Impulse & Momentum

You’ll be able to

Define linear momentum p = mv and express Newton's second law as ΣF = dp/dt
Define impulse as the time integral J = ∫F dt and relate it to the change in momentum
Apply the impulse-momentum theorem, including reading impulse as area under an F-t graph

Momentum and the second law

Linear momentum is p = mv, a vector pointing along the velocity. Newton's second law in its original form says the net force equals the rate of change of momentum, ΣF = dp/dt. For constant mass this is just ma, but the momentum form is the one that generalizes to collisions and to systems whose mass changes.

Momentum and Newton's second law

p = mv ΣF = dp/dt

Momentum is a vector (kg·m/s). The net force is the time rate of change of momentum; when m is constant this reduces to ΣF = ma.

Impulse is the time integral of force

Rearrange ΣF = dp/dt into dp = F dt and integrate over the interval of contact: the accumulated F dt is the impulse J, and it equals the change in momentum. This is the impulse-momentum theorem. On a force-versus-time graph, the impulse is simply the area under the curve — a powerful way to handle collisions where the force spikes briefly and is never known instant by instant.

Impulse-momentum theorem

J = ∫ F dt = Δp = m v_f − m v_i

Impulse (N·s) equals the change in momentum. Graphically it is the area under a force-versus-time curve; an average force gives J = F_avg Δt.

Worked example

A 0.2 kg ball hits a wall moving at 10 m/s and rebounds at 8 m/s in the opposite direction. The contact lasts 0.05 s. Find the impulse on the ball and the average force the wall exerts.

1.Choose the rebound direction as positive: v_i = −10 m/s (toward the wall), v_f = +8 m/s (away).
2.Impulse equals the change in momentum: J = m(v_f − v_i) = 0.2(8 − (−10)) = 0.2(18) = 3.6 N·s.
3.The average force follows from J = F_avg Δt: F_avg = J/Δt = 3.6/0.05 = 72 N.
4.The large force arises because a big momentum change is delivered in a very short time.

Answer: J = Δp = 3.6 N·s (away from the wall); F_avg = J/Δt = 72 N

Checkpoint

A 0.5 kg ball moving at 4 m/s is brought to rest. What is the magnitude of the impulse delivered to it?

Checkpoint

A time-varying force acts on an object during a collision. The impulse it delivers is equal to:

Tip

For a fixed change in momentum, the average force and the contact time trade off: J = F_avg Δt. Extending the contact time — a longer follow-through, a crumple zone, a padded glove — lowers the peak force. This is the physics behind most safety design.

Answer the 2 checkpoints as you read.