Impulse and force–time graphs

Newton stated his second law for momentum, Fₙₑₜ = dp/dt, and that is the form that integrates. Multiply by dt and add up the pieces from t₁ to t₂: ∫Fₙₑₜ dt = p(t₂) − p(t₁). The left side is the net impulse Jₙₑₜ, so the law becomes a ledger: whatever area of net force acts, that much momentum appears. The units agree, N s = (kg m s⁻²)(s) = kg m s⁻¹. Impulse also belongs to each force on its own — the floor's push, the weight and the friction on a sliding block each deliver one — and these add as vectors to give Jₙₑₜ. Nothing in the result depends on how the force varied inside the interval. A steady 40 N held for 0.25 s delivers 10 N s; so does 400 N for 0.025 s, and so does any other history with the same integral.

Plot one component of a force against time and its impulse over any interval is the area between the curve and the time axis, positive above the axis and negative below. Straight-line traces break into rectangles and triangles: a force that holds 300 N for 4.0 ms and then falls linearly to zero over 6.0 ms encloses 300 N × 0.0040 s + ½ × 300 N × 0.0060 s = 1.2 + 0.90 = 2.1 N s. Sensor data arrive as samples, not formulae, and the trapezoid rule J ≈ Σ ½(Fₖ + Fₖ₊₁)Δt gives the same area; on printed graph paper, count squares and multiply by what one square is worth, such as 50 N × 1 ms = 0.05 N s. Keep the signs: a force that pushes with 6.0 N s of area above the axis and then pulls with 1.5 N s below it delivers a net 4.5 N s, not 7.5 N s. The height of the curve says how fast momentum is changing at that instant; the area says how much it has changed.

Δp = m v_f − m vᵢ subtracts vectors, and the commonest numerical slip in this topic is subtracting speeds instead. Choose an axis and give every velocity its sign. A 0.30 kg ball meets a wall at 8.0 m s⁻¹ and leaves at 5.0 m s⁻¹; with +x out of the wall, Δpₓ = 0.30 kg × (+5.0 − (−8.0)) m s⁻¹ = +3.9 N s. Subtracting speeds gives 0.90 N s, more than four times too small. A rebound always takes more impulse than a dead stop, 3.9 N s here against 2.4 N s, because the wall must cancel the incoming momentum and then build the outgoing momentum. In two dimensions, apply the theorem component by component. A frictionless wall pushes only along its normal, so only the normal velocity component changes, the component along the wall passes through untouched, and the impulse on the ball points straight out of the wall whatever angle it arrived at.

Divide the impulse by the contact time and you have the average force, F_avg = Δp/Δt: the height of the rectangle with the same area as the real pulse over the same interval. Once the velocities fix Δp, contact time is the only lever left. A 0.20 kg phone that hits a tiled floor at 4.0 m s⁻¹ and stops in 1.0 ms feels F_avg = 0.80 N s ÷ 0.0010 s = 800 N; a case that stretches the stop to 5.0 ms gives 160 N. The average still flatters the pulse. A real contact force starts and ends at zero, so it must rise above its average somewhere in between: a symmetric triangle peaks at 2F_avg, a half-sine at (π/2)F_avg ≈ 1.57F_avg. Cracked screens, bruises and overloaded sensors respond to the peak, which is why a good cushion does two jobs at once: it lengthens Δt, and it flattens the pulse towards the rectangle.

The theorem balances momentum against the net impulse, so every force acting during the contact belongs in the integral. Often one force dominates so completely that the rest can be dropped — the impulse approximation — but that is a claim to test, not to assume: compare mgΔt with |Δp|. A 0.60 kg basketball that hits the floor at 5.0 m s⁻¹ and leaves at 4.0 m s⁻¹ after 12 ms has |Δp| = 5.4 N s, while gravity's impulse over the contact is 0.60 × 9.81 × 0.012 = 0.071 N s, about 1.3%: drop it. Now catch a 4.0 kg medicine ball falling at 3.0 m s⁻¹, bringing it to rest over 0.40 s. Here Δp = 12 N s, but mgΔt = 15.7 N s, larger than the momentum change itself. With up positive, (F_avg − mg)Δt = Δp gives F_avg = (12 + 15.7) N s ÷ 0.40 s = 69 N, not the 30 N the approximation predicts. The longer the contact, the more the weight matters.

When the force is a known function of time, the impulse is an ordinary definite integral, and the running integral J(t) = ∫ F dt′, taken from 0 to t, gives the momentum at every instant of the contact: p(t) = pᵢ + J(t). Model a ball on a stiff wall as a half-sine, F(t) = Fₘₐₓ sin(πt/τ). Then J(t) = (Fₘₐₓ τ/π)(1 − cos(πt/τ)), and the whole pulse delivers 2Fₘₐₓ τ/π. A 0.40 kg ball arriving at 6.0 m s⁻¹ and leaving at 3.0 m s⁻¹ after τ = 8.0 ms takes J = 0.40 × 9.0 = 3.6 N s, so Fₘₐₓ = πJ/(2τ) = 7.1 × 10² N. The ball is momentarily at rest when J(t) reaches its incoming 2.4 N s, where cos(πt/τ) = −1/3, at t = 4.9 ms: after the force has peaked, because cancelling the larger incoming momentum uses more of the pulse than building the smaller outgoing one. If a force is known as a function of position instead, ∫F dx gives work, not impulse; pick the integral that matches the variable.