Impulse and Momentum Revisited

At GCSE you met momentum as the product of mass and velocity: p = mv. You learned that momentum is conserved in collisions and explosions. Now, at A-Level, we need to go deeper — particularly into the concept of impulse and how forces change momentum over time. This lesson equips you with the tools to handle force-time graphs, variable forces, and real-world impact scenarios that appear across Papers 1 and 3.

Momentum Recap

Momentum is a vector quantity, measured in kg m s⁻¹ (or equivalently, N s). It has both magnitude and direction. For an object of mass m travelling at velocity v:

$p = mv$

Object	Mass (kg)	Speed (m s⁻¹)	Momentum (kg m s⁻¹)
Electron in copper wire	9.11 × 10⁻³¹	1.0 × 10⁶	9.1 × 10⁻²⁵
Cricket ball (bowled)	0.16	40	6.4
Sprinter	70	10	700
Family car at 30 mph	1200	13.4	16 100
Lorry at motorway speed	20 000	31	620 000
Oil tanker (loaded, cruising)	3.0 × 10⁸	8.0	2.4 × 10⁹

The sprinter has far more momentum than the cricket ball — they are much harder to stop. The oil tanker, despite its modest speed, has an enormous momentum and requires several kilometres to stop.

Newton's Second Law in Terms of Momentum

You are used to writing Newton's second law as F = ma. But Newton himself actually expressed it in terms of momentum. The resultant force acting on an object equals the rate of change of momentum:

$F = \frac{\Delta p}{\Delta t}$

When mass is constant, this reduces to the familiar F = ma, because Δp = mΔv and so Δp/Δt = m(Δv/Δt) = ma.

However, the momentum form is more general. It applies even when mass changes — for example, a rocket ejecting fuel, or rain falling into a moving cart.

flowchart TD
    A["Is the mass of the system constant?"] -->|Yes| B["F = ma is sufficient\n(constant-mass problems)"]
    A -->|No| C["Use F = Δp/Δt directly\n(variable-mass problems)"]
    C --> D["Examples:\n• Rocket ejecting fuel\n• Rain falling into cart\n• Sand on conveyor belt\n• Hose hitting a wall"]
    B --> E["Examples:\n• Car braking\n• Ball kicked from rest\n• Object on incline"]

Impulse

Rearranging F = Δp/Δt gives us the definition of impulse:

$\text{Impulse} = F\Delta t = \Delta p$

Impulse is the product of force and the time for which it acts. It equals the change in momentum of the object. Impulse is also a vector quantity, measured in N s (which is dimensionally identical to kg m s⁻¹).

Worked Example 1

A 0.42 kg football is kicked from rest and leaves the boot at 25 m s⁻¹. The foot is in contact with the ball for 12 ms.

Change in momentum: $\Delta p = mv - mu = 0.42 \times 25 - 0.42 \times 0 = 10.5 \text{ N s}$

Average force: $F = \frac{\Delta p}{\Delta t} = \frac{10.5}{0.012} = 875 \text{ N}$

That is roughly the weight of an 89 kg person — a substantial force, but applied for only 12 milliseconds.

Worked Example 2: Direction Reversal

A 0.058 kg tennis ball travelling at 30 m s⁻¹ to the right is hit by a racket and returns at 50 m s⁻¹ to the left. The contact time is 4.0 ms.

Taking rightward as positive: u = +30 m s⁻¹, v = −50 m s⁻¹.

$\Delta p = m(v - u) = 0.058 \times (-50 - 30) = 0.058 \times (-80) = -4.64 \text{ N s}$

The magnitude of impulse is 4.64 N s. Note that the full velocity change is 80 m s⁻¹, not 20 m s⁻¹ — both the stopping and reversing contribute.

$F = \frac{4.64}{0.004} = 1160 \text{ N}$

Common mistake: Forgetting the sign change when an object reverses direction. If a ball arrives at +30 m s⁻¹ and leaves at −50 m s⁻¹, the change in velocity is −80 m s⁻¹, not −20 m s⁻¹. Always define a positive direction and use consistent signs.

Force–Time Graphs

When a force varies with time, the impulse is found from the area under the force–time graph. This is directly analogous to how displacement is the area under a velocity–time graph.

For a constant force, the area is simply a rectangle: F × Δt.

For a variable force — such as during a collision — the graph typically rises to a peak and then falls. The area under this curve gives the total impulse, and therefore the total change in momentum.

Worked Example 3

A tennis ball of mass 0.058 kg strikes a racket and the force–time graph shows a triangular pulse with peak force 400 N lasting 5.0 ms.

