Aim for FivePhysics 1

Unit 4 · Topic 4.2

4.2 Change in Momentum and Impulse

Impulse is a force multiplied by the time it acts, J = F_avg Δt, and it equals the change in momentum it causes: J = Δp. That's why airbags and bent knees protect you: by stretching out the time it takes to stop, they reduce the average force. On a force–time graph, impulse is the area under the curve.

Key terms

  • impulse
  • impulse–momentum theorem
  • change in momentum
  • force–time graph
  • net external force

Impulse

Impulse is the product of the average force on a system and the time interval during which it acts: J = F_avg Δt. It's a vector in the same direction as the average force. Its unit is N·s, which equals kg·m/s, the unit of momentum.

A small force acting for a long time can give the same impulse as a large force acting briefly. Pushing a stalled car for 10 s with 300 N gives the same impulse as 3000 N for 1 s.

The impulse–momentum theorem

The net impulse on a system equals its change in momentum: J = Δp = p_f − p_i = m(v_f − v_i). The rate of change of momentum equals the net external force: F_net = Δp/Δt.

For a constant mass, Δp/Δt = mΔv/Δt = ma, so this is just Newton's second law (2.5) written another way. AP Physics 1 doesn't ask you to calculate situations where the mass changes over time, such as a rocket burning fuel.

Graphs

On a graph of net force against time, the impulse is the area under the curve. Real collisions usually give a hump-shaped curve: the force rises quickly, peaks and falls. You can estimate the area with triangles or rectangles. Area below the time axis is negative impulse.

On a graph of momentum against time, the slope is the net force. A flat stretch means no net force; the steepest part shows the largest force.

Stopping safely

To stop a moving object, you need a fixed impulse: its whole momentum. Since F_avg = Δp/Δt, the longer the stop takes, the smaller the average force. That's the idea behind airbags, crumple zones, padded gym floors, bending your knees when you land and moving your hands back as you catch a ball. None of them reduce the change in momentum; they spread it over more time.

Bouncing takes more impulse than stopping

If an object reverses direction, its momentum changes by more than if it just stops. A ball hitting a wall at 5 m/s and stopping has Δp of size m(5). If it bounces back at 5 m/s, Δp has size m(10), twice as much, so the wall must exert twice the impulse. Momentum signs matter here: Δp = m(v_f − v_i) with v_f and v_i signed.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Sticking versus bouncing (classic trap)

    A 0.50 kg ball moving at 6.0 m/s hits a wall. Contact lasts 0.010 s. Find the average force the wall exerts if the ball (a) stops dead and (b) bounces straight back at 6.0 m/s. Take the ball's initial direction as positive.

    Show the solution
    1. Step 1: (a) Δp = m(v_f − v_i) = 0.50(0 − 6.0) = −3.0 kg·m/s. F_avg = Δp/Δt = −3.0 ÷ 0.010 = −300 N.
    2. Step 2: (b) Δp = 0.50(−6.0 − 6.0) = −6.0 kg·m/s. F_avg = −6.0 ÷ 0.010 = −600 N.
    3. Step 3: Both forces point away from the wall. The bounce needs twice the force. The trap is computing Δp for a bounce as zero, or as m × 6.0, because the speed didn't change.

    Answer: (a) 300 N; (b) 600 N, both directed away from the wall

  2. Example 2Calculator allowed

    Impulse from a force–time graph

    A hockey stick hits a 0.16 kg puck at rest. The force on the puck rises steadily from 0 to 800 N and falls steadily back to 0, forming a triangle that lasts 0.010 s in all. Find the impulse and the puck's speed afterward.

    Show the solution
    1. Step 1: Impulse = area of the triangle = ½(0.010 s)(800 N) = 4.0 N·s.
    2. Step 2: J = Δp = mv_f − 0, so v_f = 4.0 ÷ 0.16 = 25 m/s.
    3. Step 3: The average force was 400 N, half the peak, since the area equals F_avg × Δt.

    Answer: 4.0 N·s; 25 m/s

  3. Example 3Calculator allowed

    Why airbags work

    A 70 kg driver moving at 15 m/s comes to a stop in a crash. Find the average force on the driver if the stop takes (a) 0.010 s against a hard dashboard and (b) 0.10 s with an airbag. Use g = 9.8 m/s² to compare with the driver's weight.

    Show the solution
    1. Step 1: The driver's change in momentum is the same either way: Δp = 70(0 − 15) = −1050 kg·m/s.
    2. Step 2: (a) F_avg = 1050 ÷ 0.010 = 105,000 N in size.
    3. Step 3: (b) F_avg = 1050 ÷ 0.10 = 10,500 N in size.
    4. Step 4: Ten times the stopping time means one-tenth the force. The driver's weight is (70)(9.8) = 686 N, so the airbag force is about 15 times the weight instead of about 150.

    Answer: (a) about 1.1 × 10⁵ N; (b) about 1.1 × 10⁴ N

Common mistakes

  • Forgetting the sign change when an object bounces, which underestimates Δp.
  • Claiming that an airbag reduces the change in momentum. It reduces the force by increasing the time; Δp is the same.
  • Using the peak force instead of the average force in J = F_avg Δt. Use the area under the graph instead.
  • Confusing impulse with force. Impulse includes time and has units of N·s.

On the exam

  • Expect force–time graphs: find the area for the impulse, then the change in velocity. Count areas below the axis as negative.
  • Qualitative questions often ask why a padded surface or a longer follow-through changes the outcome. Answer with Δp = F_avg Δt and say which quantity stays fixed.
  • In lab questions, a force sensor gives F against t. The area under it can be compared with mΔv measured by a motion sensor.

Connected topics

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Check yourself

4 questions on 4.2 Change in Momentum and Impulse. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A bat exerts an average force of 1200 N on a 0.15 kg ball for 0.0050 s. The ball starts from rest. What is its speed when it leaves the bat? Ignore other forces during the hit.

A 0.50 kg ball moving horizontally at 6.0 m/s hits a wall head-on and bounces straight back at 4.0 m/s. The ball is in contact with the wall for 0.010 s.

Described situation

Question 2 of 4Calculator allowed

What is the magnitude of the impulse the wall exerts on the ball?

Question 3 of 4Calculator allowed

What is the magnitude of the average force the wall exerts on the ball?

Question 4 of 4Calculator allowed

A lump of clay with the same mass hits the wall at the same 6.0 m/s and sticks. How does the impulse on the clay compare with the impulse on the ball?

0 of 4 answered