Impulse from Force–Time Graphs & Variable Forces

You’ll be able to

Compute impulse as the integral of force over time
Read impulse as the area under a force–time graph
Explain why extending collision time reduces peak force

Impulse is an integral over time

Where work integrates force over distance, impulse integrates the same force over time: J = ∫F dt, and it equals the change in momentum. Real collision forces are never constant — they rise sharply, peak and fall over milliseconds — so the integral is the honest definition and J = FΔt is only the constant-force approximation. On a force–time graph, impulse is the area under the curve, exactly parallel to work being the area under a force–position curve.

The impulse–momentum theorem

J = ∫F dt = Δp = mΔv F_avg = J/Δt

Impulse is a vector with the direction of the net force. The average force is defined so that F_avg Δt gives the same area as the true varying force — it is a derived quantity, not a measured one.

Why the change in momentum is what is fixed

In a collision the change in momentum is usually determined by the situation — a car must stop, a ball must reverse — and is not something a designer can alter. What can be altered is the time over which that change occurs. Since J = F_avg Δt is fixed, lengthening Δt lowers F_avg proportionally. This is the entire engineering principle behind airbags, crumple zones, helmet padding, landing by bending your knees, and a boxer riding a punch. The impulse is identical in every case; only the peak force differs.

Reading a realistic collision pulse

A measured collision force looks roughly triangular or bell-shaped: zero at first contact, rising to a peak at maximum deformation, falling back to zero at separation. The area gives the impulse and therefore Δp; the height gives the peak force that determines whether something breaks. Two collisions with identical areas can have wildly different peaks, which is why safety engineering targets the shape of the pulse rather than its area.

Worked example

A 0.50 kg ball at rest is struck by a force that rises linearly from 0 to 500 N over 0.050 s and falls linearly back to 0 over the next 0.050 s. Find the impulse, the ball's final speed, and the average force.

1.The graph is a triangle of base 0.100 s and height 500 N.
2.Impulse is the area: J = ½(0.100)(500) = 25 N·s.
3.By the impulse–momentum theorem, Δv = J/m = 25/0.50 = 50 m/s, so the ball leaves at 50 m/s.
4.Average force: F_avg = J/Δt = 25/0.100 = 250 N — exactly half the peak, as expected for a symmetric triangle.

Answer: J = 25 N·s, final speed 50 m/s, and F_avg = 250 N — half the 500 N peak

Checkpoint

A car is designed with a crumple zone that doubles the duration of a collision. Compared with a rigid car in the same crash, the crumple zone —

Tip

Keep the pair straight by what each integrates over. Work = ∫F dx changes kinetic energy, a scalar. Impulse = ∫F dt changes momentum, a vector. Same force, different variable of integration, different conserved quantity — and questions frequently pair them to see whether you can tell which is being asked for.

Checkpoint

On a force–time graph for a collision, the impulse delivered is represented by —

Answer the 2 checkpoints as you read.