Recover unknown masses or incoming velocities from an impact
Rearrange conservation for an unknown mass or velocity, substitute the result and reject claims that require a non-positive physical mass.
Section 4.3 · Momentum
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
- Rearrange a linear equation with an unknown on both sides.
- Interpret positive masses and signed velocities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In 12 − 2m = 2 + m, what value of m satisfies the equation?
What you will learn
- Solve inverse coalescence problems for a mass or incoming velocity.
- Use positivity and weighted-velocity bounds to assess whether supplied data fit the model.
An unknown mass contributes on both sides
A 2 kg particle at +6 m s⁻¹ sticks to a resting particle of unknown mass m kg, and the common velocity is +1.5 m s⁻¹. Conservation gives 12 = (2 + m) × 1.5, so the combined mass is 8 kg and m = 6 kg. The unknown initially has zero momentum, but it still appears in the combined final mass. Substitute m = 6 into both sides to check the recovered value.
If the unknown body is moving, keep its signed initial contribution too. For a 2 kg particle at +6 and an m kg particle at −2 coalescing at +1 m s⁻¹, write 12 − 2m = 2 + m. This gives m = 10/3 kg. Dividing the final momentum by a guessed initial mass or removing the negative incoming sign would corrupt the inverse equation.
A mathematical result must fit the physical data
Positive masses make the common velocity lie between the two incoming signed velocities. If a 2 kg particle at +6 sticks to a resting body and allegedly leaves at +8, the equation 12 = (2 + m)8 gives m = −0.5 kg. That is not a valid physical mass: the reported outcome cannot fit the stated model with a positive second mass. Do not replace the negative result by its magnitude and call the problem solved.
An unknown initial velocity can be recovered without being positive. For a 2 kg body initially at +5 and a 3 kg body with velocity u, coalescing at −1 gives 10 + 3u = 5(−1), so u = −5 m s⁻¹. This is a valid negative direction rather than invalid negative mass. Distinguish mathematical signs by quantity, check the original momentum equation and state precisely which modelling condition fails if a result is impossible.
Pause and explain
An impact calculation requires m = −0.5 kg for a supposedly positive-mass body. What should be concluded?
Worked example
A 2 kg particle at +6 m s⁻¹ sticks to a resting particle of mass m kg. Their common velocity is +1.5 m s⁻¹. Find m with negligible external effects along the line.
Show the worked solution
- Conservation gives 2 × 6 + m × 0 = (2 + m) × 1.5.
- Divide by 1.5 to get 2 + m = 8, hence m = 6 kg.
- The mass is positive and final momentum 8 × 1.5 = 12 kg m s⁻¹ matches the initial total.
Answer Unknown mass m = 6 kg.
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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 3 kg particle moving at +4 m s⁻¹ sticks to a resting body of unknown mass. Their common velocity is +2 m s⁻¹. Find the unknown mass under the conservation model.
Give me a hint
The combined final mass equals incoming momentum divided by common velocity.
Compare my reasoning
- Initial momentum is 3 × 4 = 12 kg m s⁻¹.
- Write 12 = (3 + m) × 2, so 3 + m = 6.
- Hence m = 3 kg; final momentum 6 × 2 = 12 checks the result.
Unknown mass 3 kg.
Look for these in your work
- Subtract the known mass from the combined mass.
- Check the recovered mass is positive and satisfies the original balance.
A 2 kg particle initially has velocity +5 m s⁻¹, while a 3 kg particle has unknown velocity u. They coalesce at −1 m s⁻¹. Find u with negligible external effects along the line.
Give me a hint
The final combined momentum is negative, so the unknown initial contribution must outweigh the positive one.
Compare my reasoning
- Final momentum is (2 + 3)(−1) = −5 kg m s⁻¹.
- Conservation gives 10 + 3u = −5, hence 3u = −15.
- Thus u = −5 m s⁻¹; the final velocity −1 lies between −5 and +5.
Unknown initial velocity u = −5 m s⁻¹.
Look for these in your work
- Use the negative common final velocity.
- Interpret a negative velocity as a direction rather than rejecting it as a negative mass.
A 2 kg particle at +6 m s⁻¹ sticks to a particle of mass m kg moving at −2 m s⁻¹. The common velocity is +1 m s⁻¹. Find m and verify both momentum and incoming-velocity bounds.
Give me a hint
The unknown moving mass contributes a negative incoming momentum and a positive final momentum.
Compare my reasoning
- Write 12 − 2m = (2 + m) × 1.
- Rearranging gives 10 = 3m, so m = 10/3 kg and combined mass is 16/3 kg.
- Incoming momentum is 12 − 20/3 = 16/3 kg m s⁻¹, matching final momentum; +1 lies between −2 and +6.
m = 10/3 kg; both momenta 16/3 kg m s⁻¹; common velocity +1 lies between −2 and +6.
Look for these in your work
- Retain both appearances of the unknown mass.
- Check the exact result, positivity and signed velocity bounds.
Common mistakes
- A resting body still belongs to the final combined mass.
- An unknown moving mass appears in both incoming and final momentum.
- Negative velocity can be valid; negative mass cannot be repaired by taking its magnitude.
What can you explain now?
A particle at +6 m s⁻¹ sticks to a resting body of positive mass. Can their common velocity be +8 m s⁻¹ under this model?
Compare with the explanation
No; +8 lies outside the incoming velocity range from 0 to +6.
A positive-mass weighted average cannot exceed both incoming signed velocities.
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Make it stick
Place unknown contributions on both sides, solve exactly, substitute and distinguish valid negative directions from invalid masses.
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.