Find the common velocity when bodies coalesce
Model bodies that stick together with their combined mass, retain incoming direction signs and interpret a zero or negative common velocity.
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
- Calculate signed total momentum.
- Use the combined mass of two bodies and solve a linear equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
After 2 kg and 3 kg bodies coalesce with velocity V, which expression is their total momentum?
What you will learn
- Set up conservation with one common final velocity for coalescing bodies.
- Interpret the direction and check the result against positive-mass weighted averages.
Sticking together supplies the extra velocity condition
When two bodies coalesce on impact, they move together immediately afterwards. Their final velocities are equal, so the final momentum is (mₐ + mᵦ)V. Under the stated conservation model, V = (mₐuₐ + mᵦuᵦ)/(mₐ + mᵦ). Include both masses in the denominator, even if one body was initially at rest. Coalescence does not imply that the combined body is stationary.
For masses 2 kg and 3 kg with velocities +6 and −2 m s⁻¹, total incoming momentum is 12 − 6 = 6 kg m s⁻¹. The combined mass is 5 kg, so V = +6/5 = +1.2 m s⁻¹. The positive direction is set by the net momentum. Adding the incoming speed magnitudes or dividing by only one mass would produce an incorrect common velocity.
A common velocity is a positive-mass weighted average
With positive masses, V lies between the two incoming signed velocities; it is not necessarily their ordinary arithmetic mean. Equal masses give the unweighted average, but unequal masses weight the larger mass more strongly. A 2 kg particle at +5 m s⁻¹ sticking to a resting 3 kg particle gives V = 2 m s⁻¹, rather than the arithmetic mean 2.5. This interval check can reveal a sign or denominator mistake before substitution.
Zero total incoming momentum gives V = 0 after coalescence. That conclusion requires the sticking condition; zero total alone did not make the incoming bodies stationary. A negative V means the combined body moves in the chosen negative direction, and its speed is |V|. Momentum can be conserved while kinetic energy decreases, with energy transferred to deformation, heating or other forms; do not impose kinetic-energy conservation for a sticking impact.
Pause and explain
For positive masses, what is a useful check on a coalesced velocity V?
Worked example
A 2 kg particle with velocity +6 m s⁻¹ collides with a 3 kg particle at −2 m s⁻¹ and they stick together. External effects along the line are negligible. Find their common velocity.
Show the worked solution
- Initial momentum is 2 × 6 + 3 × (−2) = +6 kg m s⁻¹.
- The common final mass is 2 + 3 = 5 kg, so 5V = 6.
- V = +1.2 m s⁻¹ lies between −2 and +6 and gives final momentum 5 × 1.2 = 6.
Answer Common velocity +1.2 m s⁻¹; speed 1.2 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 2 kg particle at +5 m s⁻¹ hits a resting 3 kg particle and they stick together. Find the common velocity under the conservation model.
Give me a hint
The resting particle contributes zero incoming momentum but still adds to final mass.
Compare my reasoning
- Initial momentum is 2 × 5 + 3 × 0 = 10 kg m s⁻¹.
- Combined mass is 5 kg and 5V = 10.
- V = +2 m s⁻¹ lies between the incoming velocities 0 and 5.
Common velocity +2 m s⁻¹.
Look for these in your work
- Include the initially stationary mass after impact.
- Check the velocity interval and substitute into final momentum.
A 3 kg particle at −4 m s⁻¹ and a 1 kg particle at +2 m s⁻¹ coalesce. Find their common velocity, speed and direction with negligible external effects along the line.
Give me a hint
A negative net momentum sets a negative common velocity.
Compare my reasoning
- Incoming momentum is 3 × (−4) + 1 × 2 = −10 kg m s⁻¹.
- Combined mass is 4 kg, so 4V = −10 and V = −2.5 m s⁻¹.
- Speed is |−2.5| = 2.5 m s⁻¹, in the chosen negative direction.
Common velocity −2.5 m s⁻¹; speed 2.5 m s⁻¹ in the negative direction.
Look for these in your work
- Retain the negative momentum of the 3 kg body.
- Separate signed velocity from non-negative speed.
Particles of masses 400 g and 600 g have velocities +15 and −5 m s⁻¹ and stick together on impact. Find the common velocity in SI units and check its incoming-velocity bounds.
Give me a hint
Convert both masses and take a signed weighted average.
Compare my reasoning
- Masses are 0.4 and 0.6 kg; incoming momentum is 0.4 × 15 + 0.6 × (−5) = 3 kg m s⁻¹.
- Combined mass is 1.0 kg, giving V = +3 m s⁻¹.
- This lies between −5 and +15, and final momentum 1.0 × 3 matches the incoming total.
Common velocity +3 m s⁻¹, between −5 and +15 m s⁻¹.
Look for these in your work
- Convert both gram values consistently.
- Check signed bounds rather than averaging speed magnitudes.
Common mistakes
- Use the sum of both masses after coalescence.
- Opposite incoming directions require opposite velocity signs.
- Do not assume a sticking impact conserves kinetic energy.
What can you explain now?
A 1 kg particle at +6 m s⁻¹ and a 2 kg particle at −3 m s⁻¹ stick together. What is their common velocity?
Compare with the explanation
Zero: the combined body is at rest immediately after impact.
Incoming momentum is +6 − 6 = 0, and the sticking condition gives one common velocity.
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Make it stick
Sticking gives one final velocity: divide the signed incoming momentum by combined mass and check direction and bounds.
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