PhysicsLesson 15 of 18About 45 minutes

Momentum and impulse

A soccer ball flying toward a player's head reverses direction in a fraction of a second. How hard did the head push, and for how long? Momentum and impulse are the tools for collisions like this, where forces are large and brief.

By the end of this lesson you can

calculate momentum and compare the momentum of different objects;
explain why momentum is a vector;
use the impulse-momentum theorem to relate force, time and change in momentum;
explain how airbags, crumple zones and padding reduce forces.

A change in motion

Think first

A player heads a moving soccer ball. Name two things about the ball's motion that change. What do you think decides how big the change is?

Compare your answer

Its speed and its direction both change. The size of the change depends on how hard the head pushes and on how long the head and ball stay in contact.

Heading a soccer ball

Simulation

Try this

Press Head the ball. The simulation slows down during the contact. Read how long the head touches the ball and how large the peak force is.

Momentum

Definition

The momentum of an object is its mass times its velocity:

$$\vec p = m\vec v$$

Momentum is a vector that points the same way as the velocity. Its SI unit is the kilogram meter per second (kg·m/s).

In everyday speech, a team "gains momentum" when it is hard to stop. The physics meaning is similar: an object with a lot of momentum takes a large force, a long time, or both, to bring to rest. Momentum is sometimes called the "quantity of motion".

Momentum meter

Simulation

Try this

Raise the mass, then the velocity. How does the purple momentum arrow respond to each?
Make the velocity negative. What happens to the momentum arrow?
Press Move, then try to stop the object. Which objects take longest to stop with the same force?

Comparing momentum

Roll a bowling ball and a playground ball into pins at the same speed. The bowling ball scatters them far more, because with the same velocity, more mass means more momentum. But mass is only half the story: a small object moving very fast can have as much momentum as a large slow one.

Two balls, two rows of pins

Simulation

Try this

Run all three cases. In "Equal momentum", half the mass moves at twice the speed. What happens to the pins?

Worked example 1

A 2000 kg SUV travels east at 20.0 m/s. (a) What is its momentum? (b) How fast would an 1100 kg car have to move to have the same momentum?

(a) Momentum of the SUV. $p = mv = (2000\ \text{kg})(20.0\ \text{m/s}) = 4.00 \times 10^4\ \text{kg·m/s}$, east.
(b) Set the car's momentum equal. $v = \dfrac{p}{m} = \dfrac{4.00 \times 10^4}{1100}$
(a) $4.00 \times 10^4\ \text{kg·m/s}$ east. (b) 36.4 m/s east.

Vehicles: same momentum, different speeds

Simulation

Try this

Choose each vehicle, press Move, then press the Stop it button (the same 10 000 N force every time). Compare the stopping times of the SUV and the fast small car. What do they have in common?

Impulse

Newton's second law can be written in terms of momentum. Since $\vec a = \Delta \vec v/\Delta t$, the law $\vec F = m\vec a$ becomes $\vec F = \dfrac{m\Delta \vec v}{\Delta t} = \dfrac{\Delta \vec p}{\Delta t}$. Multiply both sides by $\Delta t$:

Impulse-momentum theorem

$$\vec F\Delta t = \Delta \vec p = m\vec v_f - m\vec v_i$$

The product $\vec F\Delta t$ is called the impulse. A net force acting for a time changes an object's momentum by exactly the impulse. Units: N·s, which equal kg·m/s.

In real collisions the force rises and falls, so $\vec F$ here means the average force during the contact. On a graph of force against time, the impulse is the area under the curve.

The theorem tells you that the same change in momentum can come from a large force for a short time or a small force for a long time. That is why a fast ball stings your hand if you hold it stiff, but not if you let your hand move back with the ball.

Catching with a stiff or a soft hand

Simulation

Try this

Catch with a stiff hand, then with a soft hand. Compare the stopping times and the peak forces.
Compare the areas under the two curves. Why must they be the same?
Switch to Follow-through push. How does a longer contact time change the ball's final speed?

Worked example 2

In a crash test, a 1200 kg car moving at 18.0 m/s hits a pole and comes to rest in 0.250 s. What average force does the pole exert on the car?

Choose a direction. Take the car's direction of travel as positive: $v_i = +18.0\ \text{m/s}$, $v_f = 0$.
Change in momentum. $\Delta p = m(v_f - v_i) = (1200)(0 - 18.0) = -2.16 \times 10^4\ \text{kg·m/s}$
Average force. $$F = \frac{\Delta p}{\Delta t} = \frac{-2.16 \times 10^4}{0.250} = -8.64 \times 10^4\ \text{N}$$
$8.64 \times 10^4\ \text{N}$, directed backward (opposite the car's motion)

Crash test with a crumple zone

Simulation

Try this

Run the crash test with the numbers from Worked example 2. The dashed rectangle shows the average force. Check that it matches your answer, and notice how much higher the peak force is.

