Change in Momentum and Impulse: Study Guide (Physics C: Mechanics 4.2)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 4.2 with this title. This guide is the calculus-based one. It defines impulse as an integral, uses forces that change with time, and treats systems that gain mass at constant velocity. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

In Topic 4.1 you defined momentum, p = mv. This topic asks what changes it, and by how much.

Net force is the rate of change of momentum

The most general form of Newton’s second law is written with momentum:

F_net = dp/dt

Here F_net is the net external force on the system you have chosen, and p is the system’s total momentum. Internal forces between parts of the system come in third-law pairs and cancel in the sum, so they cannot change the total.

Read it as a graph statement: the net external force is the slope of the momentum–time graph. A flat p–t graph means zero net force. A steep one means a large net force. In two dimensions it holds for each component: F_net,x = dp_x/dt and F_net,y = dp_y/dt.

Impulse

Rearrange to dp = F_net dt and add up the small changes over an interval. The quantity on the right is called impulse. For any force F acting from t_i to t_f:

J = ∫ F dt (from t_i to t_f)

• Unit: N·s, the same as kg·m/s.
• Impulse is a vector. The impulse of the net force points the same way as the net force. If the force keeps one direction, so does the impulse.
• On a graph of force against time, the impulse is the signed area between the curve and the time axis. Area below the axis counts as negative.
• The average force over the interval is defined by F_avg = J/Δt: the constant force that would give the same impulse in the same time.

Two graph rules

Every impulse question with a graph uses one of two rules. Decide which graph you have before you start.

Graph	Read the…	It gives	Unit
Net force against time	signed area under it	impulse J = Δp	N·s
Momentum against time	slope (tangent for a curve)	net external force F_net	N

The two graphs are linked in the same way as velocity and acceleration graphs in Topic 1.2: integrate the force graph to build the momentum graph (adding the initial momentum), or differentiate the momentum graph to get the force graph.

The impulse–momentum theorem

The change in momentum is final minus initial, with signs:

Δp = p_f − p_i

Integrating F_net = dp/dt over the interval gives the impulse–momentum theorem:

J_net = ∫ F_net dt = Δp

The impulse of the net external force equals the change in momentum. Two consequences are worth stating plainly:

• Same Δp, longer time, smaller average force. A crumple zone or a padded landing does not reduce the impulse needed to stop an object. It stretches the time, so F_avg = Δp/Δt falls.
• Reversals need bigger impulses. To bounce something back, the force must first remove its momentum and then give it momentum the other way.

Worked example 1: a kick measured by a force sensor

Question. Take +x in the direction of the kick. A 0.40 kg ball rolls towards a kicker at 5.0 m/s. A force plate shows that the net force on the ball during the 8.0 ms contact is F_x(t) = kt(T − t), with T = 8.0 ms and a peak of 1500 N at t = 4.0 ms. Find (a) k, (b) the impulse, (c) the ball’s velocity after the kick and (d) the average force.

1. (a) At t = T/2 the force is kT²/4 = 1500 N, so k = 1500 ÷ (0.0040)² ≈ 9.4 × 10⁷ N/s² (9.375 × 10⁷).
2. (b) J = ∫₀ᵀ k(Tt − t²) dt = k(T³/2 − T³/3) = kT³/6. Substituting gives J = +8.0 N·s. A neat check: kT³/6 = (2/3) × 1500 N × 0.0080 s = 8.0 N·s.
3. (c) Initial momentum: p_i = 0.40 × (−5.0) = −2.0 kg·m/s. Then p_f = p_i + J = −2.0 + 8.0 = +6.0 kg·m/s, so v_f = +15 m/s.
4. (d) F_avg = J ÷ Δt = 8.0 ÷ 0.0080 = 1000 N, two-thirds of the peak.
5. Momentum during contact. p_x(t) = −2.0 + k(Tt²/2 − t³/3) in kg·m/s, with t in s. Its slope is F_x(t), so it is steepest at 4.0 ms. Setting p_x = 0 and solving numerically gives t ≈ 2.6 ms.

Check. The ball’s weight is 0.40 × 9.8 ≈ 3.9 N, under 0.3% of the peak force. Treating the force plate’s force as the net force is the collision model of Topic 4.1. Figure 1 shows the momentum passing through zero at about 2.6 ms: the ball is stopped and then driven the other way within one smooth push.

