In Activity 5-1-1 you learned that momentum is p = mv. Now ask what changes it. From Unit 2 you know that a net force gives an object an acceleration, F = ma. Acceleration is the change in velocity per unit time, a = Δv ÷ Δt. Put them together:
F = m Δv ÷ Δt → F Δt = m Δv = Δp
The product F × Δt is called impulse, with the symbol J and the unit newton-second (N·s). The equation says that the impulse on an object equals its change in momentum. That is the impulse-momentum theorem:
J = F Δt = Δp = m(v_f − v_i)
💡Since 1 N = 1 kg·m/s², a newton-second works out to kg·m/s. Impulse and momentum are measured in the same unit, because one is the change in the other.
BMath
Using the Theorem
Take the direction of the initial motion as positive, and give signs to forces and to changes in momentum. A force that slows an object down is negative.
Example 1: find the final speed
Given
A 2.0 kg cart starts from rest
A steady 4.0 N force pushes it for 5.0 s
Solve
J = F Δt = 4.0 × 5.0 = 20 N·s
J = m(v_f − v_i) → 20 = 2.0 × (v_f − 0)
v_f = 20 ÷ 2.0
v_f = 10 m/s
✓ The cart ends up moving at 10 m/s.
Example 2: an airbag
Given
A 70 kg driver moving at 15 m/s is brought to rest
Without an airbag the stop takes 0.020 s, and with one it takes 0.10 s
Solve
Δp = m(v_f − v_i) = 70 × (0 − 15) = −1,050 kg·m/s
No airbag: F = Δp ÷ Δt = −1,050 ÷ 0.020 = −52,500 N
Airbag: F = Δp ÷ Δt = −1,050 ÷ 0.10 = −10,500 N
✓ The airbag takes the same momentum change but spreads it over 5 times as long, so the force is 5 times smaller.
Example 3: a rebound
Given
A 0.50 kg ball hits a wall at +8.0 m/s and rebounds at −6.0 m/s
If it had just stuck: Δp = 0.50 × (0 − 8.0) = −4.0 kg·m/s, F = −200 N
✓ The wall pushes on the bouncing ball with an average of 350 N, compared with 200 N for a ball that stuck.
CConcept
Why Stretching the Time Helps
For a given stop, the change in momentum is fixed by the object’s mass and speed. You can’t change it. But the impulse equation, F = Δp ÷ Δt, lets you choose how it’s split between force and time. Make the time longer and the force drops in proportion. That is the whole idea behind a long list of safety devices:
Safety device
What it lengthens
Airbag
The time it takes your body to stop in a crash
Crumple zone
The distance, and so the time, the car takes to stop
Bending your knees to land
The time your body takes to slow down on landing
Catching a ball with “give”
The time your hand takes to stop the ball
Foam helmet liner
The time your head takes to stop in an impact
Gym landing mat
The time a gymnast takes to stop after a fall
🔑This is the answer to the unit’s opening question. A padded bumper and a rigid one bring a car to rest from the same speed, so the change in momentum is identical. The padded one just takes longer to do it, so the force on the car, and on you, is smaller.
DExample
Force-Time Graphs: Area Is Impulse
On a graph of force against time, the area under the curve is the impulse. A steady force makes a rectangle. A real impact, with a force that builds and then fades, makes a hill. Either way, the area is the change in momentum. Try it.
Real impacts rarely have a perfectly steady force. They rise and fall. Pick a pulse shape and set its peak force and duration. The area under the curve is the impulse, and the impulse is the change in momentum.
Pulse shape:
Impulse = area = 200 × 0.020 = 4.00 N·s Δp = 4.00 kg·m/s · for a 0.058 kg egg, Δv = 69.0 m/s
For a steady force, the impulse is just force times time.
