Momentum and Impulse

Imagine stopping a golf ball travelling at 50 m/s with your hand, it stings. Now imagine stopping a bowling ball (5 kg) moving at just 1 m/s, it barely stings at all, but it is much harder to stop. Both require roughly the same effort, yet their masses and speeds are wildly different. The property they share is the product $mv$, and stopping either one requires the same impulse.

Momentum is a vector, direction is essential. Always define a positive direction before calculating. Objects moving in the positive direction have positive momentum.

Unit: kg m/s (same as N s).

Momentum: $p = mv$ (kg m/s); momentum is a vector, always define a positive direction first; when an object bounces and reverses direction, $\Delta p = mv_f - mv_i$ uses signed velocities, giving a much larger magnitude than if the object simply stopped.

Pause, copy the highlighted definition and sign convention into your book before moving on.

We just saw that momentum is the product of mass and velocity and requires a sign convention. That raises a question: what causes momentum to change, and how does the magnitude of the change relate to force and time? This card answers it → impulse $J = F\Delta t = \Delta p$; for the same $\Delta p$, shorter contact time means larger average force.

Impulse is the "dose" of force delivered over time. The same impulse can be delivered by a large force for a short time, or a small force for a long time, and both produce the same change in momentum.

Unit: N s = kg m/s.

Rearranged for average force: $F = \Delta p/\Delta t$. This means for the same momentum change, halving the contact time doubles the force, which is why hitting a wall is more dangerous than braking.

Impulse-momentum theorem: $J = F\Delta t = \Delta p = mv_f - mv_i$ (N s = kg m/s); average force $F = \Delta p/\Delta t$; for direction-reversal problems always subtract signed velocities ($\Delta v = v_f - v_i$, not the difference of magnitudes); convert ms to s before calculating.

Pause, copy the highlighted theorem and the direction-reversal warning into your book before moving on.

We just saw that $J = F\Delta t = \Delta p$ for a constant force. That raises a question: in a real collision the force is never constant, it rises as contact begins and falls as contact ends, so what does $F\Delta t$ even mean then? This card answers it → impulse always equals the area under a force-time (F-t) graph, whether the force is constant (a rectangle) or variable (a triangle or other shape).

Plot force (y-axis, N) against time (x-axis, s) for a typical collision, a car crash test, a bat striking a ball, a foot landing on the ground. The force starts at zero, rises rapidly to a peak as the objects compress into each other, then falls back to zero as they separate. This rise-then-fall shape is far more realistic than a single constant force.

For a constant force, the F-t graph is a horizontal line, its area is a rectangle, $F \times \Delta t$, exactly matching the impulse formula already learned. For a real, changing force, the same rule still holds: impulse equals whatever area lies under the curve, whether that is a triangle, a trapezoid, or a shape counted using grid squares on the graph.

Impulse always equals the area under a force-time graph; for a constant force this is a rectangle ($J = F\Delta t$); for a real, variable force this is approximated as a triangle ($J \approx \frac{1}{2} \times \text{base} \times \text{height}$) or found by counting grid squares; for a fixed impulse, a shorter contact time means a taller (larger peak force) graph, the principle behind airbags and crumple zones.

Pause, copy the highlighted F-t graph rule into your book before moving on.