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5.7.3 Changes in momentum

5.7.3 Changes in momentum

Force and change in momentum

Definition

Momentum

Momentum is the product of an object's mass and velocity, p=mvp = mvp=mv, measured in kg m/s.

  1. A force on an object that is moving, or able to move, causes it to accelerate.
  2. Acceleration is a change in velocity, so the object's momentum changes too.
  3. For a constant mass the change in momentum is Δp=mΔv\Delta p = m\Delta vΔp=mΔv, where Δp\Delta pΔp is in kg m/s, mmm in kg and Δv\Delta vΔv in m/s.
  4. The velocity change can be written v−uv - uv−u, so Δp=m(v−u)\Delta p = m(v - u)Δp=m(v−u), where uuu is the initial and vvv the final velocity.
  5. Momentum has direction, so a negative change in momentum acts opposite to the positive direction.

Force as the rate of change of momentum

  1. Newton's second law is F=maF = maF=ma, and acceleration is a=(v−u)/Δt=Δv/Δta = (v - u)/\Delta t = \Delta v/\Delta ta=(v−u)/Δt=Δv/Δt.
  2. Substituting the acceleration gives F=mΔv/ΔtF = m\Delta v/\Delta tF=mΔv/Δt.
  3. Because mΔvm\Delta vmΔv is the change in momentum, this becomes F=Δp/ΔtF = \Delta p/\Delta tF=Δp/Δt.
  4. For a fixed mass, a larger velocity change in the same time gives a larger force.
  5. For a fixed velocity change, a longer time gives a smaller force.
  6. For the same velocity change and time, a greater mass gives a greater change in momentum and a greater force.
Key Idea

The resultant force on an object equals its rate of change of momentum, so a larger change in momentum, or the same change in a shorter time, gives a larger force.

Example

An air bag brings a 70kg70 kg70kg passenger travelling at 15m/s15 m/s15m/s to rest in 0.30s0.30 s0.30s. Find the average resultant force.

  1. State the equation: F=m(v−u)/ΔtF = m(v - u)/\Delta tF=m(v−u)/Δt.
  2. Find the change in momentum: m(v−u)=70×(0−15)=−1050m(v - u) = 70 \times (0 - 15) = -1050m(v−u)=70×(0−15)=−1050 kg m/s.
  3. Substitute and calculate: F=−1050/0.30=−3500F = -1050/0.30 = -3500F=−1050/0.30=−3500 N, a magnitude of 350035003500 N opposite to the motion.
  4. With a shorter stopping time of 0.0300.0300.030 s: F=−1050/0.030=−35,000F = -1050/0.030 = -35{,}000F=−1050/0.030=−35,000 N, ten times larger.

Safety features and stopping time

  1. During a collision a person or object must lose momentum as it comes to rest.
  2. Safety features reduce the force by increasing the time over which the momentum change happens.
  3. Air bags inflate and compress, increasing the time taken for a passenger to stop.
  4. Seat belts stretch slightly, increasing the stopping time compared with hitting the vehicle.
  5. Crumple zones at the front and rear of a car are designed to crush and deform in a collision, increasing the time over which the car and its occupants come to rest.
  6. Gymnasium crash mats compress as a person lands, increasing the time to stop.
  7. Cycle helmets contain material that compresses on impact, increasing the stopping time of the head.
  8. Cushioned playground surfaces deform on impact, increasing the stopping time of a falling person.
  9. In each case the change in momentum is spread over a longer time, lowering the rate of change of momentum and so the average force, reducing the risk of injury.
Common Mistake
  • A safety feature does not reduce the total change in momentum: the same initial velocity brought to rest gives the same change in momentum.
  • Do not say a safety feature absorbs the force; say it increases the stopping time, which reduces the rate of change of momentum and therefore the average force.
Exam technique
  • Build the chain: the person is brought to rest, the feature increases the stopping time, the rate of change of momentum falls, and since F=Δp/ΔtF = \Delta p/\Delta tF=Δp/Δt the average force falls, reducing injury.
  • Calculate change in momentum with Δp=m(v−u)\Delta p = m(v - u)Δp=m(v−u) and watch the sign, since a negative force acts opposite to the motion.
Self review
  • What equation links resultant force, change in momentum and time?
  • Why does increasing stopping time reduce the average force?
  • How do you calculate a change in momentum?
  • Why can a negative force appear in a stopping calculation?
  • Name three safety features that increase stopping time.
  • How does mass affect force for a fixed velocity change and time?
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Momentum is the product of an object's mass and velocity. It is calculated using p=mvp = mvp=mv, where ppp is momentum in kg m/s\text{kg m/s}kg m/s, mmm is mass in kg\text{kg}kg and vvv is velocity in m/s\text{m/s}m/s.

Momentum is a vector quantity, so it has both magnitude and direction. Choosing a positive direction means that motion or momentum in the opposite direction is represented by a negative value.

A force causes acceleration, which is a change in velocity. For constant mass, a change in velocity produces a change in momentum:

Δp=mΔv=m(v−u) \Delta p = m\Delta v = m(v-u) Δp=mΔv=m(v−u)

Here, uuu is the initial velocity and vvv is the final velocity.

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What equation gives momentum from mass and velocity?

5.7.3 Changes in momentum Revision Guide

  1. GCSE
  2. /Physics
  3. /5.7.3 Changes in momentum

Revision notes for AQA GCSE Physics 5.7.3 Changes in momentum: explanations and worked examples.