Impulse — force acting over time — Physics

Before this

This leads to

Explosions and recoil
Work done by a variable force · coming soon

Same ball, same speed, very different sting

A cricket ball of mass 0.16 kg flies towards you at 30 m/s. You catch it with a bare hand held stiffly, and it stings badly. Next time you wear a padded glove and let your hand move back as you catch. It hardly hurts. What changed?

Not the ball's momentum. In both catches the ball goes from 30 m/s to rest, so its momentum changes by $0.16 \times 30 = 4.8$ kg m/s each time. What changed is the time the stop took. The last lecture compared momentum just before and just after a collision. This lecture looks inside the collision itself: how big the force is, how long it lasts, and why making it last longer can save lives.

Force times time

The first momentum lecture wrote Newton's second law in its momentum form: the resultant force equals the rate of change of momentum.

F = \frac{Δ p}{Δ t}

Multiply both sides by $Δ t$ :

F Δ t = Δ p

The left-hand side, force multiplied by the time it acts, is called the impulse. Many books use the letter $J$ for it. (Do not confuse it with J for joules: one is a quantity, the other a unit.) The right-hand side is the change in momentum, $m v - m u$ .

Impulse

Impulse $= F Δ t = Δ p = m v - m u$

The impulse of a force equals the change in momentum it causes. Like momentum, impulse has a direction, so give it a sign.

What about units? Force times time gives newton seconds, written N s. Momentum is measured in kg m/s. The first momentum lecture said these are the same unit, and promised to show why. A newton is the force that gives 1 kg an acceleration of 1 m/s², so $1 N = 1 kg m/s^{2}$ . Multiply by seconds:

1 N s = 1 kg m/s^{2} \times s = 1 kg m/s

They are the same unit, written two ways. This lecture uses N s for impulse and kg m/s for momentum, but you can swap them freely.

Why the glove works

Now rearrange the other way: $F = Δ p \div Δ t$ . For the cricket ball, $Δ p$ is fixed at 4.8 kg m/s by the ball's mass and speed. You cannot change it. But you can change $Δ t$ . Suppose the stiff bare hand stops the ball in 0.01 s, and the glove, with the hand moving back, stops it in 0.05 s. (These times are illustrative, not measured.)

Bare hand: average force $= 4.8 \div 0.01 = 480$ N.
Glove: average force $= 4.8 \div 0.05 = 96$ N.

Five times as long means one fifth of the force. That is it. That is the whole idea. For the same change in momentum, a longer stop means a smaller force. Explore it below.

Explore — same impulse, different stopping times

The 0.16 kg cricket ball at 30 m/s must lose 4.8 kg m/s of momentum. Choose how long the stop takes. The shaded area is the impulse. Watch the force change while the area stays the same.

Stopping time: 0.01 s

The force needed to stop the ball (impulse 4.8 N s every time), assuming the force rises steadily to a peak and falls back, so the graph is a triangle:

Stopping time	Average force	Peak force
0.005 s	960 N	1920 N
0.01 s	480 N	960 N
0.05 s	96 N	192 N
0.1 s	48 N	96 N

The dashed grey line in the explorer is the average force. A real force in a catch rises and falls, so its peak is larger than its average. For a triangle-shaped graph, the peak is exactly twice the average. Injuries depend on the peak, so spreading the stop over a longer time protects you twice over: the average falls, and the peak falls with it.

Stop and think

A seat belt that stretches a little during a crash keeps the passenger safer. Does it reduce the passenger's change in momentum?

Show the answer

No. The passenger starts at the car's speed and ends at rest, whatever they hit. So their change in momentum, and the impulse they receive, are fixed. What the belt changes is the time. By stretching, it makes the stop last longer, so the same impulse is delivered by a smaller force. Without the belt, the passenger keeps moving (Newton's first law) until something stops them very quickly, such as the windscreen.

Crumple zones, airbags and helmets

In a crash, each car must lose its momentum. Recall the two cars from the last lecture that locked together and stopped dead. Each car's change in momentum was fixed by its mass and speed. Car designers cannot change that. What they can change is how long the stop takes.

A car's front end is built to crumple in a controlled way, so the car takes longer to stop than if its front were rigid. An airbag inflates and then lets its gas out slowly, so the passenger's head slows down over a longer time and a larger area. A helmet's foam liner crushes on impact, stretching out the time the head takes to stop. Gymnasts land with bent knees and roll; judo students are taught to fall with a slap and a roll. All of these use the same rule: $F = Δ p \div Δ t$ .

