Solve separate final velocities from a supplied collision outcome
Use conservation plus a stated final velocity and identify cases with too little information.
Section P1.6 · Momentum
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Write a signed two-object momentum total.
- Rearrange a linear equation for one unknown.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 0.40 × 3 + 0.60 × (−1)?
What you will learn
- Solve a one-dimensional collision with one supplied final velocity.
- Explain why one conservation equation may leave two unknowns.
Separate objects need separate final terms
When two objects do not stick, write m1u1 + m2u2 = m1v1 + m2v2 under negligible net external impulse. The masses remain positive while each velocity retains its sign. A given final velocity supplies the extra information needed to find the other; do not use the combined-mass shared-velocity expression for an unstated sticking outcome.
For original values, 0.40 kg at +3 m/s and 0.60 kg at −1 m/s have initial total +0.60 kg m/s. If the first finishes at −1 m/s, its final contribution is −0.40 kg m/s. Therefore −0.40 + 0.60v2 = +0.60, giving v2 = +5/3 m/s or about 1.67 m/s. The two final contributions −0.40 and +1.00 still total +0.60.
One equation cannot invent a second outcome
For two 1 kg carts initially at +4 and −2 m/s, total momentum is +2 kg m/s. If the first finishes at −1 m/s, conservation gives −1 + v2 = 2 and the second finishes at +3 m/s. Without that supplied final velocity or another valid outcome condition, the equation only says v1 + v2 = 2.
Pairs (0,+2) and (−1,+3) both satisfy that total and describe distinct possible final states under suitable collision outcomes. Conservation alone does not select one pair, require equal final velocities, or establish an elastic collision. Ask for the missing physical condition rather than adding an unsupported energy or equal-speed assumption.
Sources for this lesson
Pause and explain
Two final velocities are unknown and only total momentum is supplied. Must both be equal?
Worked example
Masses 0.40 kg and 0.60 kg start at +3 and −1 m/s. They separate after collision with first velocity −1 m/s and negligible external impulse. Find the second velocity.
Show the worked solution
- Initial momentum = 0.40 × 3 + 0.60 × (−1) = +0.60 kg m/s.
- Final equation: −0.40 + 0.60v2 = +0.60.
- v2 = 1.00/0.60 = +5/3 m/s, approximately 1.67 m/s.
Answer +5/3 m/s, approximately 1.67 m/s right.
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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 v2 for 0.40 kg initially +3 m/s and 0.60 kg initially −1 m/s, with first final velocity −1 m/s and negligible external impulse.
Give me a hint
Subtract the known first final momentum from the conserved total.
Compare my reasoning
- Initial total = +0.60 kg m/s.
- Known final first contribution = −0.40 kg m/s, so second contribution = +1.00 kg m/s.
- v2 = 1.00/0.60 = +5/3 m/s.
+5/3 m/s right.
Look for these in your work
- Retain the negative final first term.
- Check the final total.
Two 1 kg carts start at +4 and −2 m/s. With negligible external impulse, the first finishes at −1 m/s. Find the second velocity.
Give me a hint
Use the conserved signed total and the supplied first final state.
Compare my reasoning
- Initial total = 4 − 2 = +2 kg m/s.
- Final equation is −1 + v2 = 2.
- v2 = +3 m/s.
+3 m/s.
Look for these in your work
- Use one consistent direction.
- Include the known final negative velocity.
The same equal-mass pair has total +2 kg m/s, but neither final velocity is supplied and sticking is not stated. Explain what can and cannot be determined.
Give me a hint
Write the remaining equation and test two distinct pairs.
Compare my reasoning
- Conservation gives v1 + v2 = 2.
- Both (0,+2) and (−1,+3) satisfy the total.
- Another valid outcome condition is needed to select unique final velocities.
Only v1 + v2 = 2 is determined; another collision condition is needed for unique velocities.
Look for these in your work
- Demonstrate non-uniqueness.
- Avoid inventing sticking or elasticity.
Common mistakes
- Keep two separate final momentum terms.
- Use only supplied or justified collision conditions.
- A momentum equation alone may leave two unknown velocities.
What can you explain now?
With two equal 1 kg carts and total +2 kg m/s, does v1 + v2 = 2 uniquely fix both velocities?
Compare with the explanation
No.
Different final pairs can have the same total unless another condition is supplied.
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Make it stick
Write a two-final-velocity collision equation and identify the extra supplied condition that closes it.
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.