Change in Momentum and Impulse: Study Guide (Physics 1 4.2)

This guide is for the algebra-based Physics 1 course. In Topic 4.1 you defined momentum, p = mv. Here you find out what changes it. The answer is the same thing that changes velocity: a net force. The new idea is that the time for which the force acts matters just as much as its size.

Change in momentum

The change in momentum of an object or system is its final momentum minus its initial momentum:

Δp = p − p₀

For one object of constant mass, this is Δp = mv − mv₀ = m(v − v₀).

In one dimension, choose an axis and use signed velocities. Take +x towards a wall. A 0.20 kg ball that hits the wall at +5.0 m/s and stops has Δp = 0.20 × (0 − 5.0) = −1.0 kg·m/s. If it bounces back at −5.0 m/s instead, Δp = 0.20 × (−5.0 − 5.0) = −2.0 kg·m/s. Bouncing back needs twice the change in momentum of stopping, even though the final speed is the same as the starting speed.

Force is the rate of change of momentum

Newton’s second law, as you met it in Topic 2.5, says F_net = ma. There is a more general form:

F_net = Δp / Δt

In words: the net external force on an object or system equals the rate at which its momentum changes. A large net force changes momentum quickly. Zero net force means momentum does not change.

Only external forces count. Forces between parts of a system cannot change the momentum of the system as a whole; Topic 4.3 develops that idea.

Impulse

Rearranging F_net = Δp / Δt gives a product that is worth naming. Impulse, J, is the average force on a system multiplied by the time interval over which it acts:

J = F_avg Δt

• Units: newton-seconds, N·s. Since 1 N = 1 kg·m/s², 1 N·s = 1 kg·m/s, the same unit as momentum.
• Impulse is a vector. It points in the same direction as the net force.

The impulse–momentum theorem

Put the two ideas together and you get the central result of this topic:

J = Δp

The impulse delivered to a system by the net external force equals the system’s change in momentum. You can read it either way. A known force for a known time tells you the change in momentum. A measured change in momentum tells you the impulse, and dividing by the contact time gives the average net force.

Impulse as an area

Real forces are rarely constant. A kick, a bat or a bumper produces a force that rises and falls in a fraction of a second. You do not need calculus to handle this. The impulse equals the area between a net force–time graph and the time axis. Split the area into rectangles and triangles.

Net force as a slope

Turn the theorem around. Since F_net = Δp / Δt, the slope of a momentum–time graph is the net external force. A steep momentum–time graph means a large net force. A flat section means zero net force.

Newton’s second law as a special case

If the mass of the system does not change, Δp = mΔv. Substitute into F_net = Δp / Δt:

F_net = mΔv / Δt = ma

So F_net = ma is what the impulse–momentum theorem becomes for a system of constant mass. Check it on Figure 2: the sled’s velocity rises from 4.0 ÷ 2.0 = 2.0 m/s to 10.0 ÷ 2.0 = 5.0 m/s in 3.0 s, so a = 1.0 m/s² and ma = 2.0 N, the same as the slope.

The momentum form is the more general starting point, but it needs care when an object gains or loses mass. For a rocket expelling gas, the theorem works for a system that includes both the rocket and the gas it has thrown out, not for the rocket alone. (A cart leaking sand straight down keeps the same velocity with no net force, even though its own mass, and so its own mv, is falling.) This course does not ask you to calculate with systems whose mass changes with time, but you may be asked to reason about them in words.

Worked example 1: impulse from a force–time graph

Question. Take +x in the direction of the push. A 0.60 kg cart starts at rest on a level, frictionless track. A force sensor records the net force shown in Figure 1. Find (a) the impulse, (b) the cart’s final velocity and (c) the average net force during the push.

1. Impulse = area. Left triangle: ½ × 0.05 s × 12 N = 0.30 N·s. Rectangle: (0.15 − 0.05) s × 12 N = 1.2 N·s. Right triangle: 0.30 N·s. Total J = +1.8 N·s.
2. Impulse–momentum theorem: J = Δp = m(v − v₀), so 1.8 = 0.60 × (v − 0), giving v = +3.0 m/s.
3. Average force: F_avg = J ÷ Δt = 1.8 N·s ÷ 0.20 s = +9.0 N.

Check. The average force (9.0 N) is less than the peak (12 N), as it must be, because the force spends part of the time rising and falling. Using the peak for the whole 0.20 s would give 2.4 N·s and a speed of 4.0 m/s, which is too big.

