Calculate momentum changes including rebounds
Subtract initial momentum from final momentum instead of subtracting unsigned speeds.
Section 1.6 · Momentum
Exam scope and assessment notes
Extended only (Supplement) · Physics 0625, 2026–2028, section 1.6. Define and use signed momentum p = mv, impulse FΔt = Δ(mv), conservation in simple one-dimensional systems, and resultant force as momentum change per unit time. F is the relevant resultant, with a constant force or stated average over the interval. Conservation requires negligible net external impulse on the chosen system; collision outcomes need enough supplied information. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Use momentum with signed velocity.
- Subtract negative numbers correctly.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For right positive, what momentum does 0.20 kg at 4 m/s left have?
What you will learn
- Calculate Δp = pfinal − pinitial.
- Compare stopping with reversal under the same sign convention.
The order of subtraction matters
Change in momentum is final momentum minus initial momentum. For a fixed mass, Δp = m(v − u). Taking right positive, a 0.20 kg ball changes from +6 m/s to −4 m/s during a horizontal rebound. Its momenta are +1.2 and −0.8 kg m/s, so Δp = −0.8 − 1.2 = −2.0 kg m/s. The change points left.
Subtracting the speed magnitudes as 4 − 6 would miss the reversal. A ball that instead stops from the same initial +6 m/s has Δp = 0 − 1.2 = −1.2 kg m/s. For the stated final leftward speed, the rebound therefore has a larger momentum-change magnitude than stopping, even though the final speed is smaller than the initial speed.
State change and final state separately
A 0.50 kg cart slowing from +8 to +2 m/s changes momentum by 0.50(2 − 8) = −3.0 kg m/s. Its final momentum is nevertheless +1.0 kg m/s, still rightward. A leftward change in momentum need not mean that the final motion has reversed.
For the same ball model, a change from −4 to +6 m/s gives +2.0 kg m/s rather than −2.0. The direction convention and the chronological initial and final states both matter. If the body's mass changes, calculate the actual final and initial products separately; do not silently use one unchanged m in m(v − u).
Pause and explain
A cart still moves right but slows down. Must its momentum change be rightward?
Worked example
A 0.20 kg ball rebounds horizontally from 6 m/s right to 4 m/s left. Find Δp with right positive.
Show the worked solution
- Initial momentum = 0.20 × 6 = +1.2 kg m/s.
- Final momentum = 0.20 × (−4) = −0.8 kg m/s.
- Change = −0.8 − 1.2 = −2.0 kg m/s.
Answer −2.0 kg m/s, directed left.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find Δp for 0.20 kg changing from +6 to −4 m/s.
Give me a hint
Write both signed momenta, then subtract in chronological order.
Compare my reasoning
- pinitial = +1.2 kg m/s.
- pfinal = −0.8 kg m/s.
- Δp = −0.8 − 1.2 = −2.0 kg m/s.
−2.0 kg m/s.
Look for these in your work
- Use the negative final velocity.
- Subtract initial from final.
Find Δp and final momentum for a 0.50 kg cart slowing from +8 to +2 m/s.
Give me a hint
Calculate the change and the final value as separate requested quantities.
Compare my reasoning
- Δp = 0.50(2 − 8) = −3.0 kg m/s.
- Final momentum = 0.50 × 2 = +1.0 kg m/s.
- The final motion remains rightward for right positive.
Δp = −3.0 kg m/s; final momentum +1.0 kg m/s.
Look for these in your work
- Separate change from final value.
- State why the direction need not reverse.
Compare stopping and rebounding for the same 0.20 kg ball initially at +6 m/s, with rebound final velocity −4 m/s.
Give me a hint
Use the same initial momentum but different final states.
Compare my reasoning
- Stopping gives Δp = 0 − 1.2 = −1.2 kg m/s.
- Rebounding gives Δp = −0.8 − 1.2 = −2.0 kg m/s.
- The rebound has the greater change magnitude: 2.0 rather than 1.2 kg m/s.
Stopping −1.2 kg m/s; rebound −2.0 kg m/s; rebound has the larger magnitude.
Look for these in your work
- Keep one initial state.
- Compare magnitudes only after calculating signed changes.
Common mistakes
- Use final minus initial.
- Give reversed velocity its opposite sign.
- A negative change does not automatically imply negative final velocity.
What can you explain now?
What is Δp if the same 0.20 kg ball stops from +6 m/s?
Compare with the explanation
−1.2 kg m/s.
Final momentum is zero, so subtract the initial +1.2.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain why a rebound requires signed subtraction, then give a slowing example with a negative change but positive final momentum.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.