times \Delta t \times F_{\text{peak}}$ with $\Delta t = 8.0$ ms.
(b) M1	AO1.1b	Rearranging: $F_{\text{peak}} = 2J / \Delta t = 2 \times 2.90 / 0.0080$ .
(b) A1	AO1.1b	$F_{\text{peak}} = 725$ N (3 sf).
(c)	AO3.1a	Any reasonable assumption: the rise and fall are perfectly linear; air resistance during contact is negligible; the ball-racket interaction can be treated as one-dimensional.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a typical Paper 1 impulse item: a vector-aware momentum change feeding into a graphical area calculation, with an AO3 reflective tail. Candidates who skip the sign change in (a) (writing $J = m \times 6 = 0.348$ N s) lose both M1 and A1 and propagate the slip into part (b).