EMath
Estimating a Stop from Its Distance
It’s hard to time an impact that lasts a few thousandths of a second. It’s much easier to measure how far an object sinks into the thing that stops it. If we assume a steady stopping force, the object slows evenly from v to 0 over a crush distance d. Its average speed is v ÷ 2, so:
Δt = d ÷ (v ÷ 2) = 2d ÷ v
Put that into F = Δp ÷ Δt = mv ÷ Δt and you get:
F = m v² ÷ (2d)
This should look familiar. It’s exactly what the work-energy theorem gives you from Unit 4: F × d = ½ m v². The two big ideas agree. Impulse-momentum counts the force over time, and work-energy counts it over distance.
An egg dropped from 3.0 m
Given
Egg mass m = 0.058 kg, dropped from h = 3.0 m
Compare a crush distance of 2.0 cm with 5.0 cm
Solve
v = √(2gh) = √(2 × 9.8 × 3.0) = 7.67 m/s
p = m v = 0.058 × 7.67 = 0.445 kg·m/s
2.0 cm: Δt = 2d ÷ v = 2(0.020) ÷ 7.67 = 5.2 ms, F = 0.445 ÷ 0.0052 = 85 N
5.0 cm: Δt = 2(0.050) ÷ 7.67 = 13.0 ms, F = 0.445 ÷ 0.0130 = 34 N
✓ Going from 2 cm to 5 cm of cushioning cuts the force from about 85 N to 34 N, since both the momentum change and the impact speed stayed the same.
⚠️This gives an average force. Real impacts have a peak force that’s higher, as the force-time graph above showed. Still, the average is a good first estimate, and a design that lowers the average usually lowers the peak too.
FMath
Bounces Take More Impulse
When an object bounces, its velocity doesn’t just go to zero. It reverses. Taking up as positive, a ball arrives at −v_before and leaves at +v_after, so:
Δp = m(v_after + v_before)
A perfect bounce, with v_after equal to v_before, would take twice the impulse of a ball that just stops. Real balls fall between the two. You can get each speed from the heights using v = √(2gh).
A bouncing ball
Given
m = 0.10 kg, dropped from h₁ = 1.0 m
It rebounds to h₂ = 0.64 m
Solve
v_before = √(2 × 9.8 × 1.0) = 4.43 m/s
v_after = √(2 × 9.8 × 0.64) = 3.54 m/s
Δp = 0.10 × (3.54 + 4.43) = 0.797 kg·m/s
If it had stuck: Δp = 0.10 × 4.43 = 0.443 kg·m/s
Ratio = 0.797 ÷ 0.443 = 1.8
✓ The bounce needed 1.8 times the impulse of a stop, and the force on the ball is larger for the same contact time.
GWatch Out
Three Common Mix-Ups
⚠️“A cushion reduces the change in momentum.” It doesn’t. The change in momentum depends only on the mass and the speeds before and after. A cushion changes how long the change takes, and so the force.
⚠️“A bigger change in momentum always means a bigger force.” Only if the time stays the same. F = Δp ÷ Δt, so a big Δp spread over a long time can give a small force.
⚠️“Impulse is a kind of force.” It isn’t. Impulse is force times time, measured in N·s, and it equals a change in momentum. A force is measured in newtons.
HConcept
Quick Check
Four questions. They are not graded. Answer them to find out what still needs work.
Q1A cart’s momentum changes from +4.0 to +10.0 kg·m/s because of a steady 3.0 N force. For how long did the force act?
Q2Two identical cars at the same speed are brought to rest. In crash A the stop takes 0.1 s, and in crash B it takes 0.5 s. How do the average forces compare?
Q3A ball hits the floor at the same speed in two trials. In one it stops dead, and in the other it bounces back at the same speed. Which has the bigger change in momentum?
Q4An egg lands with a momentum of 0.45 kg·m/s. To keep the average force at or below 45 N, the stop must take at least:
Score: 0 of 4
IConcept
Looking Ahead
You now know how a force over time changes one object’s momentum. Next you’ll see what happens when two objects push on each other.