Mechanics · 26 September 2026 · 6 min read
M1 momentum and impulse: collisions, rebounds and force–time graphs
A ball returning along the same line can have almost the same speed but a large change in momentum. A collision can conserve total momentum while substantially reducing kinetic energy. These are not exceptions to remember separately; they follow from treating momentum as a signed quantity and defining the system carefully.
The original examples below cover one-dimensional momentum and impulse for Pearson Edexcel International A-Level Mathematics M1, WME01. They use idealised short collisions and state when external impulse is neglected. M1 is a Mathematics unit, and the academy’s M1 tuition is online.
Choose the system and direction before the equation
For two colliding particles, the forces they exert on each other are internal to the combined system. Their impulses are equal and opposite, so they cancel when calculating the total momentum change of that system. External forces may still exist, but their impulse during the short interaction must be negligible for the usual conservation model.
For one particle alone, the collision force is external to that smaller system and changes its momentum. This is why total momentum can remain constant while each particle’s momentum changes. Write a before-and-after line with mass and signed velocity for every particle before substituting.
Worked example: two particles stick together
Take right as positive. A 0.80 kg particle moves at +5.0 m s⁻¹ and collides with a 1.20 kg particle moving at −1.0 m s⁻¹. They stick together. Initial momentum is 0.80 × 5.0 + 1.20 × (−1.0) = 2.8 kg m s⁻¹. Their combined mass is 2.0 kg, so the shared velocity is +1.4 m s⁻¹.
The positive answer means motion to the right. Adding both initial speeds as positive would describe a different situation and give the wrong total momentum. The phrase “stick together” supplies the common final velocity; it does not imply that the final velocity is zero.
Concept check: momentum and kinetic energy are different
The following energy comparison explains the distinction; work–energy and restitution methods are not added M1 requirements. Before this collision, total kinetic energy is ½ × 0.80 × 5.0² + ½ × 1.20 × 1.0² = 10.6 J. Afterwards it is ½ × 2.0 × 1.4² = 1.96 J. The reduction is 8.64 J. The speed is squared in kinetic energy, so the negative sign of an initial velocity does not make its kinetic energy negative.
The lost kinetic energy is transferred into other forms such as internal energy and deformation in the collision model. Total energy is not destroyed. An elastic collision has additional kinetic-energy conservation, but a sticking collision should not be solved by assuming both momentum and kinetic energy stay unchanged.
Worked example: impulse during a rebound
A 0.15 kg ball travels towards a wall at +12 m s⁻¹ and rebounds at −8.0 m s⁻¹, where positive is towards the wall. Its impulse is m(v − u) = 0.15(−8.0 − 12) = −3.0 N s. The magnitude is 3.0 N s and the direction is away from the wall.
If the interaction lasts 0.020 s, the average resultant force is impulse divided by time: −150 N. An actual collision force can vary greatly during contact, so this is an average, not necessarily the peak. Neglecting other impulses during the short contact must be appropriate to the stated model.
Worked example: read an impulse from a graph
A force–time pulse is triangular, rising from zero to 200 N and returning to zero over 0.030 s. Its impulse is the signed area, ½ × 0.030 × 200 = 3.0 N s. For a 0.50 kg particle initially at rest, this gives a final velocity of 6.0 m s⁻¹ in the positive force direction, assuming this is the resultant force.
If the same impulse is delivered over twice as long, the average force is halved. That does not by itself determine the exact peak of an arbitrary pulse; the shape matters. For a graph crossing below the time axis, combine positive and negative signed areas to find the net impulse.
Check equal and opposite impulses on the two particles
In the sticking example, the first particle’s impulse is 0.80(1.4 − 5.0) = −2.88 N s. The second’s is 1.20(1.4 − (−1.0)) = +2.88 N s. Equal magnitudes and opposite directions provide an independent check on the calculation and the interaction model.
Before marking, check signs, system choice, common final velocity where appropriate, force versus impulse units, and whether energy conservation was justified. Practise a sticking collision, a rebound and a graph question separately, then mix them. The first decision should identify the physical principle, not simply the formula containing the numbers you can see.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
Can momentum be negative in an M1 question?
Yes. In one dimension, its sign indicates direction relative to your chosen positive direction. Momentum is mass times signed velocity. Kinetic energy is different: it depends on speed squared and is not negative because the object moves in the negative direction.
When can I use conservation of momentum?
Use it for a clearly defined system when the external impulse over the interaction is zero or negligible in the model. Internal collision forces cancel in the total-system calculation. Momentum of an individual colliding particle usually changes even when the combined system’s momentum is conserved.
What is the difference between impulse and force?
Impulse is the change in momentum and equals the signed area beneath a force–time graph. Its unit is N s. Force has unit N; average resultant force equals impulse divided by interaction time. A varying force’s peak cannot be found from impulse alone without further information.
Why do rebound calculations seem to add the speeds?
The velocities are in opposite directions. If the initial velocity is +u and final velocity is −v, the change is −v − u, whose magnitude is u + v. Writing signed velocities first explains the apparent addition and also gives the correct impulse direction.
Is kinetic energy conserved when particles stick together?
Generally not. In the ideal short-collision model, total momentum can be conserved while kinetic energy decreases and energy is transferred into other forms. Sticking supplies a common final velocity. Do not add kinetic-energy conservation unless the question’s physical model justifies it.