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This lesson covers AQA Spec 3.4.1.6 (Momentum) in full depth, including elastic and inelastic collisions, explosions, 2D momentum, and the impulse-momentum theorem.
Spec mapping (AQA 7408 §3.4.1.6): This lesson develops linear momentum as the product p = mv, Newton's second law in its general form F = Δp/Δt, conservation of momentum in isolated systems, the distinction between elastic and inelastic collisions, the impulse-momentum theorem J = FΔt = Δp, and the application to vehicle safety features that increase collision time. Two-dimensional momentum conservation (independent component conservation) extends the framework where the spec requires it. (Refer to the official AQA specification document for exact wording.)
Synoptic links: Momentum is the gateway to conservation thinking in physics. (i) Newton's second law (Lesson 3, §3.4.1.4) is the prerequisite — F = Δp/Δt reduces to F = ma when mass is constant, and the impulse equation is just the second law integrated over time. (ii) Work and energy (Lesson 5, §3.4.1.7) runs in parallel: in an elastic collision both momentum and kinetic energy are conserved, while in an inelastic collision only momentum is conserved. Many extended questions combine the two conservation laws to solve for unknowns. (iii) Nuclear physics (Year 13, §3.8.1) uses momentum conservation in alpha decay and electron-positron creation; the "kick" from emitting a particle is a direct application of conservation of momentum in an isolated system. (iv) Particle physics in the same paper applies momentum conservation in collisions to deduce the existence of unobserved particles (the neutrino was originally inferred from missing momentum in beta decay).
Key Definition: The momentum of an object is the product of its mass and velocity: p = mv
Momentum is a vector quantity. SI unit: kg m s⁻¹ (equivalent to N s).
Newton's Second Law in its general form:
F = Δp / Δt
This is more fundamental than F = ma because it applies even when mass is changing (e.g., rockets, conveyor belts).
For constant mass: F = Δ(mv)/Δt = mΔv/Δt = ma (since Δv/Δt = a).
Principle: The total momentum of a system remains constant, provided no external resultant force acts on the system.
This applies to all types of collisions and explosions.
Mathematically: Σp_before = Σp_after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
| Type | Momentum Conserved? | KE Conserved? | Example |
|---|---|---|---|
| Perfectly elastic | Yes | Yes | Atomic/molecular collisions |
| Inelastic | Yes | No (KE decreases) | Most real collisions |
| Perfectly inelastic | Yes | No (max KE loss) | Objects stick together |
| Explosion | Yes (Σp = 0 if from rest) | No (KE increases) | Bullet fired from gun |
Calculate the total kinetic energy before and after. If KE_before = KE_after, the collision is elastic. Otherwise, it is inelastic.
A 2.0 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 4.0 kg trolley. They stick together. Find (a) the velocity after collision, (b) the kinetic energy lost.
Solution:
(a) Conservation of momentum:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
2.0 × 3.0 + 4.0 × 0 = (2.0 + 4.0)v
6.0 = 6.0v → v = 1.0 m s⁻¹
(b) KE before = ½ × 2.0 × 3.0² = 9.0 J
KE after = ½ × 6.0 × 1.0² = 3.0 J
KE lost = 9.0 − 3.0 = 6.0 J (lost as heat, sound, and deformation)
A 3.0 kg ball moving at 4.0 m s⁻¹ collides head-on with a 1.0 kg ball moving at −2.0 m s⁻¹. After the collision, the 3.0 kg ball moves at 1.0 m s⁻¹ and the 1.0 kg ball moves at 7.0 m s⁻¹. Is this collision elastic?
Solution:
Momentum before: 3.0 × 4.0 + 1.0 × (−2.0) = 12.0 − 2.0 = 10.0 kg m s⁻¹
Momentum after: 3.0 × 1.0 + 1.0 × 7.0 = 3.0 + 7.0 = 10.0 kg m s⁻¹ ✓ Conserved.
KE before = ½ × 3.0 × 4.0² + ½ × 1.0 × 2.0² = 24.0 + 2.0 = 26.0 J
KE after = ½ × 3.0 × 1.0² + ½ × 1.0 × 7.0² = 1.5 + 24.5 = 26.0 J
KE before = KE after → Yes, the collision is elastic.
A 5.0 kg trolley at rest explodes into two pieces: a 2.0 kg piece moves to the right at 6.0 m s⁻¹. Find the velocity of the 3.0 kg piece.
Solution:
Total momentum before = 0 (at rest). By conservation:
0 = 2.0 × 6.0 + 3.0 × v
v = −12.0/3.0 = −4.0 m s⁻¹ (i.e., 4.0 m s⁻¹ to the left)
KE released = ½ × 2.0 × 6.0² + ½ × 3.0 × 4.0² = 36 + 24 = 60 J
Key Definition: Impulse is the change in momentum of an object.
Impulse = FΔt = Δp = mv − mu
Unit: N s (equivalent to kg m s⁻¹)
On a force–time graph, the impulse equals the area under the curve.
A 0.15 kg cricket ball is bowled at 30 m s⁻¹ and is hit back by a bat at 40 m s⁻¹. The bat is in contact with the ball for 0.005 s. Find (a) the impulse and (b) the average force.
Solution: Take the direction of the hit as positive.
u = −30 m s⁻¹ (towards bat), v = +40 m s⁻¹ (away from bat)
(a) Impulse = mv − mu = 0.15 × 40 − 0.15 × (−30) = 6.0 + 4.5 = 10.5 N s
(b) F = Impulse / Δt = 10.5 / 0.005 = 2100 N
Exam Tip: Be extremely careful with signs in impulse and momentum calculations. Define a positive direction first and stick to it. A ball bouncing back has its velocity sign change.
