3.4 Momentum — Learn

Prerequisites: Section 3.1, especially velocity; Section 3.2, especially $F = ma$ and Newton's third law; rearranging a formula.

By the end of this section you can: define and calculate momentum; use conservation of momentum in a collision; calculate a force from the rate of change of momentum; and explain how safety features reduce the force in a crash.

1. Momentum

Momentum is a property of a moving object. It is the product of the mass and the velocity:

p = m v

where $p$ is the momentum in kilogram metres per second (kg m/s), $m$ is the mass in kilograms, and $v$ is the velocity in metres per second.

Momentum is a vector, so its direction matters. A car moving north and an identical car moving south at the same speed have equal and opposite momenta.

Worked example 1. A car of mass 1200 kg travels at 15 m/s. Calculate its momentum.

\begin{aligned} p &= m v \\ &= 1200 \times 15 \\ &= 18000\,\text{kg m/s} \\ &= 1.8 \times 10^4\,\text{kg m/s} \end{aligned}

Worked example 2. In the chapter problem the bus has a mass of 12000 kg and travels at 12 m/s. Calculate its momentum.

\begin{aligned} p &= m v \\ &= 12000 \times 12 \\ &= 144000\,\text{kg m/s} \\ &= 1.44 \times 10^5\,\text{kg m/s} \end{aligned}

2. Conservation of momentum

A closed system is a system that no mass and no energy enter or leave. In a closed system, the total momentum before an event equals the total momentum after the event. This is the conservation of momentum.

In a collision, the objects push on each other. By Newton's third law the two forces are equal and opposite, so one object gains the momentum that the other one loses. The total momentum does not change.

To use conservation of momentum:

add the momenta before the event, with a sign for the direction;
add the momenta after the event, with a sign for the direction;
set the two totals equal.

Worked example 3. In the chapter problem the bus of mass 12000 kg travels at 12 m/s. It hits a stationary van of mass 8000 kg, and the two move off together. Calculate their common velocity.

The total momentum before the collision is the momentum of the bus:

p = 12000 \times 12 = 144000\,\text{kg m/s}

After the collision the two vehicles move together with a combined mass of $12000 + 8000 = 20000\,\text{kg}$ .

\begin{aligned} \text{total momentum before} &= \text{total momentum after} \\ 144000 &= 20000 \times v \\ v &= \frac{144000}{20000} \\ &= 7.2\,\text{m/s} \end{aligned}

Worked example 4. A trolley of mass 4.0 kg moves at 3.0 m/s and hits a stationary trolley of mass 2.0 kg. The two trolleys stick together. Calculate their common velocity.

\begin{aligned} 4.0 \times 3.0 &= (4.0 + 2.0) \times v \\ 12 &= 6.0 v \\ v &= 2.0\,\text{m/s} \end{aligned}

3. The force from a change in momentum

When a force acts on a moving object, the momentum of the object changes. Newton's second law links the force to the rate of change of momentum.

F = \frac{m \Delta v}{\Delta t}

where $m\Delta v$ is the change in momentum and $\Delta t$ is the time taken. The force equals the rate of change of momentum.

You can see where this comes from. Write $F = ma$ , and replace the acceleration $a$ with $\dfrac{\Delta v}{\Delta t}$ .

Worked example 5. A car of mass 1200 kg travels at 15 m/s and stops in 0.30 s. Calculate the force on the car.

\begin{aligned} F &= \frac{m \Delta v}{\Delta t} \\ &= \frac{1200 \times (15 - 0)}{0.30} \\ &= \frac{18000}{0.30} \\ &= 60000\,\text{N} \end{aligned}

Worked example 6. In a crash test, the same car stops in 0.60 s because the front of the car crumples. Calculate the force now.

\begin{aligned} F &= \frac{m \Delta v}{\Delta t} \\ &= \frac{1200 \times 15}{0.60} \\ &= \frac{18000}{0.60} \\ &= 30000\,\text{N} \end{aligned}

The change in momentum is the same in both crashes, but the force is halved when the time is doubled. A longer time means a smaller force.

Try it: the force in a crash

The figure shows a vehicle in a collision. The mass and the speed are fixed, so the change in momentum is fixed. You can change how long the collision lasts.

Predict first. Do not move the slider yet. If the collision lasts twice as long, what happens to the force? Write down your prediction.

Test. Set the collision time to 0.10 s and read the force. Now set it to 0.20 s, and then to 0.40 s, and read the force each time.

Explain. The change in momentum is the same on every setting. Why does a longer collision time give a smaller force? Write your answer in two or three sentences.

Watch out. Momentum is a vector, so opposite directions need opposite signs. If one object moves to the right and another moves to the left, give one momentum a positive sign and the other a negative sign before you add them. A common error is to add the sizes and lose the direction. Also keep the unit as kg m/s, and check that the mass is in kilograms before you multiply.

4. Safety features

A safety feature works by making the change in momentum take longer, so the force is smaller. The change in momentum itself is fixed by the mass and the speed, so the only way to reduce the force is to increase the time.

Safety feature	How it helps
Seat belt	Stretches a little, so the wearer stops over a longer time and over a larger area.
Air bag	Inflates to give a soft, slow stop for the head and chest.
Crumple zone	The front of the car crumples, so the whole car stops over a longer time.
Cycle helmet	The foam crushes, so the head stops over a longer time.
Crash mat	Compresses when a gymnast lands, so the body stops over a longer time.

Each feature increases $\Delta t$ in $F = \dfrac{m \Delta v}{\Delta t}$ . A smaller force means a smaller chance of injury.

Progress check

Return to the chapter problem. The bus has a mass of 12000 kg and travels at 12 m/s. It hits the stationary 8000 kg van, and the two move off together. Work out the total momentum before the collision, then work out the velocity of the two vehicles afterwards.

You should now be able to

define momentum and calculate it from mass and velocity;
use the conservation of momentum in a collision or an explosion;
calculate the force from the rate of change of momentum, using $F = \dfrac{m \Delta v}{\Delta t}$ ;
explain how seat belts, air bags, crumple zones, cycle helmets and crash mats reduce the force in a crash.