Impulse-Momentum Theorem Explained (AP Physics 1)

What the Impulse-Momentum Theorem Says

The impulse-momentum theorem states that the impulse delivered to an object equals its change in momentum. On the AP Physics 1 equation sheet it appears as $J = F_{avg} \Delta t = \Delta p$, where $p = mv$ is linear momentum.

Impulse is what a force accomplishes over a time interval. Push harder, or push for longer, and you change the object's momentum by more. The units line up too: a newton second (N s) is exactly a kilogram meter per second (kg m/s), so impulse and momentum change are measured in the same unit.

The theorem is Newton's second law in disguise. Start with $F_{net} = ma$, write acceleration as $\Delta v / \Delta t$, and multiply both sides by $\Delta t$: you get $F_{net} \Delta t = m \Delta v = \Delta p$. It is the time analog of the work-energy theorem: force acting through a distance changes kinetic energy, while force acting over a time changes momentum.

Three Ways to Find Impulse

How you find impulse depends on what the problem gives you.

• Force and time known: use $J = F_{avg} \Delta t$. A 200 N average force acting for 0.30 s delivers $J = 200 \times 0.30 = 60$ N s.
• Velocities known: use $J = \Delta p = m v_f - m v_i$. This works even when you know nothing about the force itself.
• Force-time graph given: impulse is the area under the curve.

The power of the theorem is switching between these views. In a collision you almost never know the force directly, but you can measure velocities before and after. Compute $\Delta p$, divide by the contact time, and you have the average force. That is exactly how problems ask you to estimate the force on a ball, a car, or a person: momentum change first, then force.

Force-Time Graphs: Area Equals Impulse

On a graph of net force versus time, the impulse over any interval equals the area between the curve and the time axis. This is the graph skill AP loves to test.

For straight-line segments, break the area into rectangles and triangles. A constant 400 N for 0.05 s is a rectangle: $J = 400 \times 0.05 = 20$ N s. A force ramping linearly from 0 up to 400 N over 0.05 s is a triangle: $J = \frac{1}{2}(0.05)(400) = 10$ N s.

Two details matter. First, area below the time axis counts as negative impulse, so the momentum change can shrink or even reverse direction. Second, real collision forces are spike-shaped, not constant. The average force $F_{avg}$ is the height of the rectangle that has the same area (the same impulse) over the same time interval as the actual spike.

Watch the Signs: Bouncing Beats Sticking

Impulse and momentum are vectors, and sign errors are the top way to lose points here. Pick a positive direction before plugging in numbers, and keep it for the whole problem.

Compare two 0.50 kg balls that each hit a wall at 4.0 m/s. One sticks (think clay), one bounces straight back at 4.0 m/s. Take away from the wall as positive, so both arrive with $v_i = -4.0$ m/s.

• Sticks: $\Delta p = 0.50(0 - (-4.0)) = +2.0$ kg m/s
• Bounces: $\Delta p = 0.50(4.0 - (-4.0)) = +4.0$ kg m/s

The bouncing ball's momentum change is twice as large, so the wall exerts twice the impulse on it, and (for equal contact times) twice the average force. Reversing a velocity always requires more impulse than merely stopping it.

Why Airbags and Crumple Zones Work

Airbags and crumple zones do not reduce your momentum change. If you are moving at 15 m/s and end at rest, $\Delta p$ is fixed the moment the crash starts. What safety features change is the time over which that momentum is removed.

Rearrange the theorem: $F_{avg} = \Delta p / \Delta t$. With $\Delta p$ locked in, average force is inversely proportional to stopping time. Triple the stopping time and you cut the average force to one third.

A crumple zone folds the front of the car so the passenger compartment decelerates over roughly 0.1 to 0.2 s instead of a few milliseconds. An airbag then spreads the force over a larger area of your body and stretches your personal stopping time further. The worked example below puts real numbers on this: extending a stop from 0.05 s to 0.30 s drops the average force from 21000 N to 3500 N.

The Bridge to Conservation of Momentum

The impulse-momentum theorem explains why momentum is conserved in collisions. When two carts collide, Newton's third law says cart A pushes on cart B exactly as hard as B pushes on A, in the opposite direction, for exactly the same contact time. Equal and opposite forces over equal times mean equal and opposite impulses, so whatever momentum one cart gains, the other loses. The total never changes.

That single idea powers everything in conservation of momentum: elastic collisions, perfectly inelastic collisions, explosions, and recoil. If you want to check your before-and-after numbers on a specific problem, the momentum and collision calculator handles one-dimensional collisions of both types. For the full topic list and how impulse fits into the rest of the unit, see Unit 4: Linear Momentum.