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.
- A force on an object that is moving, or able to move, causes it to accelerate.
- Acceleration is a change in velocity, so the object's momentum changes too.
- 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.
- 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.
- Momentum has direction, so a negative change in momentum acts opposite to the positive direction.
Force as the rate of change of momentum
- 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.
- Substituting the acceleration gives F=mΔv/ΔtF = m\Delta v/\Delta tF=mΔv/Δt.
- Because mΔvm\Delta vmΔv is the change in momentum, this becomes F=Δp/ΔtF = \Delta p/\Delta tF=Δp/Δt.
- For a fixed mass, a larger velocity change in the same time gives a larger force.
- For a fixed velocity change, a longer time gives a smaller force.
- 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.
- State the equation: F=m(v−u)/ΔtF = m(v - u)/\Delta tF=m(v−u)/Δt.
- 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.
- 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.
- 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
- During a collision a person or object must lose momentum as it comes to rest.
- Safety features reduce the force by increasing the time over which the momentum change happens.
- Air bags inflate and compress, increasing the time taken for a passenger to stop.
- Seat belts stretch slightly, increasing the stopping time compared with hitting the vehicle.
- 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.
- Gymnasium crash mats compress as a person lands, increasing the time to stop.
- Cycle helmets contain material that compresses on impact, increasing the stopping time of the head.
- Cushioned playground surfaces deform on impact, increasing the stopping time of a falling person.
- 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?