Worked example 2: a glancing bounce

Question. Take +x perpendicular to a wall, pointing away from it, and +y along the wall. A 0.16 kg ball hits the wall at 9.0 m/s, travelling at 35° to the normal, and leaves at the same speed and the same angle on the other side of the normal. Contact lasts 15 ms and friction from the wall is negligible. Find (a) the impulse from the wall and (b) the average force. (c) Compare with a head-on bounce at the same speed.

1. Velocity components: before, v = (−9.0 cos 35°, +9.0 sin 35°) = (−7.37, +5.16) m/s; after, v = (+7.37, +5.16) m/s.
2. (a) Δp_x = 0.16 × (7.37 − (−7.37)) = +2.36 N·s; Δp_y = 0.16 × (5.16 − 5.16) = 0. So J = 2.4 N·s, perpendicular to the wall, away from it. Symbolically, J = 2mv cos θ.
3. (b) F_avg = 2.36 ÷ 0.015 ≈ 160 N (157 N), also perpendicular to the wall.
4. (c) Head-on, θ = 0 and J = 2mv = 2 × 0.16 × 9.0 = 2.9 N·s. The glancing impulse is cos 35° ≈ 0.82 of this.

Interpretation. The impulse points the way of the net force (the wall’s push), not the way the ball moves. The ball’s momentum along the wall never changes, because no force acts along the wall.

Newton’s second law as a special case

Apply the product rule to p = mv:

F_net = dp/dt = m dv/dt + v dm/dt

• Constant mass (dm/dt = 0): F_net = m dv/dt = ma. Newton’s second law in its familiar form is the impulse–momentum idea applied to a system whose mass does not change.
• Constant velocity, changing mass (dv/dt = 0): F_net = v dm/dt. A force is still needed, because new mass must be given momentum as it joins the moving system. This form assumes the added mass arrives with no momentum along the direction of motion.

Only these two cases are in this topic. Problems where mass and velocity both change, such as a rocket, use the same law but need a carefully chosen system.

Worked example 3: gravel on a conveyor belt

Question. Take +x along the belt. A hopper drops gravel straight down onto a horizontal conveyor belt at 15 kg/s. A motor keeps the belt moving at a constant 1.2 m/s. Find (a) the extra horizontal force the motor must supply, (b) the extra power and (c) the rate at which the gravel gains kinetic energy. Explain the difference between (b) and (c).

1. (a) The gravel arrives with no horizontal momentum and leaves the loading zone moving at 1.2 m/s. F = v dm/dt = 1.2 × 15 = 18 N in +x.
2. (b) Power (Topic 3.5): P = Fv = 18 × 1.2 ≈ 22 W (21.6 W).
3. (c) dK/dt = ½v² dm/dt = ½ × 1.2² × 15 ≈ 11 W (10.8 W).
4. Explanation. Each lump of gravel slides on the belt before it reaches belt speed. Kinetic friction between gravel and belt turns the other half of the motor’s output, 10.8 W, into thermal energy. The result does not depend on the friction coefficient: a lower coefficient means a longer slide, but the same energy loss.

Check. In 10 s the belt picks up 150 kg of gravel, which gains 150 × 1.2 = 180 kg·m/s of momentum. The impulse of 18 N for 10 s is 180 N·s. They match.

Common misconceptions

• “Impulse is a kind of force.” Impulse is force integrated over time, in N·s. A small force for a long time can give the same impulse as a large force briefly.
• Using speeds instead of signed velocities in Δp. A reversal from −5.0 to +15 m/s is a change of 20 m/s, not 10 m/s.
• “Impulse points the way the object moves.” It points the way the net force acts (Worked example 2).
• Treating one force’s impulse as Δp when other forces matter. Only the net force’s impulse equals Δp.
• Using the peak force as the average. For a smooth pulse, F_avg = J/Δt is smaller than the peak.
• Using F = ma when mass changes. Start from F_net = dp/dt.
• “Zero net force means the momentum graph is at zero.” It means the graph is flat, at any value.

Where this leads

Topic 4.3 applies F_net = dp/dt to a whole system: if the net external force is zero, the system’s momentum stays constant, and the impulses objects exert on each other are equal and opposite. Continue with the Conservation of Linear Momentum study guide. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.