You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
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.
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
A 70 kg sprinter running at 10 m s⁻¹ has momentum 700 kg m s⁻¹. A 0.16 kg cricket ball bowled at 40 m s⁻¹ has momentum 6.4 kg m s⁻¹. The sprinter has far more momentum — they are much harder to stop.
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=ΔtΔp
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.
Rearranging F = Δp/Δt gives us the definition of impulse:
Impulse=FΔt=Δp
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).
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: Δp=mv−mu=0.42×25−0.42×0=10.5 N s
Average force: F=ΔtΔp=0.01210.5=875 N
That is roughly the weight of an 89 kg person — a substantial force, but applied for only 12 milliseconds.
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.
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: Impulse=21×base×height=21×0.005×400=1.0 N s
If the ball was initially travelling at 20 m s⁻¹ towards the racket and the impulse reverses its direction:
Δp=1.0 N s
Taking the initial direction as negative (towards the racket): mv−mu=1.0 0.058v−(0.058×(−20))=1.0 0.058v+1.16=1.0 v=0.0581.0−1.16=−2.8 m s−1
Wait — the negative sign means the ball is still heading towards the racket. Let us reconsider: if the impulse from the racket is in the positive direction (away from racket), and the ball initially moves at −20 m s⁻¹:
0.058v=0.058×(−20)+1.0=−1.16+1.0=−0.16 v=−2.8 m s−1
This means 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.
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.
This is the same principle behind:
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): Δp=75×0−75×6.0=−450 N s
(a) Stiff-legged: F=0.005450=90,000 N
This is 122 times the person's body weight — easily enough to fracture bones.
(b) Bent knees: F=0.200450=2,250 N
This is about 3 times body weight — uncomfortable but safe.