Predict, then reveal — an egg on concrete and on a cushion

Two identical eggs are dropped from the same height. One lands on concrete and breaks. The other lands on a thick cushion and stops without bouncing. Which statement is true?

Both eggs receive the same impulse, because both go from the same speed to rest. The cushion makes the stop last much longer, so the force on that egg is much smaller. The concrete stops its egg in a tiny time with a large force, which breaks it.

The area under a force–time graph

When the force is constant, impulse is simply $F \times Δ t$ . On a graph of force against time, that is the area of a rectangle. When the force changes, as it does in every real collision, split the time into many tiny steps. In each step the force is almost constant, so each step adds a thin rectangle of area. Add them all up and you get the area under the curve. So, whatever the shape: impulse = area under the force–time graph.

Worked examples

Worked Example 1 — A ball that bounces

A 0.5 kg ball travelling at 20 m/s hits a wall and bounces straight back at 15 m/s. The contact lasts 0.02 s. Find the impulse on the ball and the average force on it.

Take the ball's original direction as positive. Then $u = + 20$ m/s and $v = - 15$ m/s.

Impulse = m v - m u = 0.5 (- 15) - 0.5 (20) = - 7.5 - 10 = - 17.5 N s

The minus sign means the impulse points back the way the ball came: the wall pushes it back. Average force $= - 17.5 \div 0.02 = - 875$ N, so 875 N away from the wall. By Newton's third law, the ball pushes the wall with 875 N the other way.

Notice that the bounce makes the impulse bigger than a simple stop would. Stopping the ball would need only 10 N s. Sending it back needs 7.5 N s more.

Worked Example 2 — Reading a force–time graph

A 2 kg object starts from rest. A force acts on it as shown: it rises steadily from 0 to 600 N in 0.1 s, stays at 600 N until 0.3 s, then stops. Find the object's final speed.

The impulse is the area under the graph: a triangle of ½ × 0.1 × 600 = 30 N s, then a rectangle of 600 × 0.2 = 120 N s. The total is 150 N s.

Impulse $= 30 + 120 = 150$ N s. This equals the change in momentum: $150 = 2 v - 2 \times 0$ , so $v = 75$ m/s.

Worked Example 3 — A passenger in a crash

A 70 kg passenger is in a car travelling at 15 m/s when it crashes. Compare the average force on the passenger if a seat belt and airbag stop them in 0.1 s, with the force if they hit the dashboard and stop in 0.01 s. (The times are illustrative.)

Change in momentum: $70 \times 15 = 1050$ kg m/s in both cases.

With belt and airbag: $F = 1050 \div 0.1 = 10500$ N. Against the dashboard: $F = 1050 \div 0.01 = 105000$ N. Ten times the time, one tenth of the force. The belt also spreads the force over the strong parts of the body, the hips and the chest, instead of the head.

Impulse inside a collision

Go back to the first worked example of the last lecture: a 3 kg cart at 6 m/s hits a 2 kg cart at rest, and they stick together at 3.6 m/s. Now look at the impulses. Cart A's momentum fell from 18 to $3 \times 3.6 = 10.8$ kg m/s, so it received an impulse of $- 7.2$ N s. Cart B's momentum rose from 0 to $2 \times 3.6 = 7.2$ kg m/s, so it received $+ 7.2$ N s. Equal and opposite, exactly as the third law requires.

Impulse also tells you the force, which conservation of momentum alone could not. If the carts were in contact for 0.05 s, the average force between them was $7.2 \div 0.05 = 144$ N. If a springy buffer made the contact last 0.2 s, the force would drop to 36 N. The final velocity, 3.6 m/s, would be the same either way. Momentum tells you where the carts end up; impulse tells you how hard the journey was.

Stop and think

The same idea works in reverse. Suppose a space probe's engine can push with a force of only 0.1 N. How could such a tiny force ever change the probe's speed much?

Show the answer

By acting for a very long time. Impulse is force times time, so a small force can deliver a large impulse if it keeps going. Firing for one day, 86 400 s, gives $0.1 \times 86400 = 8640$ N s. For a probe of mass 1000 kg, that changes its speed by $8640 \div 1000 = 8.64$ m/s. Keep firing for weeks and the change becomes large. (These numbers are illustrative.) Safety equipment stretches the time to make a force smaller; a gentle engine stretches the time to make an impulse bigger.

