Momentum & Collisions
Specification: OCR A H556 | Section: 3.5.1(b–e), 3.5.2 | Teaching time: approx. 5 hours
- 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
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).
where p is momentum (kg m s⁻¹ or equivalently N s), m is mass (kg) and v is velocity (m s⁻¹).
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.
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.
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 has units of N s (equivalent to kg m s⁻¹). It equals the change in momentum of the object.
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.
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.
Knowledge Check — Momentum & Impulse
A tennis ball of mass 58 g is served at 55 m s⁻¹. Calculate its momentum.
- p = mv = 0.058 × 55 ✓
- p = 3.2 kg m s⁻¹ ✓
A 900 kg car travelling at 20 m s⁻¹ brakes to a halt in 6.0 s. Calculate the average braking force.
- 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) ✓
Use the idea of impulse to explain why an airbag reduces the risk of injury to a driver in a collision.
- 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 ✓
Conservation of Momentum
The principle of conservation of momentum states:
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.
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.
8.0 = 5.0v
v = 1.6 m s⁻¹
Elastic & Inelastic Collisions
Momentum is always conserved in collisions (assuming a closed system). However, kinetic energy may or may not be conserved.
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.
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?
Ek after = ½(5.0)(1.6)² = 6.4 J
Energy lost = 9.6 J → collision is inelastic ✓
Collision Simulator
3.0
4.0
3.0
−2.0
Exam-Style Questions
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]
(a)
- Take the initial direction as positive ✓
- Impulse = Δp = m(v − u) = 0.15(−8.0 − 12) = 0.15 × (−20) ✓
- 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 ✓
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]
(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)(78/45)² = 46.8 + 67.6 ✓
- Ek total = 114.4 J before rounding = 1.1 × 10² J (2 significant figures) ✓
- 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 ✓
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]
(a)
- Conservation of momentum: (1.5 × 3.0) + (2.5 × 0) = (1.5 × 0.60) + 2.5v ✓
- 4.5 = 0.90 + 2.5v → 3.6 = 2.5v ✓
- v = 1.44 m s⁻¹ in the same direction ✓
(b)
- Ek before = ½(1.5)(3.0)² = 6.75 J; Ek after = ½(1.5)(0.60)² + ½(2.5)(1.44)² = 0.27 + 2.59 = 2.86 J ✓
- Since 2.86 J ≠ 6.75 J, kinetic energy is not conserved → inelastic ✓
(c)
- Converted to thermal energy (heat) and some energy is used in deforming the trolleys / sound ✓
The graph below shows the force acting on a golf ball during impact with a club.
(The force rises linearly from 0 to a maximum of 2.4 kN over 0.15 ms, then falls linearly back to 0 over a further 0.35 ms.)
(a) Estimate the impulse delivered to the ball. [3 marks]
(b) The ball has mass 46 g and is initially at rest. Calculate its speed immediately after impact. [2 marks]
(a)
- Impulse = area under F–t graph (triangle) ✓
- Area = ½ × base × height = ½ × (0.15 + 0.35) × 10⁻³ × 2400 ✓
- Impulse = ½ × 0.50 × 10⁻³ × 2400 = 0.60 N s ✓
(b)
- Impulse = Δp = mv − 0 → v = impulse/m = 0.60 / 0.046 ✓
- v = 13 m s⁻¹ ✓
Topic Summary
Momentum
A vector: p = mv. Units: kg m s⁻¹ or N s. Always state direction.
Newton's 2nd Law
F = Δp/Δt. More general than F = ma (which assumes constant mass).
Impulse
FΔt = Δp. Equals the area under a force–time graph. Explains car safety features.
Conservation
Total momentum is conserved in all collisions in a closed system: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
Elastic
Both momentum and kinetic energy conserved. Objects bounce apart.
Inelastic
Momentum conserved but Ek is not. Lost energy → heat, sound, deformation. Objects may stick together.