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How do you use momentum, impulse and the conservation of momentum to analyse collisions?

Use momentum and impulse: calculate momentum as mass times velocity, find impulse as change in momentum, and apply conservation of momentum to collisions in a straight line.

A CCEA GCSE Further Mathematics answer on momentum and impulse, covering momentum as mass times velocity, impulse as the change in momentum, and the conservation of momentum in collisions and when objects coalesce in the Mechanics unit.

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  1. What this dot point is asking
  2. Momentum
  3. Impulse
  4. Conservation of momentum
  5. Coalescing objects
  6. Why this matters

What this dot point is asking

Momentum measures the motion stored in a moving mass, and CCEA GCSE Further Mathematics uses it to analyse collisions. You must calculate momentum as mass times velocity, find impulse as the change in momentum, and apply the principle of conservation of momentum to collisions along a straight line, including the case where two objects join and move together. Keeping a consistent positive direction is essential because momentum is directed.

Momentum

Momentum captures how hard it is to stop a moving object: it grows with both mass and velocity. Because velocity has direction, so does momentum, and you must assign signs consistently.

Impulse

When a force acts on an object for a time, it changes the object's momentum, and that change is the impulse. Impulse gives a way to handle short, hard interactions such as a bat hitting a ball, where the exact force is hard to measure but the change in velocity is known.

For a rebound, the velocity reverses sign, so the change in momentum is larger than for an object simply brought to rest, which is why a bounced ball receives a bigger impulse than one that is caught.

Conservation of momentum

When two objects collide and no external horizontal force acts, the total momentum of the system is unchanged. This single principle solves most collision problems.

Coalescing objects

A common case is when two objects stick together on impact, such as railway trucks coupling. They then move with one common velocity, so the momentum after is the combined mass times that velocity. Setting total momentum before equal to the combined momentum after gives the common velocity in a single step.

Why this matters

Momentum and impulse give a powerful shortcut for interactions that would be awkward to handle with forces and acceleration directly, because conservation lets you skip the details of the collision itself. The topic reinforces the directed nature of velocity from the vectors work, and impulse links back to Newton's second law, since force times time is the change in momentum. It rounds out the Mechanics unit's account of how moving bodies interact.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 2 (style)3 marksA ball of mass 0.20.2 kg moving at 1515 m/s is struck and rebounds at 99 m/s in the opposite direction. Find the magnitude of the impulse on the ball.
Show worked answer β†’

Impulse is the change in momentum. Take the original direction as positive, so the final velocity is βˆ’9-9 m/s.

Initial momentum: 0.2Γ—15=30.2 \times 15 = 3 kg m/s. Final momentum: 0.2Γ—(βˆ’9)=βˆ’1.80.2 \times (-9) = -1.8 kg m/s.

Impulse == final βˆ’- initial =βˆ’1.8βˆ’3=βˆ’4.8= -1.8 - 3 = -4.8 kg m/s, so the magnitude is 4.84.8 N s.

Marks are for using a sign change on rebound, the change in momentum, and the magnitude. Forgetting the sign change is the main error.

CCEA Unit 2 (style)4 marksA 33 kg trolley moving at 44 m/s collides with a stationary 22 kg trolley and they move off together. Find their common velocity.
Show worked answer β†’

Momentum is conserved in the collision. Total momentum before =3Γ—4+2Γ—0=12= 3 \times 4 + 2 \times 0 = 12 kg m/s.

After the collision the combined mass is 3+2=53 + 2 = 5 kg moving at a common velocity vv, so the momentum after is 5v5v.

Set before equal to after: 12=5v12 = 5v, so v=2.4v = 2.4 m/s.

Marks are for the total momentum before, the combined-mass expression after, and the common velocity. The objects coalesce, so they share one velocity.

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