Momentum (Higher Tier)

This lesson covers momentum and the conservation of momentum as required by AQA GCSE Combined Science Trilogy (8464), section 6.5.3. Momentum is a Higher Tier topic — it will only appear on Higher Tier papers. It builds on your knowledge of Newton's laws and is tested through both calculation and explanation questions.

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What Is Momentum?

Momentum is a measure of how much "motion" an object has. It depends on both the mass and the velocity of the object.

$p = m \times v$p=m×v

Where:

• p = momentum (kg m/s)
• m = mass (kg)
• v = velocity (m/s)

Key facts about momentum:

• Momentum is a vector quantity — it has both magnitude and direction.
• The direction of momentum is the same as the direction of velocity.
• A stationary object has zero momentum (because $v = 0$v=0).
• A heavier object or a faster object has more momentum.

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Worked Examples — Calculating Momentum

Example 1

A car of mass 1200 kg travels at 15 m/s to the right. Calculate its momentum.

Solution:

$p = m \times v = 1200 \times 15 = 18\,000$p=m×v=1200×15=18000 kg m/s to the right

Example 2

A tennis ball of mass 0.06 kg is hit at 50 m/s. Calculate its momentum.

Solution:

$p = 0.06 \times 50 = 3$p=0.06×50=3 kg m/s (in the direction of travel)

Example 3 — Finding Velocity from Momentum

A lorry has a momentum of 150,000 kg m/s and a mass of 6000 kg. Calculate its velocity.

Solution:

$v = \frac{p}{m} = \frac{150\,000}{6000} = 25$v=mp​=6000150000​=25 m/s

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Conservation of Momentum

The law of conservation of momentum: In a closed system (no external forces), the total momentum before an event equals the total momentum after the event.

$\text{Total momentum before} = \text{Total momentum after}$Total momentum before=Total momentum after

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$m1​u1​+m2​u2​=m1​v1​+m2​v2​

Where:

• $m_1$m1​, $m_2$m2​ = masses of the two objects
• $u_1$u1​, $u_2$u2​ = velocities before the event
• $v_1$v1​, $v_2$v2​ = velocities after the event

This applies to collisions, explosions, and any interaction in a closed system.

flowchart LR
    subgraph "Before Collision"
        A1["Object A: mass m₁, velocity u₁"] --> C["COLLISION"]
        B1["Object B: mass m₂, velocity u₂"] --> C
    end
    subgraph "After Collision"
        C --> A2["Object A: mass m₁, velocity v₁"]
        C --> B2["Object B: mass m₂, velocity v₂"]
    end

    style C fill:#e74c3c,color:#fff
    style A1 fill:#2980b9,color:#fff
    style B1 fill:#2980b9,color:#fff
    style A2 fill:#27ae60,color:#fff
    style B2 fill:#27ae60,color:#fff

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Worked Examples — Conservation of Momentum

Example 1 — Collision Where Objects Stick Together

A 2 kg trolley moving at 3 m/s collides with a stationary 1 kg trolley. After the collision, they stick together. Calculate their velocity after the collision.

Solution:

Total momentum before = $m_1 u_1 + m_2 u_2 = (2 \times 3) + (1 \times 0) = 6$m1​u1​+m2​u2​=(2×3)+(1×0)=6 kg m/s

Total momentum after = $(m_1 + m_2) \times v = 3 \times v$(m1​+m2​)×v=3×v

By conservation: $6 = 3v$6=3v

$v = \frac{6}{3} = 2$v=36​=2 m/s (in the original direction)

Example 2 — Collision Where Objects Bounce Apart

A 4 kg ball moving at 5 m/s to the right hits a 2 kg ball that is stationary. After the collision, the 4 kg ball moves at 2 m/s to the right. Find the velocity of the 2 kg ball.

