Physics

1.16 - Momentum - Conservation

8 min

Momentum

Conservation of momentum

Recoil and explosions

note 1 of 5

Momentum

Every moving object has momentum. The greater an object's mass and the greater its velocity, the more momentum it has.
Momentum is a vector quantity - it has both a magnitude (size) and a direction.

Formula for momentum

$p = m v$

Where:

$p$ = Momentum, in kilogram metres per second (kg m/s)
$m$ = Mass, in kilograms (kg)
$v$ = Velocity, in metres per second (m/s)

note 2 of 5

Worked example - Calculating momentum

A car of mass 1,100 kg is travelling at 15 m/s. Calculate its momentum.

Step 1 - Formula

$p = m v$

Step 2 - Substitution and calculation

$p = 1,100 \times 15 = 16,500 kg m/s$

note 3 of 5

Conservation of momentum

In a closed system (where no external forces act), the total momentum before an event equals the total momentum after it. This is the principle of conservation of momentum.
It applies to both collisions and explosions.

Collisions

When objects collide, momentum is transferred between them, but the total momentum remains the same.
When two objects collide and lock together, they move off at a lower velocity because the combined mass is larger.

Explosions

In an explosion, the total momentum before is zero, so the total momentum after is also zero.
The parts move in opposite directions so their momenta cancel out.

Why momentum is conserved

Conservation of momentum follows from Newton's third law.
Colliding objects exert equal and opposite forces on each other for the same amount of time, so their changes in momentum are equal and opposite and cancel out.

note 4 of 5

Worked example - Velocity after a collision

Trolley A (mass 0.8 kg) moves at 3.0 m/s towards trolley B (mass 2.0 kg), which moves at 0.6 m/s in the same direction. They collide. After the collision trolley A is stationary and trolley B moves away. Calculate the velocity of trolley B after the collision.

Step 1 - Total momentum before

The trolleys move the same way, so add their momenta:

$\begin{matrix} p_{before} & = (0.8 \times 3.0) + (2.0 \times 0.6) \\ = 3.6 kg m/s \end{matrix}$

Step 2 - Apply conservation of momentum

Momentum is conserved, so the total after equals the total before:

$p_{after} = 3.6 kg m/s$

Step 3 - Write the total after in terms of the unknown

Trolley A is stationary ( $v_{A} = 0$ ):

$3.6 = (0.8 \times 0) + (2.0 \times v_{B})$

Step 4 - Solve for the velocity

$v_{B} = \frac{3.6}{2.0} = 1.8 m/s$

note 5 of 5

Worked example - Head-on collision, opposite directions

Two dodgem cars approach head-on, collide and lock together. Car P (mass 240 kg) moves at 2.5 m/s and car Q (mass 160 kg) moves at 2 m/s in the opposite direction. Calculate the velocity they move off with after the collision.

Step 1 - Choose a positive direction

Take 'to the right' (car P's direction) as positive, so car Q's velocity is negative.

Step 2 - Total momentum before

$\begin{matrix} p_{before} & = (240 \times 2.5) + (160 \times (- 2)) \\ = 600 - 320 \\ = 280 kg m/s \end{matrix}$

Step 3 - Total momentum after

The cars move together as a combined mass of 240 + 160 = 400 kg at velocity $v$ :

$p_{after} = 400 v$

Step 4 - Apply conservation of momentum and solve

$\begin{matrix} 400 v & = 280 \\ \Rightarrow v & = \frac{280}{400} \\ = 0.7 m/s to the right \end{matrix}$

key terms

TermDefinition

MomentumThe product of an object's mass and velocity.

Closed systemA system where no matter or energy can enter or leave, and no external forces act on it.

Principle of conservation of momentumIn a closed system, the total momentum before an event is equal to the total momentum after the event.

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Key points to remember

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