Worked example 2: rebound against stopping

Question. Take +x towards a buffer. A 0.50 kg trolley hits a buffer at 1.2 m/s and rebounds at 0.80 m/s. The contact lasts 0.10 s. Find (a) the change in momentum, (b) the average force from the buffer on the trolley. (c) Compare with a trolley that sticks to the buffer, with the same contact time. Friction is negligible.

1. Signed velocities: v₀ = +1.2 m/s, v = −0.80 m/s.
2. Δp = m(v − v₀) = 0.50 × (−0.80 − 1.2) = −1.0 kg·m/s.
3. The buffer force is the only horizontal force, so it is the net force: F_avg = Δp ÷ Δt = −1.0 ÷ 0.10 = −10 N (10 N away from the buffer).
4. Sticking: Δp = 0.50 × (0 − 1.2) = −0.60 kg·m/s, so F_avg = −6.0 N.

Interpretation. Bouncing needs a larger impulse than stopping, because the buffer must first remove the trolley’s momentum and then give it momentum in the opposite direction.

Check. If you used speeds without signs, you would get Δp = 0.50 × (0.80 − 1.2) = −0.20 kg·m/s, five times too small. Reversals are where sign errors cost the most.

Worked example 3: why a soft landing protects

Question. Take +y upward. A 0.18 kg phone falls and hits the floor at 4.0 m/s, then stops. On a tile floor it stops in 0.0020 s. On a thick cushion it stops in 0.040 s. Find the average net force in each case, and the average force from the surface.

1. Δp = m(v − v₀) = 0.18 × (0 − (−4.0)) = +0.72 kg·m/s in both cases.
2. Tile: F_net = 0.72 ÷ 0.0020 = +360 N. Cushion: F_net = 0.72 ÷ 0.040 = +18 N.
3. During contact, two forces act: the surface pushes up, and gravity pulls down with mg = 0.18 × 9.8 = 1.8 N. So F_surface = F_net + mg: about 362 N on tile and about 20 N on the cushion.

Interpretation. The change in momentum is fixed by how fast the phone arrives and that it stops. The cushion stretches the stopping time by a factor of 20, so the average force falls by a factor of 20. Crumple zones in cars, bending your knees when you land, and “giving” with your hands when you catch a water balloon all work this way.

Check. On tile, the weight is less than 1% of the surface force. During a short, hard impact, gravity barely matters, which is why the collision model of Topic 4.1 ignores it.

Testing the theorem in the lab

A typical procedure for testing J = Δp:

1. Measure the cart’s mass on a balance.
2. Mount a force sensor on the cart, or on a fixed post, so that it records the force during a short push or a collision with a bumper.
3. Place a motion sensor at the end of the track to record the cart’s velocity just before and just after the push.
4. Repeat with pushes of different sizes.
5. For each run, find the impulse as the area under the force–time graph (the software can do this) and the change in momentum as m(v − v₀).

To analyse the data, make a graph that should be a straight line if the theorem holds. For example, plot Δv on the vertical axis against J on the horizontal axis. Since Δv = J ÷ m, the points should lie on a line through the origin with slope 1/m. A line that misses the origin suggests a systematic error, such as friction or an offset in the force sensor.

Common misconceptions

• “Impulse is just a big force.” Impulse is force × time. Time matters as much as force (Worked example 3).
• “Change in momentum is the change in speed times mass.” Use signed velocities. A rebound changes momentum by more than the change in speed suggests (Worked example 2).
• “The peak force times the contact time gives the impulse.” Only if the force is constant. Otherwise use the area or the average force (Worked example 1).
• “A soft landing reduces the impulse.” The impulse is the same; it is spread over a longer time, so the average force is smaller.
• “Any force on a system changes its momentum.” Only the net external force does.
• “A flat momentum–time graph means the object is at rest.” It means constant momentum and zero net force; the object can be moving.
• “Impulse has its own unit, unrelated to momentum.” 1 N·s = 1 kg·m/s.

Where this leads

Topic 4.3, Conservation of Linear Momentum, applies the theorem to both objects in an interaction at once. By Newton’s third law, the impulses they exert on each other are equal and opposite, which leads to conservation of momentum. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also go back to Topic 4.1, Linear Momentum, or return to the course roadmap.