Define linear momentum and recall that it is a vector quantity
Use p = mv in calculations involving momentum
State and apply Newton's second law in the form F = Δp / Δt
Define impulse and relate it to the area under a force–time graph
Apply the principle of conservation of momentum to collisions and explosions
Distinguish between elastic and inelastic collisions using energy considerations
Part 1
Linear Momentum
The momentum of a body is defined as the product of its mass and its velocity. Because velocity is a vector, momentum is also a vector — it has both magnitude and direction. In one-dimensional problems the direction is indicated by the sign (positive or negative).
Momentum
p = mv
where p is momentum (kg m s⁻¹ or equivalently N s), m is mass (kg) and v is velocity (m s⁻¹).
⚡ Key Point
Momentum is a vector. A 2 kg object moving right at 3 m s⁻¹ has momentum +6 kg m s⁻¹. The same object moving left at 3 m s⁻¹ has momentum −6 kg m s⁻¹. Always choose and state a positive direction.
A heavy lorry moving slowly can have the same momentum as a light car moving quickly. Momentum captures the idea that both mass and speed matter when we consider how difficult it is to stop something.
Different masses and velocities can give the same momentum.
Part 2
Newton's Second Law & Impulse
Newton's second law can be expressed in its most general form: the net force acting on an object equals the rate of change of momentum.
Newton's Second Law (general form)
F = Δp / Δt
If mass is constant, this simplifies to F = ma because Δp = mΔv and so Δp/Δt = m(Δv/Δt) = ma. The F = ma form is a special case of the more general momentum equation.
Rearranging gives the definition of impulse: the product of force and the time over which it acts.
Impulse
Impulse = FΔt = Δp = mv − mu
Impulse has units of N s (equivalent to kg m s⁻¹). It equals the change in momentum of the object.
⚡ Exam Tip
Impulse is the area under a force–time graph. For a constant force this is a rectangle. For a varying force you may need to count squares or estimate the area. This is a very common exam question — always check if you can find impulse from a graph rather than calculating it directly.
Impulse equals the area under a force–time graph.
Real-world application: Car safety features increase the time over which a collision happens (Δt increases), which reduces the force for the same change in momentum. Airbags, crumple zones and seatbelts all work on this principle.
Same change in momentum (Δp), but a longer collision time means a smaller force.
Check your understanding
Knowledge Check — Momentum & Impulse
1
A tennis ball of mass 58 g is served at 55 m s⁻¹. Calculate its momentum.
2 marks
p = mv = 0.058 × 55 ✓
p = 3.2 kg m s⁻¹ ✓
2
A 900 kg car travelling at 20 m s⁻¹ brakes to a halt in 6.0 s. Calculate the average braking force.
3 marks
Δp = mv − mu = 0 − (900 × 20) = −18 000 kg m s⁻¹ ✓
F = Δp/Δt = −18 000 / 6.0 ✓
F = −3000 N (the negative sign indicates the force opposes the motion) ✓
3
Use the idea of impulse to explain why an airbag reduces the risk of injury to a driver in a collision.
3 marks
The driver's change in momentum (impulse) is the same with or without the airbag ✓
The airbag increases the time Δt over which the driver decelerates ✓
Since impulse = FΔt, a larger Δt means a smaller average force F on the driver, reducing injury ✓
Part 3
Conservation of Momentum
The principle of conservation of momentum states:
⚡ Fundamental Principle
In a closed system (no external forces), the total momentum before a collision or explosion equals the total momentum after it.
This applies to any interaction between objects, including collisions, explosions, and objects joining together. It works because it is a direct consequence of Newton's second and third laws.
Momentum is conserved in all collisions (provided no external forces act).
Worked example: A trolley of mass 2.0 kg moving at 4.0 m s⁻¹ collides with a stationary trolley of mass 3.0 kg. They stick together. Find their common velocity.
