AP Physics C: Mechanics · Unit 4: Linear Momentum · Lesson 4.2

Deep Dive: Change in Momentum and Impulse

🔬 Deep Dive

This is your textbook for this topic. Take your time. Read it more than once.

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Force, Momentum, and the Impulse Integral

The rate of change of a system's momentum equals the net external force exerted on it:

F⃗(net) = dp⃗/dt

Integrating this relationship over a time interval defines impulse— the total effect a force has when it acts for some length of time:

J⃗ = ∫(t1 to t2) F⃗(net)(t) dt

🔑Impulse is a vector, pointing in the same direction as the net force that produced it. And since it's defined by an integral, impulse is exactly — geometrically — the area under a force-vs-time graph, the same bell curve shown throughout this unit's header art.

Explore this directly below.

A bell-shaped force pulse, exactly like this unit's header art — a bat striking a ball, or two carts colliding. Set the interval boundaries — the shaded area equals the impulse delivered over that interval.

t₁1.0 s

t₂4.0 s

J = ∫[1.0 to 4.0] F(t) dt ≈ 1897.9 N·s

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Force as the Slope of a Momentum-Time Graph

Run the same relationship in reverse: since F(net) = dp/dt, the net external force at any instant is simply the slope of a momentum-vs-time graph at that point.

💡This is exactly the same slope-and-area relationship you used for position, velocity, and acceleration graphs back in Lesson 1.3 — just relabeled. Steep sections of a p(t) graph mean large net force; flat sections mean zero net force.

Explore this directly below.

A momentum-time graph with three distinct phases — steep rise, gentle rise, then decline. Drag through time — the tangent line's slope is exactly the net force at that instant.

time t2.00 s

dp/dt ≈ 5.0 N = F(net) at this instant

Notice the steepest phase (early on) corresponds to the largest net force — exactly the same slope-reading skill from position-velocity-acceleration graphs back in Lesson 1.3, just applied to momentum and force instead.

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The Impulse-Momentum Theorem

A system's change in momentum is simply its final momentum minus its initial momentum:

Δp⃗ = p⃗ − p⃗₀

Putting everything in this lesson together gives the impulse-momentum theorem: the impulse delivered to an object equals its change in momentum.

J⃗ = ∫F⃗(net) dt = Δp⃗

🔑This theorem isn't a brand-new law sitting alongside Newton's second law — it contains Newton's second law as a special case. For a system of constant mass, F(net) = dp/dt = m(dv/dt) = ma, exactly the F=ma you've used since Unit 2.

But the impulse-momentum theorem reaches further than F=ma alone can. For a system with constant velocity but changing mass — a rocket burning fuel, a raindrop accumulating water as it falls — the theorem still applies directly:

F⃗(net) = dp⃗/dt = v⃗(dm/dt)

ExampleGuided Example — Applying the Impulse-Momentum Theorem

A 0.5 kg ball moving at 8 m/s is struck by a bat, reversing its direction to 12 m/s the other way. The bat is in contact with the ball for 0.02 s. Find the average force the bat exerts on the ball.

Step 1 — Set up a sign convention

Let the ball's initial direction be positive.

ExampleWorked Example — A Variable-Mass System

A cart of mass 2 kg moves at a constant 3 m/s while sand pours into it at a rate of 0.4 kg/s. Find the net external force required to keep the cart's velocity constant.

← Back to Lesson 4.2Ready for 4.3? Conservation of Linear Momentum turns this lesson's ideas into perhaps the single most powerful tool in the unit.