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Edexcel A-level Physics (9PH0) · Topic 6: Further Mechanics
Mini-Lesson

Further Mechanics

Topic 6 has two halves: momentum and impulse (collisions, explosions, force–time graphs) and circular motion (angular velocity, centripetal acceleration and force).

p = mv · impulse = FΔt = Δp · a = v²/r = ω²r · F = mv²/rmomentum is a vector — direction and sign matter in every collision question

The rule that never fails: in a closed system, momentum is always conserved — in every collision and every explosion. Kinetic energy is only conserved in a perfectly elastic collision. Get those two straight and half the topic is done.

Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.

Topic 6 · momentum

Momentum and Newton's second law

p = mv  ·  F = Δp / Δtmomentum in kg m s⁻¹ (or equivalently N s) — and it is a VECTOR

Newton's second law in its full form is: the resultant force equals the rate of change of momentum. Only when mass is constant does this simplify to F = ma.

  • Take one direction as positive. A body moving the other way has negative momentum.
  • In two dimensions, momentum is conserved independently in each perpendicular direction.
  • The principle of conservation of linear momentum: for a system with no external resultant force, total momentum before = total momentum after.

Where it comes from: Newton's third law says the forces on two colliding bodies are equal and opposite, and they act for the same time. So the impulses are equal and opposite, so the momentum changes are equal and opposite — and the total is unchanged.

Calculate

Your turn — a sticky collision

1A trolley of mass 2.0 kg moving at 6.0 m s⁻¹ collides with a stationary trolley of mass 4.0 kg. They stick together. Calculate their common velocity after the collision, in m s⁻¹.
m s⁻¹
Hint: Momentum before = 2.0 × 6.0 + 4.0 × 0 = 12 kg m s⁻¹. After, the combined mass is 6.0 kg. So v = 12 ÷ 6.0.
Calculate

Your turn — kinetic energy lost

2For that same collision, calculate the kinetic energy lost. Give your answer in J.
J
Hint: KE before = ½ × 2.0 × 6.0² = 36 J. KE after = ½ × 6.0 × 2.0² = 12 J. Lost = 36 − 12.
Quick check

What is conserved?

?Two cars crash and crumple, ending up as a single wreck. Which quantity or quantities are conserved?
Topic 6 · impulse

Impulse and the force–time graph

impulse = FΔt = Δp = mv − muunit: N s, which is identical to kg m s⁻¹

The impulse of a force is the change of momentum it produces. On a force–time graph, impulse is the area under the curve — which is how you handle a force that varies during an impact.

Rearranged, F = Δp/Δt says: for a fixed change in momentum, extending the contact time reduces the force. That single idea explains:

  • Crumple zones and airbags — extend Δt, so the force on the passenger falls.
  • Bending your knees when you land, and catching a ball by drawing your hands back.
  • Rocket propulsion — the rocket expels gas backwards; the gas gains backward momentum, so the rocket gains an equal forward momentum.
Calculate

Your turn — impulse

3A constant force of 12 N acts on a body for 0.25 s. Calculate the impulse delivered. Give your answer in N s.
N s
Hint: Impulse = FΔt = 12 × 0.25.
Quick check

Reading the graph

?A ball bounces off a wall. The force on it varies with time during the contact. What does the area under the force–time graph represent?
Sort it

Elastic, inelastic, or explosion?

Tap a statement, then tap the type of interaction it describes. Momentum is conserved in all three.

⚡ Elastic collision

💥 Inelastic collision

🚀 Explosion

Topic 6 · circular motion

Angular velocity and centripetal acceleration

ω = 2π/T = 2πf  ·  v = rωω = angular velocity in rad s⁻¹ · v = linear (tangential) speed in m s⁻¹

An object moving in a circle at constant speed is still accelerating, because its velocity — a vector — is constantly changing direction. That acceleration points towards the centre:

a = v²/r = ω²r  ·  F = mv²/r = mω²rcentripetal acceleration and centripetal force — both directed to the CENTRE
  • Centripetal force is not a new force. It is the name for whatever existing force is doing the job: tension in a string, friction on tyres, gravity on a satellite, the normal contact force on a wall of death.
  • It acts perpendicular to the velocity, so it does no work and the speed never changes.
  • Remove it and the object flies off along a tangent — it does not fly radially outwards.

There is no centrifugal force. The outward push you feel on a roundabout is your inertia: your body tries to keep going in a straight line while the seat pushes you inwards.

Calculate

Your turn — angular velocity

4A wheel completes one revolution every 0.50 s. Calculate its angular velocity. Give your answer in rad s⁻¹ to 3 significant figures.
rad s⁻¹
Hint: ω = 2π ÷ T = 2π ÷ 0.50 = 4π.
Calculate

Your turn — centripetal force

5A ball of mass 0.20 kg is whirled on a string of radius 0.80 m at a constant speed of 4.0 m s⁻¹. Calculate the centripetal force. Give your answer in N.
N
Hint: F = mv²/r = (0.20 × 4.0²) ÷ 0.80 = (0.20 × 16) ÷ 0.80.
Quick check

Accelerating at constant speed

?A car goes round a roundabout at a constant speed of 8 m s⁻¹. Which statement is correct?
Quick check

Cut the string

?A ball is whirled in a horizontal circle on a string. The string suddenly snaps. Which path does the ball take (viewed from above)?
Match it

Match the quantity to its equation

Tap an item on the left, then its partner on the right.

Quantity
Equation
Recap

The big ideas to know

Momentum: p = mv, a vector; F = Δp/Δt is the full form of Newton's second law

Conservation: momentum is conserved in every collision and explosion in a closed system

Elastic vs inelastic: kinetic energy conserved only in an elastic collision

Impulse: FΔt = Δp = area under the force–time graph; longer contact time → smaller force

Circular motion: ω = 2π/T, v = rω, a = v²/r = ω²r, F = mv²/r = mω²r

Centripetal force is not a new force — it is whatever real force points to the centre; it does no work

Cut the string and the object leaves along the tangent

That is Edexcel Topic 6 — and circular motion is exactly what you will need for orbits in Topic 12. Press Finish to see your score.

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Mini-lesson complete!

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