Track successive impacts and distinguish momentum from kinetic energy
Carry combined mass and velocity into the next impact, conserve momentum event by event and use a supplied energy formula without assuming energy conservation.
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
- Solve a two-body coalescence balance.
- Use the supplied kinetic-energy expression K = ½mv² with kg and m s⁻¹.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 1 kg particle at +6 sticks to a resting 2 kg body. What mass and velocity enter a later impact?
What you will learn
- Solve an explicitly ordered sequence of coalescing impacts.
- Check the whole-system momentum and distinguish conserved momentum from kinetic-energy changes.
After the first impact, carry both the new mass and velocity
Suppose A has mass 2 kg and velocity +6 m s⁻¹, B has mass 3 kg and is initially at rest, and C has mass 1 kg and velocity −3 m s⁻¹. They are arranged so A and B coalesce first, then their combined body coalesces with C. External effects along the line are absent during and between these events. The first pair’s momentum is 12 kg m s⁻¹, so the 5 kg combined body moves at 12/5 = 2.4 m s⁻¹.
For the second impact, incoming momentum is 5 × 2.4 + 1 × (−3) = 9 kg m s⁻¹. The new combined mass is 6 kg, giving final velocity +1.5 m s⁻¹. The initial momentum of all three bodies is also 12 + 0 − 3 = 9, matching final 6 × 1.5. C’s incoming velocity remains −3 only because the interval model supplies no force changing it; if later forces are specified, update that velocity before its impact.
Momentum conservation does not conserve kinetic energy
Using the supplied expression K = ½mv², initial kinetic energy of all three bodies is ½ × 2 × 6² + ½ × 1 × 3² = 40.5 J. Final kinetic energy is ½ × 6 × 1.5² = 6.75 J, a decrease of 33.75 J. The first impact loses 36 − 14.4 = 21.6 J; the second loses (14.4 + 4.5) − 6.75 = 12.15 J. These decreases add to the whole-system decrease and are consistent with the sticking model.
Kinetic energy uses v² and is non-negative regardless of direction, unlike signed momentum. Equal opposite momenta can sum to zero while both bodies have positive kinetic energy. In a sticking impact, kinetic energy can be converted to deformation, heating or sound; total energy accounting includes these other forms. This supplied-formula check does not replace the wider Energy, work and power sequence, which remains a separate syllabus requirement. No impulse or restitution calculation is needed here.
Pause and explain
If momentum is conserved in a sticking impact, must kinetic energy also be conserved?
Worked example
A (2 kg, +6 m s⁻¹) first coalesces with resting B (3 kg); their combined body then coalesces with C (1 kg, −3 m s⁻¹). Assume no external effects along the line during or between events. Find both common velocities and the overall kinetic-energy decrease, using K = ½mv².
Show the worked solution
- First impact: 12 = 5V₁, so V₁ = +2.4 m s⁻¹.
- Second impact: 5(2.4) − 3 = 6V₂, so V₂ = +1.5 m s⁻¹; whole-system momentum is 9 kg m s⁻¹ throughout.
- Initial K = 40.5 J and final K = 6.75 J, giving decrease 33.75 J, while momentum remains conserved.
Answer First velocity +2.4 m s⁻¹; final +1.5 m s⁻¹; kinetic-energy decrease 33.75 J; total momentum 9 kg m s⁻¹.
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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 +6 m s⁻¹ first sticks to a resting 2 kg body; that combined body then sticks to a resting 3 kg body. No external effects act along the line during or between events. Find both common velocities and the total kinetic-energy decrease using K = ½mv².
Give me a hint
At each event, add the newly involved mass and preserve the incoming signed total.
Compare my reasoning
- First common velocity is 6/(1 + 2) = +2 m s⁻¹.
- Second common velocity is 3 × 2/(3 + 3) = +1 m s⁻¹, with total momentum 6 kg m s⁻¹.
- Initial K is ½ × 1 × 6² = 18 J and final K is ½ × 6 × 1² = 3 J, a decrease of 15 J.
First velocity +2 m s⁻¹; final +1 m s⁻¹; kinetic-energy decrease 15 J.
Look for these in your work
- Use combined masses 3 kg then 6 kg.
- Check momentum at both events and compute energy separately.
A 2 kg particle at −4 m s⁻¹ first sticks to a resting 2 kg body. The resulting body then sticks to a 4 kg particle at +1 m s⁻¹. This order is stipulated, with no external effects along the line between events. Find both common velocities and overall kinetic-energy decrease using K = ½mv².
Give me a hint
The second incoming particle contributes positive momentum against the moving group’s negative momentum.
Compare my reasoning
- First common velocity is (−8)/4 = −2 m s⁻¹.
- Second balance is 4(−2) + 4(1) = 8V, giving V = −0.5 m s⁻¹.
- Initial total K is ½ × 2 × 4² + ½ × 4 × 1² = 18 J; final K is ½ × 8 × 0.5² = 1 J, so the decrease is 17 J.
First velocity −2 m s⁻¹; final −0.5 m s⁻¹; kinetic-energy decrease 17 J.
Look for these in your work
- Preserve both directed incoming momenta in the second event.
- Square velocities for energy and retain the stipulated collision order.
A 2 kg particle at +3 m s⁻¹ coalesces with a 3 kg particle at −2 m s⁻¹ under the conservation model. Find the common velocity and compare initial and final kinetic energy using K = ½mv². Explain why the incoming total momentum zero does not mean no motion or no energy.
Give me a hint
Calculate signed momentum and non-negative energy as different sums.
Compare my reasoning
- Initial momentum is 2 × 3 + 3 × (−2) = 0, so the 5 kg combined body has V = 0.
- Initial kinetic energy is ½ × 2 × 3² + ½ × 3 × 2² = 9 + 6 = 15 J; final kinetic energy is zero.
- The incoming bodies were moving with positive kinetic energy; their momenta cancelled. Coalescence converts the 15 J kinetic-energy decrease into other forms within the wider energy accounting.
Common velocity 0; initial K 15 J, final K 0, decrease 15 J; zero initial total momentum does not imply stationary incoming bodies.
Look for these in your work
- Use the sticking condition to infer the final rest state.
- Distinguish signed cancellation from a sum of positive energy contributions.
Common mistakes
- Carry the combined mass as well as velocity into the next impact.
- Use the specified event order and update velocities if forces act between events.
- Momentum can be conserved while kinetic energy decreases.
What can you explain now?
Two moving particles have total momentum zero. Must their total kinetic energy be zero before impact?
Compare with the explanation
No; moving positive-mass particles have positive kinetic energy even if their momenta cancel.
Momentum is signed and can cancel, whereas each ½mv² contribution is non-negative.
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Make it stick
Track event order, carry mass and velocity forward, verify signed system momentum and calculate any supplied energy check separately.
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.