Three classic mistakes

Trap 1 — Ignoring the sign when an object bounces

In Worked Example 1, writing $0.5 \times 15 - 0.5 \times 20 = - 2.5$ N s treats the bounce as if the ball were still moving forwards. The velocity after is $- 15$ m/s, and the impulse is $- 17.5$ N s, seven times bigger.

Trap 2 — "A cushion reduces the impulse"

A cushion, helmet or airbag does not reduce the impulse or the change in momentum. Those are fixed by the speed and mass. It increases the time, which reduces the force.

Trap 3 — Peak force × time

When the force changes, the impulse is the area under the graph, not the peak force multiplied by the time. For a triangle-shaped pulse, peak × time is twice the real impulse.

Try it yourself

Give each impulse a sign where the direction matters. Use g = 9.8 N/kg.

A 0.43 kg football at rest is kicked, leaving the boot at 25 m/s. The boot touches the ball for 0.01 s. Find the impulse and the average force.

Hint 1
Impulse $= m v - m u$ with $u = 0$ .

Hint 2
Average force = impulse ÷ time.

Full answer
Impulse $= 0.43 \times 25 - 0 = 10.75$ N s in the direction of the kick. Average force $= 10.75 \div 0.01 = 1075$ N.
A fielder catches the 0.16 kg cricket ball at 30 m/s, letting the hands move back so the catch takes 0.08 s. Find the average force. How long would the catch need to take for the average force to be only 40 N?

Hint 1
The change in momentum is 4.8 kg m/s.

Hint 2
Rearrange $F = Δ p \div Δ t$ to $Δ t = Δ p \div F$ .

Full answer
$F = 4.8 \div 0.08 = 60$ N. For 40 N: $Δ t = 4.8 \div 40 = 0.12$ s.
A 0.5 kg trolley moves at 2 m/s. A push acts on it in its direction of motion: the force rises steadily from 0 to 30 N and falls back to 0, over a total of 0.2 s (a triangle on the force–time graph). Find its new speed.

Hint 1
Impulse = area of the triangle = ½ × base × height.

Hint 2
The impulse is 3 N s. Then $3 = 0.5 v - 0.5 \times 2$ .

Full answer
Impulse $= \frac{1}{2} \times 0.2 \times 30 = 3$ N s. So $0.5 v = 3 + 1 = 4$ , and $v = 8$ m/s.
A 60 kg person jumps off a wall and lands at 5 m/s. Landing stiff-legged, they stop in 0.02 s. Bending their knees, they stop in 0.2 s. Find the average force from the ground in each case. Remember that the ground must also hold up the person's weight.

Hint 1
The change in momentum is 300 kg m/s, upwards. The resultant force is this divided by the time.

Hint 2
Resultant force = ground force − weight, so ground force = resultant force + 588 N.

Full answer
Stiff: resultant $= 300 \div 0.02 = 15000$ N, so the ground pushes with $15000 + 588 = 15588$ N, about 26.5 times body weight. Knees bent: resultant $= 300 \div 0.2 = 1500$ N, so the ground pushes with $1500 + 588 = 2088$ N, about 3.6 times body weight. Bending the knees cuts the force by a factor of more than seven.

The bigger idea

There are two ways to add up the effect of a force. Multiply it by the time it acts, and you get the impulse, which changes momentum. Multiply it by the distance it acts over, and you get the work done, which changes kinetic energy. The two measures answer different questions. In a collision, momentum is always conserved but kinetic energy usually is not, as the last lecture showed, so impulse is the measure that collisions obey.

Impulse also explains conservation of momentum in one line. In a collision, the two objects push on each other with equal and opposite forces for the same time, so they receive equal and opposite impulses. The next lecture runs this backwards: objects that start at rest and fly apart, in explosions and recoil.

Key concepts in this lesson

Impulse = FΔt = Δp = mv − mu; it has a direction, so give it a sign.
N s and kg m/s are the same unit, because 1 N = 1 kg m/s².
For a fixed change in momentum, a longer stopping time means a smaller force: F = Δp ÷ Δt.
Crumple zones, airbags, seat belts, helmets and bent knees all work by making the stop take longer.
Impulse is the area under a force–time graph, whatever its shape.