Car safety features are designed to increase the collision time, thereby reducing the maximum force on the occupants (since F = Δp/Δt — for a given Δp, increasing Δt decreases F).
| Safety Feature | How It Increases Δt |
|---|---|
| Crumple zones | Car body collapses progressively |
| Airbags | Cushion decelerates head over longer time |
| Seatbelts | Stretch slightly, preventing sudden stop |
| Cycle helmets | Foam crushes on impact |
In two-dimensional collisions, momentum is conserved independently in both the x and y directions:
A 2.0 kg ball moving at 5.0 m s⁻¹ along the x-axis strikes a stationary 2.0 kg ball. After the collision, the first ball moves at 3.0 m s⁻¹ at 30° above the x-axis. Find the speed and direction of the second ball.
Solution:
x-momentum: 2.0 × 5.0 = 2.0 × 3.0 cos 30° + 2.0 × v₂ₓ
10.0 = 2.0 × 2.598 + 2.0v₂ₓ = 5.196 + 2.0v₂ₓ
v₂ₓ = 4.804/2.0 = 2.40 m s⁻¹
y-momentum: 0 = 2.0 × 3.0 sin 30° + 2.0 × v₂ᵧ
0 = 2.0 × 1.5 + 2.0v₂ᵧ = 3.0 + 2.0v₂ᵧ
v₂ᵧ = −3.0/2.0 = −1.50 m s⁻¹ (below x-axis)
Speed: v₂ = √(2.40² + 1.50²) = √(5.76 + 2.25) = √8.01 = 2.83 m s⁻¹
Direction: θ = tan⁻¹(1.50/2.40) = 32.0° below the x-axis
When the force varies with time, the impulse is found from the area under the force–time graph. For a rectangular pulse of force F lasting time Δt, the impulse is simply FΔt. For a triangular pulse with peak force F₀ lasting time Δt, the impulse is ½F₀Δt.
A force acts on a 0.50 kg object initially at rest. The force–time graph is a triangle with peak force 200 N at t = 0.01 s, returning to zero at t = 0.02 s. Find the final velocity.
Solution:
Impulse = area = ½ × base × height = ½ × 0.02 × 200 = 2.0 N s
Δp = impulse → mv − mu = 2.0 → 0.50v − 0 = 2.0
v = 2.0 / 0.50 = 4.0 m s⁻¹
For a perfectly elastic 1D collision between mass m₁ at u₁ and mass m₂ at rest, both momentum and kinetic energy are conserved. The simultaneous equations admit a closed-form solution:
v₁ = (m₁ − m₂)/(m₁ + m₂) × u₁
v₂ = 2m₁/(m₁ + m₂) × u₁
Three limiting cases are worth memorising:
m₁ = m₂ (equal masses): v₁ = 0, v₂ = u₁. The incoming mass stops dead; the target carries away all the kinetic energy. This is the principle behind Newton's cradle and the breaking shot in pool.
m₁ ≫ m₂ (heavy hits light): v₁ ≈ u₁, v₂ ≈ 2u₁. The heavy mass barely slows; the light mass shoots off at twice the incoming speed. This is the basis for the gravitational slingshot effect used in spacecraft trajectories.
m₁ ≪ m₂ (light hits heavy): v₁ ≈ −u₁, v₂ ≈ 0. The light mass bounces back with reversed velocity; the heavy mass barely moves. This is the basis for elastic backscattering, used to probe atomic nuclei in Rutherford-style experiments.
A 3.0 kg ball moving at 6.0 m s⁻¹ collides elastically with a 1.0 kg ball at rest. Find the velocities after the collision.
Solution:
v₁ = (3.0 − 1.0)/(3.0 + 1.0) × 6.0 = (2.0/4.0) × 6.0 = 3.0 m s⁻¹ (same direction).
v₂ = 2 × 3.0 / (3.0 + 1.0) × 6.0 = (6.0/4.0) × 6.0 = 9.0 m s⁻¹ (same direction).
Check momentum: before = 3.0 × 6.0 = 18.0 kg m s⁻¹; after = 3.0 × 3.0 + 1.0 × 9.0 = 9.0 + 9.0 = 18.0 kg m s⁻¹ ✓.
Check KE: before = ½ × 3.0 × 6.0² = 54.0 J; after = ½ × 3.0 × 3.0² + ½ × 1.0 × 9.0² = 13.5 + 40.5 = 54.0 J ✓.
Note how the light target ball ends up moving faster than the original heavy projectile. This is counter-intuitive but follows directly from energy conservation: with less mass, more speed is needed to carry the same kinetic energy share.
A 12 g bullet travelling at 320 m s⁻¹ strikes a 2.5 kg wooden block resting on a frictionless surface and embeds itself in the block. Calculate (a) the velocity of the block + bullet immediately after the collision, (b) the kinetic energy converted to internal energy (heat / deformation) during the impact, (c) the fraction of the original kinetic energy lost.
Solution: This is a perfectly inelastic collision.
(a) Conservation of momentum: m_b u_b = (m_b + m_block) v.
(0.012 × 320) + 0 = (0.012 + 2.5) v
3.84 = 2.512 v → v = 3.84/2.512 = 1.528 m s⁻¹ (3 s.f.).
(b) KE before = ½ × 0.012 × 320² = ½ × 0.012 × 102 400 = 614.4 J.
KE after = ½ × 2.512 × 1.528² = ½ × 2.512 × 2.335 = 2.93 J.
KE converted to internal energy = 614.4 − 2.93 = 611.5 J (3 s.f.).
(c) Fraction lost = 611.5 / 614.4 = 0.995 = 99.5%.
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