Solution:

Momentum before = $(4 \times 5) + (2 \times 0) = 20$(4×5)+(2×0)=20 kg m/s

Momentum after = $(4 \times 2) + (2 \times v_2) = 8 + 2v_2$(4×2)+(2×v2​)=8+2v2​

By conservation: $20 = 8 + 2v_2$20=8+2v2​

$2v_2 = 12$2v2​=12

$v_2 = 6$v2​=6 m/s to the right

Example 3 — Explosion (Objects Initially Stationary)

A cannon of mass 500 kg fires a 5 kg cannonball at 100 m/s. Calculate the recoil velocity of the cannon.

Solution:

Total momentum before = 0 (both at rest)

Total momentum after = $(500 \times v_{\text{cannon}}) + (5 \times 100) = 500v_{\text{cannon}} + 500$(500×vcannon​)+(5×100)=500vcannon​+500

By conservation: $0 = 500v_{\text{cannon}} + 500$0=500vcannon​+500

$500v_{\text{cannon}} = -500$500vcannon​=−500

$v_{\text{cannon}} = -1$vcannon​=−1 m/s

The cannon recoils at 1 m/s in the opposite direction to the cannonball.

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Force and Momentum — Newton's Second Law (Higher Tier)

Newton's second law can also be expressed in terms of momentum:

$F = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{t}$F=ΔtΔp​=tmv−mu​

Where:

• F = force (N)
• $\Delta p$Δp = change in momentum (kg m/s)
• $\Delta t$Δt = time over which the force acts (s)

This shows that:

• A large force produces a large change in momentum.
• The same change in momentum can be achieved with a smaller force acting over a longer time.

Why Crash Safety Features Work

Car safety features (crumple zones, airbags, seatbelts) work by increasing the time over which the momentum change occurs during a crash. This reduces the force on the passengers.

Safety feature	How it works
Crumple zones	The front and rear of the car crumple on impact, increasing the time of deceleration
Seatbelts	Stretch slightly to increase the stopping time of the passenger
Airbags	Inflate and deflate slowly, increasing the time over which the head decelerates
Helmets	Foam lining compresses on impact, increasing the time of deceleration

flowchart TD
    C["Crash — momentum must change to zero"]
    C --> S["Short impact time"] --> LF["LARGE force on passenger — injury"]
    C --> L["Long impact time (safety features)"] --> SF["SMALL force on passenger — reduced injury"]

    style S fill:#e74c3c,color:#fff
    style LF fill:#c0392b,color:#fff
    style L fill:#27ae60,color:#fff
    style SF fill:#2ecc71,color:#fff

Worked Example

A 75 kg passenger in a car crash decelerates from 15 m/s to 0 m/s. Calculate the force on the passenger if (a) the crash lasts 0.1 s (no safety features) and (b) 0.5 s (with crumple zones and seatbelt).

Solution:

Change in momentum: $\Delta p = mv - mu = (75 \times 0) - (75 \times 15) = -1125$Δp=mv−mu=(75×0)−(75×15)=−1125 kg m/s

(a) $F = \frac{\Delta p}{\Delta t} = \frac{1125}{0.1} = 11\,250$F=ΔtΔp​=0.11125​=11250 N

(b) $F = \frac{\Delta p}{\Delta t} = \frac{1125}{0.5} = 2250$F=ΔtΔp​=0.51125​=2250 N

The safety features reduce the force by a factor of 5.

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Momentum Comparison Table

Quantity	Formula	Unit	Scalar or Vector
Momentum	$p = mv$p=mv	kg m/s	Vector
Force (from momentum)	$F = \Delta p / \Delta t$F=Δp/Δt	N	Vector
Kinetic energy	$KE = \frac{1}{2}mv^2$KE=21​mv2	J	Scalar

Exam Tip: Momentum is conserved in collisions, but kinetic energy is only conserved in elastic collisions. In most real-world collisions, some KE is converted to heat or sound — these are inelastic collisions. AQA Combined Science does not require you to calculate KE changes in collisions, but understanding the difference is useful.

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Common Mistakes