Find an unknown outgoing velocity after direct impact
Use one supplied outgoing velocity to solve the other, interpret rebounds and explain why conservation alone cannot determine two unknown velocities.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.3. Direct impacts are restricted to one dimension. Choose one positive direction for all velocities and momenta. Conservation applies to the selected system when external effects in that direction are absent or negligible during the impact; it does not make each particle’s momentum constant. Coalescing bodies share a common final velocity and combined mass. Impulse and the coefficient of restitution are not required in this section. Kinetic energy need not be conserved; where used, its expression ½mv² is supplied as an additional model check. Bodies have positive constant masses; stated impact order and interval assumptions must be respected. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Set up the signed conservation equation.
- Solve a linear equation with two mass-weighted contributions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle rebounds left at speed 3 m s⁻¹ with right positive. Which velocity enters conservation?
What you will learn
- Solve a direct-impact conservation equation for an unknown velocity.
- Interpret the sign and identify when further outcome information is required.
Associate each signed velocity with the correct mass
For a 2 kg particle initially at +6 m s⁻¹ and a 3 kg particle initially at −2 m s⁻¹, initial momentum is 12 − 6 = 6 kg m s⁻¹. If the 2 kg particle rebounds at −3 m s⁻¹, conservation gives 2(−3) + 3v = 6. Thus v = +4 m s⁻¹ for the 3 kg particle. Its direction comes from the solution, rather than being assumed from the incoming velocity.
Draw a before-and-after table with columns for mass, initial velocity and final velocity. The shared sign convention prevents an approaching particle’s speed from being treated as a positive velocity. After solving, substitute both outgoing momenta into the original equation. A negative answer is a valid directed velocity; it does not mean negative speed or negative mass. Report both its signed value and its physical direction if requested.
One conservation equation does not fix two independent outcomes
If both final velocities are unknown for that 2 kg and 3 kg pair, conservation gives only 2vₐ + 3vᵦ = 6. Both (vₐ, vᵦ) = (−3, 4) and (0, 2) satisfy it. The question must provide another relevant condition, such as one final velocity or bodies coalescing, to determine a unique result. Do not invent equal final velocities for bodies that remain separate.
For same-direction impacts, the faster particle must initially be behind the slower one to catch it; a supplied impact event establishes the relevant ordering. After a completed impact with bodies remaining separate, their separating directions or relative velocities must agree with the positions. A momentum balance alone does not impose a particular kinetic-energy behaviour. This section does not require introducing the coefficient of restitution or impulse to solve the stated problems.
Pause and explain
If 2(−3) + 3v = 6, what is the unknown velocity?
Worked example
A 2 kg particle at +6 m s⁻¹ collides directly with a 3 kg particle at −2 m s⁻¹. External effects along the line are negligible. The 2 kg particle rebounds at −3 m s⁻¹. Find the 3 kg particle’s final velocity.
Show the worked solution
- Initial momentum is 2 × 6 + 3 × (−2) = 6 kg m s⁻¹.
- Conservation gives 2 × (−3) + 3v = 6, so 3v = 12.
- Hence v = +4 m s⁻¹; substitution gives −6 + 12 = 6 kg m s⁻¹.
Answer The 3 kg particle leaves with velocity +4 m s⁻¹, in the positive direction.
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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.
A 1 kg particle at +5 m s⁻¹ hits a resting 2 kg particle. The 1 kg particle then moves at −1 m s⁻¹. Find the other final velocity with negligible external effects along the line.
Give me a hint
Move the known negative outgoing momentum to the other side of the equation.
Compare my reasoning
- Initial momentum is 1 × 5 + 2 × 0 = 5 kg m s⁻¹.
- Conservation gives −1 + 2v = 5, so 2v = 6.
- The 2 kg particle has v = +3 m s⁻¹; the final total is −1 + 6 = 5.
Other final velocity +3 m s⁻¹.
Look for these in your work
- Retain the known rebound’s negative sign.
- Substitute the solution to verify the full system total.
Particles of masses 3 kg and 2 kg approach with velocities +4 and −1 m s⁻¹. After direct impact, the 2 kg particle has velocity +3 m s⁻¹. Find the 3 kg particle’s final velocity under the conservation model.
Give me a hint
Initial total is a signed sum; the known final contribution belongs to the 2 kg body.
Compare my reasoning
- Initial momentum is 3 × 4 + 2 × (−1) = 10 kg m s⁻¹.
- Conservation gives 3v + 2 × 3 = 10, hence 3v = 4.
- The 3 kg final velocity is +4/3 m s⁻¹; its momentum 4 plus the other momentum 6 totals 10.
3 kg final velocity +4/3 m s⁻¹.
Look for these in your work
- Pair the supplied +3 final velocity with mass 2 kg.
- Keep the exact fraction before any requested rounding.
A 2 kg particle moving at +8 m s⁻¹ catches a 1 kg particle ahead moving at +2 m s⁻¹. After direct impact, the 1 kg particle has velocity +8 m s⁻¹. Find the 2 kg final velocity and check the particles can separate in their original ordering.
Give me a hint
For particles moving in the same direction, compare their outgoing velocities as well as the momentum sum.
Compare my reasoning
- Initial momentum is 2 × 8 + 1 × 2 = 18 kg m s⁻¹.
- The final balance is 2v + 8 = 18, so v = +5 m s⁻¹.
- The particle ahead leaves at 8 m s⁻¹ while the one behind leaves at 5 m s⁻¹, so their separation increases after impact.
2 kg final velocity +5 m s⁻¹; the front particle’s 8 m s⁻¹ exceeds the rear particle’s 5 m s⁻¹, so they separate.
Look for these in your work
- Retain the stated same-direction geometry.
- Check both the momentum total and post-impact separating velocities.
Common mistakes
- A rebound reverses velocity sign, not mass sign.
- Keep each final velocity with its own mass.
- Two unknown outgoing velocities need more than a single conservation equation.
What can you explain now?
Can 2vₐ + 3vᵦ = 6 alone determine a unique pair of outgoing velocities?
Compare with the explanation
No; an additional outcome condition is required.
For example, (−3, 4) and (0, 2) both satisfy the same momentum equation.
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Make it stick
Write a mass-and-velocity table, solve the signed balance, substitute the result and check what the given information can determine.
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.