Momentum and Conservation

This lesson covers the definition of momentum, the conservation of momentum in collisions, and how safety features use this principle, as required by the Edexcel GCSE Combined Science specification (1SC0). Momentum is a key concept that connects force, mass and velocity.

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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.

$p = mv$p=mv

where:

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

Momentum is a vector quantity — it has both magnitude and direction.

Worked Examples

Example 1: Calculate the momentum of a 1200 kg car travelling at 25 m/s.

$p = mv = 1200 \times 25 = 30\,000 \text{ kg m/s}$p=mv=1200×25=30000 kg m/s

Example 2: A 0.16 kg cricket ball has a momentum of 6.4 kg m/s. What is its velocity?

$v = \frac{p}{m} = \frac{6.4}{0.16} = 40 \text{ m/s}$v=mp​=0.166.4​=40 m/s

Example 3: A 0.5 kg ball travels at 8 m/s to the right. What is its momentum?

$p = 0.5 \times 8 = 4 \text{ kg m/s to the right}$p=0.5×8=4 kg m/s to the right

Exam Tip: Always include the direction when giving a momentum value, since momentum is a vector. Use positive for one direction and negative for the opposite direction in calculations.

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Conservation of Momentum

The law of conservation of momentum states:

In a closed system (where no external forces act), 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​

This applies to collisions, explosions, and any interaction where no external resultant force acts.

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Types of Collision

Collision Where Objects Stick Together

If two objects collide and stick together, they move with a common velocity after the collision.

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$m1​u1​+m2​u2​=(m1​+m2​)v

Worked Example

A 2 kg trolley moving at 3 m/s collides with a stationary 1 kg trolley. They stick together. Find the velocity after the collision.

Before: $p_{total} = 2 \times 3 + 1 \times 0 = 6 \text{ kg m/s}$ptotal​=2×3+1×0=6 kg m/s

After: $6 = (2 + 1) \times v$6=(2+1)×v

$v = \frac{6}{3} = 2 \text{ m/s}$v=36​=2 m/s

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Collisions in Opposite Directions

When objects move in opposite directions, assign one direction as positive and the other as negative.

Worked Example

A 3 kg ball moving at 4 m/s to the right collides with a 2 kg ball moving at 5 m/s to the left. They stick together. Find the velocity after the collision.

Taking right as positive:

Before: $p_{total} = 3 \times 4 + 2 \times (-5) = 12 - 10 = 2 \text{ kg m/s}$ptotal​=3×4+2×(−5)=12−10=2 kg m/s

After: $2 = (3 + 2) \times v$2=(3+2)×v

$v = \frac{2}{5} = 0.4 \text{ m/s to the right}$v=52​=0.4 m/s to the right

Exam Tip: When objects move in opposite directions, be very careful with signs. Choose positive for one direction at the start and stick with it throughout the calculation.

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Explosions and Recoil

In an explosion, the total momentum before is zero (both objects are stationary). By conservation of momentum, the total momentum after must also be zero.

$0 = m_1 v_1 + m_2 v_2$0=m1​v1​+m2​v2​

$m_1 v_1 = -m_2 v_2$m1​v1​=−m2​v2​

Worked Example

A cannon of mass 500 kg fires a 2 kg cannonball at 100 m/s. What is the recoil velocity of the cannon?

Before: total momentum = 0 (both at rest)

After: $0 = 500 \times v_{cannon} + 2 \times 100$0=500×vcannon​+2×100

$500 v_{cannon} = -200$500vcannon​=−200

$v_{cannon} = \frac{-200}{500} = -0.4 \text{ m/s}$vcannon​=500−200​=−0.4 m/s

The negative sign means the cannon recoils in the opposite direction to the cannonball.

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Force and Rate of Change of Momentum

Force is related to the rate of change of momentum:

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

where:

• F = force (N)
• Δp = change in momentum (kg m/s)
• Δt = time over which the change occurs (s)

Worked Example

A 60 kg person's velocity changes from 8 m/s to 0 m/s in 0.1 s. What force acts on them?

$\Delta p = mv - mu = 60 \times 0 - 60 \times 8 = -480 \text{ kg m/s}$Δp=mv−mu=60×0−60×8=−480 kg m/s

$F = \frac{\Delta p}{\Delta t} = \frac{-480}{0.1} = -4800 \text{ N}$F=ΔtΔp​=0.1−480​=−4800 N

The magnitude of the force is 4800 N.

Exam Tip: The equation $F = \Delta p / \Delta t$F=Δp/Δt explains why increasing the time of impact reduces the force. This is the key physics behind safety features.

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Safety Features and Momentum

Many safety features in vehicles are designed to increase the time over which a person's momentum changes during a collision. Since $F = \Delta p / \Delta t$F=Δp/Δt, increasing Δt reduces F, which reduces the risk of injury.

Safety Feature	How It Reduces Force
Crumple zones	Front and rear of car deform gradually, increasing the time of deceleration
Seat belts	Stretch slightly to increase the time over which the passenger decelerates
Airbags	Inflate and deflate gradually, increasing the time of deceleration of the head and upper body
Crash helmets	Foam lining compresses on impact, increasing the time of deceleration
Crash barriers	Deform on impact, increasing the time over which the vehicle decelerates

Worked Example

A 70 kg passenger in a crash decelerates from 15 m/s to 0. Compare the force with and without a crumple zone.

Change in momentum: $\Delta p = 0 - 70 \times 15 = -1050 \text{ kg m/s}$Δp=0−70×15=−1050 kg m/s

Without crumple zone (Δt = 0.05 s): $F = \frac{1050}{0.05} = 21\,000 \text{ N}$F=0.051050​=21000 N

With crumple zone (Δt = 0.5 s): $F = \frac{1050}{0.5} = 2100 \text{ N}$F=0.51050​=2100 N

The crumple zone reduces the force by a factor of 10.

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Momentum Is Always Conserved

Even in a crash where a car crumples and stops, momentum is conserved for the whole system (car + whatever it hits). The kinetic energy may not be conserved (it is converted to heat, sound and deformation), but the total momentum of the system is.

graph LR
    subgraph "Before collision"
        A["Car: 1000 kg at 20 m/s → p = 20,000 kg m/s"]
        B["Wall: at rest → p = 0"]
    end
    subgraph "After collision"
        C["Car + Wall system: total p = 20,000 kg m/s"]
        D["(Wall + Earth absorbs momentum)"]
    end

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Summary