Skip to main content
Northern IrelandCombined ScienceSyllabus dot point

How do forces change motion, and what does momentum tell us about collisions?

Balanced and unbalanced forces, Newton's first, second (F = m a) and third laws, momentum (p = m v) and the conservation of momentum in collisions.

A CCEA GCSE Double Award Science (Physics Unit P1) answer on balanced and unbalanced forces, Newton's three laws of motion including F equals m a, momentum p equals m v, and the conservation of momentum in collisions.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Balanced and unbalanced forces
  3. Newton's three laws
  4. Momentum and collisions
  5. Forces and safety
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA Double Award wants you to explain balanced and unbalanced (resultant) forces, state Newton's three laws, use F = m a, and use momentum p = m v with the conservation of momentum in collisions. The calculations with F=maF = ma and momentum appear in most P1 papers.

Balanced and unbalanced forces

The resultant force is the single force that has the same effect as all the forces added together, taking direction into account.

Newton's three laws

A larger force gives a larger acceleration; a larger mass gives a smaller acceleration for the same force.

Momentum and collisions

In a collision or explosion with no external resultant force, momentum is conserved: the total momentum before equals the total momentum after.

Forces and safety

Momentum links directly to vehicle safety. A force changes momentum, and the same change in momentum can be achieved with a small force over a long time or a large force over a short time. Safety features such as crumple zones, seatbelts and airbags all work by increasing the time over which a passenger's momentum is brought to zero in a crash. Because the change in momentum is the same, spreading it over a longer time means the force on the passenger is smaller, reducing injuries.

Examples in context

Example 1. A rocket
Hot gas is pushed down (action); by Newton's third law the gas pushes the rocket up (reaction), so it lifts off. The two forces are equal and opposite but act on different objects (the gas and the rocket), so they do not cancel.
Example 2. A car crash
Crumple zones make the collision last longer, reducing the force on the passengers because the same change in momentum is spread over more time. An airbag does the same job for the head and chest, while a seatbelt stops the passenger being thrown forward.
Example 3. Walking
When you walk, your foot pushes backward on the ground (action) and the ground pushes you forward (reaction). This forward reaction force from the ground is what drives you along, which is why walking on ice (low grip) is so difficult.

Try this

Q1. State Newton's second law as an equation. [1 mark]

  • Cue. F=maF = m a.

Q2. A 3.0 kg3.0\ \text{kg} mass has a resultant force of 12 N12\ \text{N} on it. Find its acceleration. [2 marks]

  • Cue. a=F/m=12/3.0=4.0 m/s2a = F/m = 12/3.0 = 4.0\ \text{m/s}^2.

Q3. Calculate the momentum of a 1500 kg1500\ \text{kg} car moving at 20 m/s20\ \text{m/s}. [2 marks]

  • Cue. p=mv=1500×20=30000 kg m/sp = m v = 1500 \times 20 = 30000\ \text{kg m/s}.

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-style3 marksA car of mass 1200 kg accelerates at 2.5 m/s squared. Calculate the resultant force on the car.
Show worked answer →

Use Newton's second law, force equals mass times acceleration.

F=ma=1200×2.5=3000 N.F = m a = 1200 \times 2.5 = 3000\ \text{N}.

So the resultant force is 3000 N.

Markers reward F=maF = ma, the substitution, and the value 3000 N.

CCEA-style4 marksA 0.50 kg trolley moving at 4.0 m/s collides with and sticks to a stationary 0.30 kg trolley. Calculate the velocity of the combined trolleys after the collision.
Show worked answer →

Momentum is conserved, so total momentum before equals total momentum after.

Before: p=mv=0.50×4.0=2.0 kg m/sp = m v = 0.50 \times 4.0 = 2.0\ \text{kg m/s} (the second trolley is at rest, so adds nothing).

After: the combined mass is 0.50+0.30=0.80 kg0.50 + 0.30 = 0.80\ \text{kg}, moving at velocity vv.

0.80×v=2.00.80 \times v = 2.0, so v=2.0/0.80=2.5 m/sv = 2.0 / 0.80 = 2.5\ \text{m/s}.

Markers reward momentum before equals momentum after, the value 2.0 kg m/s, the combined mass 0.80 kg, and the answer 2.5 m/s.

Related dot points

Sources & how we know this