Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 4: Mechanics · official syllabus
MOMENTUM LAB
Learn it by playing it
Four short missions on signed momentum and direct impacts in one dimension. Choose a direction, predict the rule, then calculate and test your answer. Keep before and after velocities separate.
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Mission 1 of 4
Give momentum a sign
Take right as positive. A particle of mass 2 kg moves left at speed 3 m/s. Its signed velocity is -3 m/s. Set its signed linear momentum. The white arrow shows velocity; the teal momentum arrow shows your chosen answer on its own labelled scale.
Reversing a particle's velocity while keeping its mass and speed fixed...
Separate scales: white is velocity; teal is your momentum. The arrow's direction shows the sign, not a second force.
Your momentum is 4.0 kg m/s. Left is the negative direction.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Momentum has direction 4.3
Linear momentum is mass multiplied by velocity: p=mv. In this section motion is in one dimension. Choose a positive direction and use signed velocities throughout the calculation. Units are kg m/s.
A 2 kg particle moving left at speed 3 m/s has velocity -3 m/s if right is positive, so its momentum is -6 kg m/s. Its mass is positive and its speed is 3 m/s; the negative sign describes the direction of velocity and momentum.
Total momentum is the signed sum of the individual momenta. Equal speeds in opposite directions cancel momentum only when the masses are equal. Do not add speed magnitudes when the particles approach each other.
Choose the system and state the impact model 4.3
For a system with no resultant external effect along the line of motion during the impact, total linear momentum just before equals total linear momentum just after. In typical brief-impact models the external effect along that line is taken as negligible; state that assumption where appropriate.
For two bodies use m1u1+m2u2=m1v1+m2v2. The forces the particles exert on one another are internal to the two-particle system. Each particle's own momentum can change even though the total is conserved.
Draw before-and-after diagrams with masses and signed velocities. Use one direction convention on both diagrams. 'Before' and 'after' mean immediately on either side of the impact, not distant times when external forces may have changed the velocities.
Bodies that coalesce have one final velocity 4.3
Coalesce means stick together. Replace the two final velocities by a common v, so m1u1+m2u2=(m1+m2)v. The final mass is the combined mass; do not divide total momentum by only one body's mass.
Worked example: a 2 kg body at +6 m/s collides with a stationary 1 kg body and they coalesce. Under the isolated-impact model, 2(6)+1(0)=3v, giving v=+4 m/s. The combined body moves in the positive direction.
With positive masses, the common velocity is a mass-weighted average of the initial signed velocities. In this example it lies between 0 and 6 m/s. This is a useful sign and size check, not a replacement for the momentum equation.
Approaching bodies: retain the negative velocity 4.3
Worked example: take right as positive. A has mass 2 kg and velocity +3 m/s; B has mass 1 kg and velocity -3 m/s. If they coalesce, 2(3)+1(-3)=3v, so v=+1 m/s. The unequal opposing momenta leave a positive total.
If the opposing momenta are equal, the combined body is instantaneously at rest after coalescence. For example, 2 kg at +3 m/s and 1 kg at -6 m/s have zero total momentum, giving common velocity zero under the same model.
A negative calculated final velocity is a physical direction, not an invalid answer. Interpret it relative to the direction chosen at the start. Check the original signed total against the final signed total.
Bodies that remain separate 4.3
Do not give the particles a common final velocity unless the question says they coalesce or otherwise supplies that condition. Use separate final velocities in the conservation equation.
Worked example: A (2 kg) initially has velocity +4 m/s; B (1 kg) initially has velocity -1 m/s. After their direct impact B has velocity +5 m/s. Conservation gives 2(4)+1(-1)=2v+1(5), so A's final velocity is +1 m/s.
If both final velocities are unknown and the bodies do not coalesce, momentum conservation supplies only one equation for two unknowns. Do not claim a unique solution without more information. Impulse and the coefficient of restitution are not required knowledge for section 4.3 of this syllabus.
Momentum conservation is not kinetic-energy conservation 4.3
Conservation of total momentum under the stated model does not imply conservation of kinetic energy. Use momentum as the governing equation in these impact questions; do not invent a kinetic-energy equality.
In the 2 kg at +6 m/s and stationary 1 kg coalescence example, kinetic energy before is (1/2)(2)(6^2)=36 J. Afterwards it is (1/2)(3)(4^2)=24 J. Total momentum is 12 kg m/s on both sides, but 12 J of kinetic energy has been transferred to other forms.
Keep the modelling assumptions and the requested quantity separate: momentum uses signed velocity, kinetic energy uses the square of speed, and a momentum result alone does not describe every detail of a collision.
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