Lessons · momentum

Impulse as area: the force–time curve

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A $0.5$ kg ball is struck by a bat: the contact force rises and falls as a brief pulse, $F(t) = F_{\max}\sin(\pi t/\tau)$ with $F_{\max} = 200$ N over $\tau = 0.02$ s. What impulse does it deliver, and how fast does the ball leave? Watch the impulse accumulate as the area under the force–time curve.

Impulse

J = \tfrac{2}{\pi}F_{\max}\tau

(the shaded area)

2.546 N·s

Change in speed

\Delta v = J/m

5.093 m/s

Peak force

F_{\max}

200 N

Average force

\bar F = \tfrac{2}{\pi}F_{\max}

127.3 N

Impulse is the area under the force–time curve

J = \int F\,dt

Impulse is the integral of force over time — the area under the $F$ – $t$ curve. For a constant force this is a rectangle ( $\bar F\,\Delta t$ ); for a real pulse it is the area under the curve.

check $\tfrac{dJ}{dt} = F$ ; check $J = \int_0^t F\,dt'$ ; evaluate the pulse area $\tfrac{2}{\pi}F_{\max}\tau$ ; collapse the constant-force case to $F\,\Delta t$

✓ The impulse's slope is the force: $J'(t) = F(t)$ — the area's rate of growth is the curve's height. [structural]
✓ The accumulated impulse is the area: $J(t) = \int_0^t F\,dt'$ . [structural]
✓ The total impulse is $J = \tfrac{2}{\pi}F_{\max}\tau = \bar F\,\tau$ — exactly the area of the pulse. [structural]
✓ For a constant force the integral collapses to $J = F\,\Delta t$ — the area is a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A bigger peak force always means a bigger change in momentum.”

Not necessarily — the change in momentum is the area under the force–time curve (the impulse $\int F\,dt$ ), not the peak height. A small force acting over a long time delivers the same impulse — and the same $\Delta v$ — as a large force over a brief time. The area, not the height, is what changes the motion.

Modeling assumptions — author-asserted, disclosed not discharged

The contact force follows a half-sine pulse $F(t) = F_{\max}\sin(\pi t/\tau)$ (an idealized but realistic collision profile).
The struck object is free (the only horizontal force during contact is the strike), starting from rest.

The F–t graph, fully annotated

A static rendering (Matplotlib): the shaded area under F is the accumulated integral J, and the slope of J is F. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):