Explain recoil when an initially resting system separates
Keep the original system boundary and calculate equal opposite momenta rather than equal speeds.
Section P1.6 · Momentum
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Add positive and negative momenta.
- Choose both separating objects as the system.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is total momentum of two carts both initially at rest?
What you will learn
- Apply momentum conservation to a resting system that separates.
- Compare recoil speeds of unequal masses.
Zero initial total still permits motion afterwards
Two carts initially at rest are pushed apart by a released internal spring, with negligible external horizontal impulse. Their initial total momentum is zero, so m1v1 + m2v2 = 0 after separation. A 1 kg cart moving right at +3 m/s has +3 kg m/s; a 2 kg cart must have −3 kg m/s and velocity −1.5 m/s left.
The final momenta have equal magnitudes and opposite directions. Their speeds are not equal because the masses differ. The more massive cart has the smaller speed magnitude for this fixed pair of opposing momenta. Internal stored energy can supply motion without supplying a non-zero external impulse or changing the initial zero system total.
Keep both parts and the supplied frame
A 0.10 kg small cart separating at +12 m/s from a 1.5 kg platform in an initially resting isolated horizontal system has +1.2 kg m/s. The platform therefore recoils at v = −1.2/1.5 = −0.80 m/s. These are final masses of the two included parts; their individual momenta change while their total remains zero.
If an external push acts during separation, or the original system already moves in the chosen frame, the initial total need not be zero. Calculate the stated initial system total and any external impulse first. Recoil does not mean momentum has appeared from nothing, nor that a spring force on one cart should be counted twice as an external contribution to the pair.
Sources for this lesson
Pause and explain
For unequal masses recoiling from rest without external impulse, what is equal in magnitude?
Worked example
A resting 2 kg cart and 1 kg cart separate under an internal spring with negligible external impulse. The 1 kg cart ends at +3 m/s. Find the 2 kg cart velocity.
Show the worked solution
- Initial total is zero.
- 2v + 1 × 3 = 0.
- v = −1.5 m/s, left for right positive.
Answer −1.5 m/s left.
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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.
Find recoil velocity of 2 kg when a 1 kg partner moves at +3 m/s from an initially resting isolated pair.
Give me a hint
Use zero initial total and retain the opposite recoil sign.
Compare my reasoning
- The 1 kg contribution is +3 kg m/s.
- 2v must equal −3 kg m/s.
- v = −1.5 m/s.
−1.5 m/s.
Look for these in your work
- Use both masses.
- Give the opposite direction.
An initially resting isolated pair separates into 0.10 kg at +12 m/s and a 1.5 kg platform. Find platform velocity.
Give me a hint
The known part supplies one signed momentum that the platform must oppose.
Compare my reasoning
- Known momentum = 0.10 × 12 = +1.2 kg m/s.
- Platform momentum = −1.2 kg m/s.
- Platform velocity = −1.2/1.5 = −0.80 m/s.
−0.80 m/s.
Look for these in your work
- Use final separated masses.
- Check that the total remains zero.
A student says the two unequal carts must recoil at equal speeds because total momentum starts at zero. Explain the error.
Give me a hint
Write equal opposite mass-velocity products rather than comparing speeds alone.
Compare my reasoning
- Conservation requires m1v1 = −m2v2.
- Different masses can have equal opposite products with different speed magnitudes.
- The heavier part has the smaller speed for those equal momentum magnitudes.
Equal opposite momenta, not equal speeds; the heavier cart moves more slowly in magnitude.
Look for these in your work
- Use the products in the conservation equation.
- Explain the role of unequal masses.
Common mistakes
- Conserve the two-part total.
- Equal opposite momenta need not mean equal speeds.
- Use zero initial total only when rest in the chosen frame is stated.
What can you explain now?
Which cart has the smaller recoil speed for equal opposite momenta, the heavier or lighter one?
Compare with the explanation
The heavier one.
Speed magnitude is momentum magnitude divided by mass.
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Make it stick
Explain how a zero-total system can separate into moving parts and why recoil speeds depend on mass.
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.