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Impulse is the formal way of measuring how a force changes an object's momentum over time. It turns Newton's Second Law from a rate equation into a product, gives us a powerful graphical interpretation (area under the force-time graph), and explains a huge range of real phenomena — from crumple zones in cars to why cricketers "follow through" when catching a ball.
For OCR A-Level Physics A, impulse is assessed in OCR Module 3.5.2 and appears in both multiple-choice calculations and extended written answers on real-world safety applications.
Impulse is the product of the resultant force on an object and the time for which it acts.
J = F Δt (for a constant force)
From Newton's Second Law (F = Δp/Δt), rearranging gives:
J = F Δt = Δp
That is, impulse equals change in momentum. An impulse of 5 N s applied to a body increases its momentum by 5 kg m s⁻¹, regardless of the body's mass, initial velocity, or direction of travel.
In words:
The change in momentum of a body equals the impulse applied to it.
In equation form:
Δp = F Δt (constant force) Δp = ∫ F dt (variable force — the area under the F-t graph)
The second form is the key insight: if the force on an object varies with time, the total impulse is the area under the F-t graph from t₁ to t₂. A-Level candidates are not expected to compute integrals, but they are expected to interpret graphs geometrically — counting squares, computing triangle/rectangle areas, or estimating areas under curves.
flowchart TD
A[Force varies with time F(t)] --> B[Plot F vs t]
B --> C[Area under curve = ∫ F dt = Δp]
C --> D[Impulse equals change in momentum]
A 3.0 kg ball experiences a constant force of 12 N for 0.50 s. Find its change in momentum and final velocity (starting from rest).
Alternative route: a = F/m = 4.0 m s⁻², v = a t = 2.0 m s⁻¹. Both methods must agree.
A 0.5 kg ball experiences a force that varies with time as follows (triangular pulse):
Find the impulse and the ball's change in velocity.
The area under the F-t graph is a triangle with base 0.20 s and peak height 50 N:
Area = ½ × base × height = ½ × 0.20 × 50 = 5.0 N s
So Δp = 5.0 kg m s⁻¹ and Δv = Δp / m = 5.0 / 0.5 = 10 m s⁻¹.
Notice that the average force over the 0.20 s interval is 25 N (half the peak), which would have given us the same result: F_avg × Δt = 25 × 0.20 = 5.0 N s. Mathematically, the average force is the height of a rectangle with the same area as the actual F-t curve.
A 0.16 kg cricket ball travels at 30 m s⁻¹ before being caught. Compare the average force on the catcher's hand if the ball is stopped in (a) 0.010 s (a rigid catch) and (b) 0.10 s (a "soft" catch with hand recoil).
By extending the catch over a ten-times-longer time, the average force drops by a factor of 10. The impulse is the same — the momentum has to be removed — but the peak force is dramatically reduced. This is the principle of every sports catching technique, every crumple zone, and every airbag.
In a crash, a passenger's momentum must be reduced to zero. The impulse Δp is fixed by the mass and initial velocity. A seat belt and crumple zone extend the time Δt over which the impulse is delivered, reducing the average force on the passenger:
F_avg = Δp / Δt
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