2.24Momentum is a property of moving objectsHigher tier
Momentum is defined by the equation:
momentum = mass × velocity p = m v
momentum, p, in kilograms metre per second, kg m/s; mass, m, in kilograms, kg; velocity, v, in metres per second, m/s.
You should be able to recall and apply this equation.
Because velocity is a vector, momentum is a vector too — it has a direction. That matters in every calculation below: momentum to the left counts as negative if momentum to the right is positive.
Worked example
Calculate the momentum of a 1200 kg car travelling at 15 m/s.
- Write the equation: p = m v.
- Substitute: p = 1200 × 15.
- = 18 000.
- Give the unit: kilograms metre per second.
Answer: 18 000 kg m/s
- Momentum
- mass × velocity, measured in kilograms metre per second; because velocity is a vector, momentum has a direction too
2.25Conservation of momentumHigher tier
In a closed system, the total momentum before an event is equal to the total momentum after the event. This is called conservation of momentum.
A closed system means nothing enters or leaves and no external force acts — so however the momentum is shared between the objects, the total is unchanged.
You should be able to use the concept of momentum as a model to describe and explain examples of momentum in an event, such as a collision. Physics only students should also be able to complete calculations involving an event, such as the collision of two objects.
Worked example
A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 4.0 kg trolley and they stick together. Calculate their common velocity afterwards.
- Total momentum before = (2.0 × 3.0) + (4.0 × 0) = 6.0 kg m/s.
- Momentum is conserved, so the total after is also 6.0 kg m/s.
- After the collision the two move together, so the combined mass is 2.0 + 4.0 = 6.0 kg.
- p = m v, so v = p ÷ m = 6.0 ÷ 6.0.
- Check it is sensible: the moving trolley has picked up extra mass, so it must slow down — and it has.
Answer: 1.0 m/s in the original direction of motion
Worked example
A stationary 60 kg skater pushes a 40 kg skater away at 3.0 m/s. Calculate the velocity of the 60 kg skater.
- Before the push both are stationary, so the total momentum is zero.
- Momentum is conserved, so the total afterwards must also be zero.
- Momentum of the 40 kg skater: 40 × 3.0 = 120 kg m/s in one direction.
- For the total to be zero, the 60 kg skater must have 120 kg m/s in the opposite direction.
- v = p ÷ m = 120 ÷ 60.
Answer: 2.0 m/s in the opposite direction — which is why the total is still zero.
Common mistake
Momentum is a vector, so opposite directions must carry opposite signs. Adding two momenta as if both were positive when the objects move towards each other gives the wrong total every time.
- Conservation of momentum
- in a closed system, the total momentum before an event is equal to the total momentum after the event
2.26Changes in momentumHigher tier
Physics only, and Higher tier only — not on the Combined Science specification
When a force acts on an object that is moving, or able to move, a change in momentum occurs.
The equations F = m × a and a = (v − u) / t combine to give the equation:
F = m ∆v / ∆t
where m∆v is the change in momentum — that is, force equals the rate of change of momentum.
This equation is given on the Physics equation sheet.
Seeing where it comes from makes it easier to use. Newton's Second Law says F = m a, and acceleration is the change in velocity divided by the time taken. Substituting one into the other gives force as the change in momentum divided by the time over which it happens.
You should be able to explain safety features such as air bags, seat belts, gymnasium crash mats, cycle helmets and cushioned surfaces for playgrounds with reference to the concept of rate of change of momentum.
Every one of them works the same way, and the explanation is worth learning as a single sentence you then apply. In a collision the change in momentum is fixed — the person is going to be brought to a stop either way. Because force is the rate of change of momentum, increasing the time over which that change happens reduces the force. Each safety feature increases the stopping time, so the force on the body is smaller and the injury is less severe.
Worked example
A 70 kg passenger travelling at 14 m/s is brought to rest in a crash. Calculate the force with a stopping time of 0.10 s, and with an air bag extending it to 0.50 s.
- Change in momentum: m∆v = 70 × (14 − 0) = 980 kg m/s in both cases.
- Without the air bag: F = 980 ÷ 0.10 = 9800 N.
- With the air bag: F = 980 ÷ 0.50 = 1960 N.
- Compare: the change in momentum is identical, but the force is five times smaller.
- That is exactly the factor by which the stopping time was increased.
Answer: 9800 N without the air bag and 1960 N with it — five times less force, because the same change in momentum is spread over five times the duration.
You should also be able to apply equations relating force, mass, velocity and acceleration to explain how the changes involved are inter-related — that is, to move between F = m a, a = ∆v / t and F = m ∆v / ∆t as the question requires.
Common mistake
Safety features do not reduce the change in momentum — the person still stops. They increase the time over which the change happens, and that is what reduces the force.
Before you practise
- momentum = mass × velocity, in kg m/s, and it is a vector.
- In a closed system the total momentum before an event equals the total momentum after it.
- Set a direction as positive before you start, and give momentum in the opposite direction a minus sign.
- force = change in momentum ÷ time taken.
- A longer impact time means a smaller force for the same change in momentum — that is what crumple zones, air bags and crash mats do.
- Two objects that stick together share one combined mass and one velocity after the collision.