Area under triangle: $\text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.005 \times 400 = 1.0 \text{ N s}$

If the ball was initially travelling at 20 m s⁻¹ towards the racket and the impulse reverses its direction:

Taking the initial direction as negative (towards the racket): $0.058v - (0.058 \times (-20)) = 1.0$ $0.058v + 1.16 = 1.0$ $v = \frac{1.0 - 1.16}{0.058} = -2.8 \text{ m s}^{-1}$

The negative sign means the ball is still heading towards the racket. The 1.0 N s impulse was not enough to reverse the ball fully — it slowed it from 20 to 2.8 m s⁻¹. In reality, the impulse from a racket is much larger, typically 2–4 N s, which fully reverses the ball.

Worked Example 4: Trapezoidal Force–Time Graph

A 2.0 kg trolley is struck by a force that ramps from 0 to 600 N in 2 ms, stays at 600 N for 3 ms, then drops back to 0 in 1 ms. Calculate the velocity gained.

$\text{Impulse} = \frac{1}{2}(0.002)(600) + (0.003)(600) + \frac{1}{2}(0.001)(600)$ $= 0.60 + 1.80 + 0.30 = 2.70 \text{ N s}$

$v = \frac{\Delta p}{m} = \frac{2.70}{2.0} = 1.35 \text{ m s}^{-1}$

Variable Force and Impulse in Real Applications

In many real situations the force is not constant. During a car crash, for example, the force builds up as the car structure deforms, reaches a peak, and then drops. Crumple zones work by increasing the collision time Δt, which for the same change in momentum Δp means the average force F = Δp/Δt is reduced.

Safety Feature	How It Works	Typical Time Extension
Crumple zone	Car body deforms progressively	5 ms → 80–120 ms
Airbag	Inflates and slowly deflates	3 ms → 50–60 ms
Seatbelt (pre-tensioned)	Allows controlled stretching	2 ms → 30–50 ms
Cycle helmet (inner foam)	Crushes on impact	1 ms → 8–12 ms
Bending knees on landing	Increases stopping distance	5 ms → 200+ ms

Worked Example 5

A 75 kg person jumps from a wall and lands on the ground. They are travelling at 6.0 m s⁻¹ just before landing. Calculate the average force on their legs if they (a) land stiff-legged and stop in 5 ms, and (b) bend their knees and stop in 200 ms.

Change in momentum (same in both cases): $\Delta p = 75 \times 0 - 75 \times 6.0 = -450 \text{ N s}$

(a) Stiff-legged: $F = \frac{450}{0.005} = 90{,}000 \text{ N}$

This is 122 times the person's body weight — easily enough to fracture bones.

(b) Bent knees: $F = \frac{450}{0.200} = 2{,}250 \text{ N}$

This is about 3 times body weight — uncomfortable but safe.

Continuous Flow Problems

A category of problems that specifically requires F = Δp/Δt is continuous flow — where mass is continuously arriving or leaving.

Worked Example 6: Water Hose

A hose delivers water at 5.0 kg s⁻¹ against a wall. The water hits the wall at 12 m s⁻¹ and does not bounce back.

Each kilogram of water loses 12 kg m s⁻¹ of momentum. Rate of momentum change = 5.0 × 12 = 60 N.

By Newton's third law, the water exerts 60 N on the wall.

If the water bounced back at 12 m s⁻¹, each kilogram would change from +12 to −12 kg m s⁻¹ — a change of 24 kg m s⁻¹. The force would double to 120 N.

Worked Example 7: Rocket Thrust

A rocket ejects 50 kg of exhaust gas per second at 2000 m s⁻¹ relative to the rocket.

$\text{Thrust} = \frac{\Delta m}{\Delta t} \times v_{\text{exhaust}} = 50 \times 2000 = 100{,}000 \text{ N} = 100 \text{ kN}$

Common Exam Mistakes

flowchart TD
    A["Impulse / Momentum\nCommon Mistakes"] --> B["Forgetting direction reversal\n• Δv = v − u, including signs\n• Ball at +30 returning at −50:\n  Δv = −80, NOT −20"]
    A --> C["Wrong time units\n• Convert ms to s BEFORE calculating\n• 5 ms = 0.005 s, NOT 5 s"]
    A --> D["Confusing impulse with force\n• Impulse = FΔt (N s)\n• Force = Δp/Δt (N)"]
    A --> E["Ignoring continuous flow\n• Use Δm/Δt for hose/rocket\n• Don’t apply F = ma when mass changes"]
    A --> F["Misreading F–t graphs\n• Area = impulse, NOT the peak force\n• Triangle area = ½ × base × height"]

Summary

Momentum: p = mv (vector, units kg m s⁻¹ or N s)
Newton's second law: F = Δp/Δt
Impulse = FΔt = Δp
The area under a force–time graph equals the impulse
Increasing collision time reduces the average force for a given momentum change
This principle underpins crumple zones, airbags, helmets, and safe landing techniques
For direction reversals, always use signed velocities: Δp = m(v − u)
Continuous flow: F = (Δm/Δt) × v for hoses, rockets, and conveyor belts