Solution
(2.0 × 4.0) + (3.0 × 0) = (2.0 + 3.0) × v
8.0 = 5.0v
v = 1.6 m s⁻¹
Part 4
Elastic & Inelastic Collisions
Momentum is always conserved in collisions (assuming a closed system). However, kinetic energy may or may not be conserved.
Kinetic Energy
Ek = ½mv²
There are two extreme types of collision:
Perfectly elastic collision: Both momentum and kinetic energy are conserved. Objects bounce off each other without any energy loss. This is an idealisation — real collisions always lose some energy, but billiard balls and Newton's cradle come close.
Inelastic collision: Momentum is conserved but kinetic energy is not conserved. Some kinetic energy is transferred to other forms (thermal, sound, deformation). When objects stick together after colliding, this is a completely inelastic collision — the maximum possible kinetic energy is lost while still conserving momentum.
⚡ Exam Tip
Never assume kinetic energy is conserved unless the question explicitly says the collision is elastic. Always check by calculating ½mv² before and after. If a question says objects "stick together", the collision is completely inelastic — kinetic energy is definitely not conserved.
Worked example — checking energy: In the earlier trolley example (2.0 kg at 4.0 m s⁻¹ hits 3.0 kg at rest, they stick together at 1.6 m s⁻¹), is kinetic energy conserved?
Check
Ek before = ½(2.0)(4.0)² + ½(3.0)(0)² = 16.0 J
Ek after = ½(5.0)(1.6)² = 6.4 J
Energy lost = 9.6 J → collision is inelastic ✓
Interactive
Collision Simulator
1D Collision Simulator
Total p: — kg m s⁻¹
Total Ek: — J
Type: —
3.0
4.0
3.0
−2.0
Exam Practice
Exam-Style Questions
1
A ball of mass 0.15 kg is thrown against a wall. It strikes the wall horizontally at 12 m s⁻¹ and rebounds at 8.0 m s⁻¹. The ball is in contact with the wall for 0.050 s.
(a) Calculate the impulse exerted on the ball. [3 marks]
(b) Calculate the average force exerted on the ball by the wall. [2 marks]
(c) State the magnitude of the force exerted on the wall by the ball. Explain your answer. [2 marks]
Impulse = −3.0 N s (the negative sign shows the impulse is in the opposite direction to the initial velocity) ✓
(b)
F = impulse / Δt = −3.0 / 0.050 ✓
F = −60 N (60 N in the opposite direction to the initial velocity) ✓
(c)
60 N (by Newton's third law, the wall exerts a force on the ball and the ball exerts an equal and opposite force on the wall) ✓
The force on the wall is in the same direction as the initial velocity of the ball ✓
2
Two ice skaters are initially at rest on frictionless ice. Skater A has mass 65 kg and skater B has mass 45 kg. Skater A pushes skater B. After the push, skater A moves backwards at 1.2 m s⁻¹.
(a) Calculate the velocity of skater B after the push. [3 marks]
(b) Calculate the total kinetic energy after the push and explain where this energy came from. [3 marks]
6 marks
(a)
Total momentum before = 0 (both at rest) ✓
0 = (65 × −1.2) + 45v → 0 = −78 + 45v ✓
v = 78/45 = 1.7 m s⁻¹ (in the opposite direction to A) ✓
(b)
Ek = ½(65)(1.2)² + ½(45)(1.7)² = 46.8 + 65.0 ✓
Ek total = 112 J ✓
This kinetic energy comes from the chemical energy in the skater's muscles (work done by the push). It is not a collision — it is an explosion-type event where internal energy is converted to kinetic energy ✓
3
A trolley of mass 1.5 kg travelling at 3.0 m s⁻¹ collides with a stationary trolley of mass 2.5 kg. After the collision, the 1.5 kg trolley moves at 0.60 m s⁻¹ in the same direction.
(a) Calculate the velocity of the 2.5 kg trolley after the collision. [3 marks]
(b) Show that the collision is inelastic. [2 marks]
(c) State what happens to the kinetic energy that is lost. [1 mark]