Worked example 3

A 0.450 kg volleyball moving at 10.0 m/s is hit straight back so that it leaves at 14.0 m/s. The hand is in contact with the ball for 0.0300 s. Find the average force on the ball.

Signs matter. Take the ball's final direction as positive: $v_i = -10.0\ \text{m/s}$, $v_f = +14.0\ \text{m/s}$.
Change in momentum. $\Delta p = 0.450\,(14.0 - (-10.0)) = 0.450 \times 24.0 = 10.8\ \text{kg·m/s}$. Reversing direction makes the change larger than either speed alone.
Average force. $F = \dfrac{10.8}{0.0300} = 360\ \text{N}$
360 N in the ball's final direction

Stopping times and distances

Rearranged, the theorem gives the time needed to stop: $\Delta t = \Delta p / F$. With the same braking force, an object with twice the momentum needs twice the time to stop. A loaded truck with twice the mass of an empty one, moving at the same speed, also travels twice as far while braking.

An empty truck and a loaded truck brake

Simulation

Try this

Predict the ratio of the stopping distances before you press Brake!. Then read the stopping times from the graph.

Worked example 4

An 1800 kg car moving west at 24.0 m/s brakes with a constant force of 6000 N toward the east until its speed drops to 6.0 m/s. (a) How long does this take? (b) How far does the car travel while braking?

Choose west as positive. $v_i = 24.0\ \text{m/s}$, $v_f = 6.0\ \text{m/s}$, $F = -6000\ \text{N}$.
(a) Time from the theorem. $$\Delta t = \frac{m(v_f - v_i)}{F} = \frac{1800(6.0 - 24.0)}{-6000} = 5.4\ \text{s}$$
(b) Distance. The force is constant, so the acceleration is too, and the average velocity is halfway between the start and the end: $$\Delta x = \tfrac12 (v_i + v_f)\Delta t = \tfrac12 (24.0 + 6.0)(5.4) = 81\ \text{m}$$
(a) 5.4 s. (b) 81 m to the west.

Braking with a constant force

Simulation

Try this

Press Brake to 6.0 m/s and compare the simulation with your answers to Worked example 4. Then press Brake to a stop and predict the time first: it should be $1800 \times 24.0 / 6000$ seconds.

Longer contact, smaller force

Firefighters catch people on inflatable cushions. Gymnasts land on thick mats. Cars have crumple zones and airbags. In every case the change in momentum is fixed: the person must go from moving to stopped. What the design can change is the time. Stretching the stopping time makes the average force smaller in the same proportion.

An egg dropped on a plate or a pillow

Simulation

Try this

Drop the egg on the plate, then on the pillow. Compare the stopping times and peak forces. Why is the change in momentum the same in both cases?

Check yourself

Question 1

A 0.160 kg hockey puck slides at 30.0 m/s. What is the size of its momentum?

Show solution

$p = (0.160)(30.0) = 4.80\ \text{kg·m/s}$

Question 2

A 70.0 kg person walks at 0.100 m/s. How fast would a 0.145 kg baseball have to move to have the same momentum?

Show solution

$p = (70.0)(0.100) = 7.00\ \text{kg·m/s}$; $v = 7.00 / 0.145 = 48.3\ \text{m/s}$.

Question 3

A racket gives a 57.0 g tennis ball, initially at rest, a speed of 50.0 m/s. The contact lasts 4.00 ms. What is the average force on the ball?

Show solution

$F = \dfrac{\Delta p}{\Delta t} = \dfrac{(0.0570)(50.0)}{0.00400} = 713\ \text{N}$

Question 4

A 1500 kg car moving at 20.0 m/s brakes with a constant force of 7500 N. How long does it take to stop?

Show solution

$\Delta t = \dfrac{\Delta p}{F} = \dfrac{(1500)(20.0)}{7500} = 4.0\ \text{s}$. It travels $\tfrac12(20.0 + 0)(4.0) = 40\ \text{m}$.

Question 5

How does an airbag reduce injuries in a crash?

It reduces the person's change in momentum.
It increases the person's change in momentum.
It makes the stopping take longer, so the average force is smaller.
It makes the stopping quicker, so the force acts for less time.

Show solution

The person's change in momentum is the same either way (from moving to stopped). The airbag spreads it over a longer time, and $F = \Delta p/\Delta t$ gets smaller.

Question 6

A particle's speed doubles. What happens to its momentum and its kinetic energy?

Both double.
Momentum doubles; kinetic energy becomes four times larger.
Momentum becomes four times larger; kinetic energy doubles.
Both become four times larger.

Show solution

$p = mv$ is proportional to $v$; $KE = \tfrac12 mv^2$ is proportional to $v^2$.

Summary

Momentum is mass times velocity: $\vec p = m\vec v$, a vector in kg·m/s.
Impulse-momentum theorem: $\vec F\Delta t = \Delta \vec p$. The impulse is the area under a force-time graph.
With the same force, more momentum takes more time and more distance to stop.
Increasing the time of an impact decreases the average force, which is how padding, airbags and crumple zones protect people.

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