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Subjects · Leaving Cert Applied Maths

Leaving Cert Applied Maths: Momentum & direct collisions

How often Momentum & direct collisions comes up on the Applied Maths papers, every year it was asked, and questions to try.

HL Asked on 2 of the last 3 Higher Level papers, most recently in 2025. most years

OL Asked on 3 of the last 3 Ordinary Level papers, most recently in 2025. banker

Momentum & direct collisions, Higher Level(6 marks)

Quick ones on Momentum & direct collisions.

(a)When is the total momentum of a system conserved in a given direction?
  1. Only when the collision is perfectly elastic
  2. When no external force acts in that direction
  3. Only when the bodies stick together
(b)Newton's experimental law for a direct collision?
  1. v₁ + v₂ = e(u₁ + u₂)
  2. v₂ − v₁ = e(u₂ − u₁)
  3. v₂ − v₁ = −e(u₂ − u₁)
(c)Possible values of the coefficient of restitution e?
  1. 0 ≤ e ≤ 1
  2. e ≥ 1
  3. −1 ≤ e ≤ 1
Show the answers

(a) When no external force acts in that direction

(b) v₂ − v₁ = −e(u₂ − u₁)

(c) 0 ≤ e ≤ 1

Your turn: Higher Level questions on Momentum & direct collisions.

Higher Level

Asked on 2 of the last 3 Higher Level papers, most recently in 2025. most years

Every paper, year by year

YearWhere it came up
2025Q4, Q7
2024Q5
2023Not asked

Links open the State Examinations Commission’s paper for that year.

More Momentum & direct collisions questions

Momentum & direct collisions, 3 marks

A sphere hits an identical sphere at rest, directly, with e = 1. What happens?

  1. Both stop
  2. Both move on at half the speed
  3. They exchange velocities
Show the answer

They exchange velocities

PCM: v₁ + v₂ = u. NEL with e = 1: v₂ − v₁ = u. So v₁ = 0 and v₂ = u: the moving sphere stops and the other takes all its velocity.

Momentum & direct collisions, 3 marks

A 2 kg sphere at 3 m/s and a 1 kg sphere at 6 m/s move towards each other and coalesce. Common velocity?

  1. 2 m/s
  2. 0 m/s
  3. 4 m/s
Show the answer

0 m/s

Take the 2 kg direction as positive: momentum = 2(3) + 1(−6) = 0. So after coalescing 3v = 0 and both stop. Forgetting the opposite direction gives 4 m/s.

Momentum & direct collisions, 3 marks

A ball hits a fixed floor directly at speed u and rebounds, coefficient e. Fraction of its KE kept?

  1. e²
  2. e
  3. 1 − e
Show the answer

e²

Rebound speed is eu, so KE after ÷ KE before = ½m(eu)² ÷ ½mu² = e². With e = 0.5 the ball keeps only a quarter of its KE, not half.

Ordinary Level

Asked on 3 of the last 3 Ordinary Level papers, most recently in 2025. banker

Every paper, year by year

YearWhere it came up
2025Q5
2024Q7
2023Q7

Links open the State Examinations Commission’s paper for that year.

More Momentum & direct collisions questions

Momentum & direct collisions, 2 marks

In a direct collision, the coefficient of restitution e = 0 means…

  1. No kinetic energy is lost
  2. The bodies separate at their approach speed
  3. The bodies move together after impact
Show the answer

The bodies move together after impact

e = 0 is a perfectly inelastic collision: there is no bounce, so the bodies coalesce. e = 1 is perfectly elastic, with no kinetic energy lost.

Momentum & direct collisions, 2 marks

The impulse on a body is equal to…

  1. Force divided by time
  2. Its change in momentum
  3. Its change in kinetic energy
Show the answer

Its change in momentum

Impulse = force × time = mv − mu, measured in N s. In a collision the two bodies get equal and opposite impulses.

Momentum & direct collisions, 2 marks

Newton's experimental law for a direct collision is…

  1. v₂ − v₁ = −e(u₂ − u₁)
  2. v₁ + v₂ = e(u₁ + u₂)
  3. m₁v₁ = e m₂v₂
Show the answer

v₂ − v₁ = −e(u₂ − u₁)

Speed of separation = e × speed of approach. The minus sign shows the bodies' relative velocity reverses in the impact.

Other Applied Maths topics

All of Leaving Cert Applied Maths