Force changes momentum over time
Definition
Rate of change of momentum
Rate of change of momentum is the change in an object's momentum divided by the time taken for that change.
Definition
Impulse
Impulse is the product of a force and the time for which it acts, and it equals the change in momentum it produces.
- Newton's second law can be written as F=Δpt=mv−mutF=\dfrac{\Delta p}{t}=\dfrac{mv-mu}{t}F=tΔp=tmv−mu for constant mass.
- FFF is the resultant force in newtons, mmm is mass in kilograms, uuu is initial velocity, vvv is final velocity and ttt is time in seconds.
- The change in momentum is Δp=pfinal−pinitial=mv−mu\Delta p=p_{\text{final}}-p_{\text{initial}}=mv-muΔp=pfinal−pinitial=mv−mu.
- The direction of the resultant force is the direction of the change in momentum, not necessarily the direction in which the object is moving.
Signs show changes in direction
- Choose a positive direction before substituting velocities.
- A negative velocity represents motion in the opposite direction, so a rebound usually gives a larger momentum change than stopping.
Example
- Stopping: a 0.20 kg0.20\ \text{kg}0.20 kg ball moving at 12 m/s12\ \text{m/s}12 m/s is brought to rest in 0.040 s0.040\ \text{s}0.040 s.
- Δp=m(v−u)=0.20(0−12)=−2.4 kg m/s\Delta p=m(v-u)=0.20(0-12)=-2.4\ \text{kg m/s}Δp=m(v−u)=0.20(0−12)=−2.4 kg m/s.
- F=−2.40.040=−60 NF=\dfrac{-2.4}{0.040}=-60\ \text{N}F=0.040−2.4=−60 N.
- The mean force has magnitude 60 N60\ \text{N}60 N and acts opposite to the ball's initial motion.
A longer stopping time reduces force
- Rearranging gives FΔt=ΔpF\Delta t=\Delta pFΔt=Δp, so the same momentum change spread over a longer time produces a smaller mean force.
- Crumple zones, airbags, seat belts, crash mats and bending the knees on landing increase the time over which momentum changes.
- The change in momentum is fixed by the initial and final velocities, so increasing time reduces the rate of change of momentum and therefore the force.
Example
- Safety comparison: a passenger's momentum changes by 420 kg m/s420\ \text{kg m/s}420 kg m/s.
- Without an airbag, F=4200.030=1.4×104 NF=\dfrac{420}{0.030}=1.4\times10^4\ \text{N}F=0.030420=1.4×104 N.
- With an airbag, F=4200.12=3.5×103 NF=\dfrac{420}{0.12}=3.5\times10^3\ \text{N}F=0.12420=3.5×103 N.
- Increasing the stopping time by a factor of four reduces the mean force by a factor of four.
Force-time graphs give impulse
- The area under a force-time graph equals impulse, J=FΔtJ=F\Delta tJ=FΔt, when the force is constant.
- For a changing force, estimate or calculate the total area under the graph to find Δp\Delta pΔp.
- The unit of impulse is N s\text{N s}N s, which is equivalent to kg m/s\text{kg m/s}kg m/s.
Common Mistake
- Do not calculate mv+mumv+mumv+mu unless the chosen velocity signs make that the correct subtraction.
- Do not use speed without direction when an object rebounds.
- Use the resultant or mean force specified by the data, not an unrelated contact force.
Multi-stage calculations need a clear route
- If force and time are given, find momentum change using Δp=FΔt\Delta p=F\Delta tΔp=FΔt.
- If momentum change and one velocity are known, use m(v−u)=Δpm(v-u)=\Delta pm(v−u)=Δp to find the missing velocity.
Exam technique
- Write Δp=m(v−u)\Delta p=m(v-u)Δp=m(v−u) before dividing by time.
- Keep signs throughout the calculation, then describe the direction of a negative answer in words.
- When explaining a safety feature, state that it increases the stopping time for the same momentum change, which reduces the mean force.
Self review
- Write Newton's second law in momentum form.
- Why can a rebound produce a large change in momentum?
- How does an airbag reduce the force on a passenger?
- What does the area under a force-time graph represent?
