Momentum, Impulse and Conservation

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.

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1 Momentum

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).

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2 Newton's Second Law in Terms of Momentum

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).

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3 Conservation of Linear Momentum

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

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4 Types of Collision

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

How to Determine Collision Type

Calculate the total kinetic energy before and after. If KE_before = KE_after, the collision is elastic. Otherwise, it is inelastic.

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5 Worked Examples — 1D Collisions

Worked Example 1 — Perfectly Inelastic Collision

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)

Worked Example 2 — Elastic Collision Check

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.

Worked Example 3 — Explosion

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

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6 Impulse

Key Definition: Impulse is the change in momentum of an object.

Impulse = FΔt = Δp = mv − mu