Conservation of Momentum | CIE A Level Physics Revision Notes 2025

The principle of conservation of momentum

The principle of conservation of linear momentum states that:

The total linear momentum before a collision is equal to the total linear momentum after a collision unless the system is acted on by a resultant external force

Therefore:

momentum before = momentum after

Momentum is a vector quantity, therefore:
- opposing vectors can cancel each other out, resulting in a net momentum of zero
- an object that collides with another object and rebounds, has a positive velocity before the collision and a negative velocity after
Momentum is always conserved

For example:
- Ball A moves with an initial velocity of $u_{A}$
- Ball A collides with Ball B which is stationary

After the collision, both balls travel in opposite directions
- Taking the direction of the initial motion of Ball A as the positive direction (to the right)
- The momentum before the collision is

$p_{b e f o r e} = m_{A} u_{A} + 0$

- The momentum after the collision is

$p_{a f t e r} = - m_{A} v_{A} + m_{B} v_{B}$

The minus sign shows that Ball A travels in the opposite direction to the initial travel
- If an object is stationary, like Ball B before the collision, then it has a momentum of zero

conversation-of-momentum

The conservation of momentum for two objects A and B colliding then moving apart

External and internal forces

External forces are forces that act on a structure from outside e.g. friction and weight
Internal forces are forces exchanged by the particles in the system e.g. tension in a string
- Which forces are internal or external will depend on the system itself
A system with no external forces acting can be described as a closed or isolated system

Internal and external forces for a mass on a spring

The spring force is internal to the system because the spring is part of the system, whereas weight (the gravitational pull of the Earth on the mass) is external to the system

Collisions in one & two dimensions

One-dimensional momentum problems

Recall that linear momentum is p = mv
Using the conversation of linear momentum, it is possible to calculate missing velocities and masses of components in a system

Elastic collisions are commonly those where objects colliding do not stick together and then move in opposite directions
Inelastic collisions are where objects collide and stick together after the collision
To find out whether a collision is elastic or inelastic, compare the amount of kinetic energy in the system before and after the collision
- If the kinetic energy is conserved, it is an elastic collision
- If the kinetic energy is not conserved, it is an inelastic collision

Worked example

Trolley A, of mass 0.80 kg.

Trolley A collides head-on with stationary Trolley B at a velocity of 3.0 m s^-1.

Trolley B has twice the mass of Trolley A. The trolleys stick together.

(a)

Using the conservation of momentum, calculate the common velocity of both trolleys after the collision.

(b)

Prove that the collision is inelastic.

Answer:

(a)

Step 1: List the known quantities

Mass of Trolley A, m_A = 0.80 kg
Velocity of Trolley A, v_A = 3.0 m s^-1
Mass of Trolley B, m_B = 2m_A = 1.60 kg

Step 2: Determine the momentum before the collision

$p_{b e f o r e} = m_{A} v_{A} + m_{B} v_{B}$

$p_{b e f o r e} = (0.80 \times 3.0) + 0$

$p_{b e f o r e} = 2.4 kg m s^{- 1}$

Since Trolly B was stationary before the collision, its momentum is zero

Step 3: Determine the momentum after the collision

$p_{a f t e r} = (m_{A} + m_{B}) \times v_{A + B}$

$p_{a f t e r} = (0.80 + 1.60) \times v_{A + B}$

$p_{a f t e r} = 2.4 \times v_{A + B}$

Step 4: Equate the momentum before and after the collision using the principle of conservation of momentum

momentum before collision = momentum after collision

$p_{b e f o r e} = p_{a f t e r}$

$2.4 = 2.4 \times v_{A + B}$

Step 5: Calculate the common velocity of both trolleys after the collision

$v_{A + B} = \frac{2.4}{2.4}$

$v_{A + B} = 1.0 m s^{- 1}$

(b)

Step 1: Calculate the kinetic energy of the system before the collision

$E_{k b e f o r e} = \frac{1}{2} m_{A} {(v_{A})}^{2} + \frac{1}{2} m_{B} {(v_{B})}^{2}$

$E_{k b e f o r e} = 0.5 \times 0.80 \times {(3.0)}^{2} + 0$

$E_{k b e f o r e} = 3.6 J$

Since trolly B was stationary before the collision, its kinetic energy is zero

Step 2: Calculate the kinetic energy of the system after the collision

$E_{k a f t e r} = \frac{1}{2} m_{A + B} {(v_{A + B})}^{2}$

$E_{k a f t e r} = 0.5 \times (0.8 + 1.6) \times {(1.0)}^{2}$

$E_{k a f t e r} = 1.2 J$

Step 3: State whether the collision was inelastic

The kinetic energy before the collision, 3.6 J, was greater than the kinetic energy after the collision, 1.2 J
Therefore, the kinetic energy of the system is not conserved, and the collision is inelastic

Examiner Tip

If an object is stationary or at rest, its velocity equals 0, therefore, the momentum and kinetic energy are also equal to 0.

When an inelastic collision occurs in which two objects are stuck together, treat the final object as a single object with a mass equal to the sum of the two individual objects.

Two-dimensional momentum problems

Since momentum is a vector, in two dimensions it can be split up into its x and y components
See Scalars & Vectors for a reminder on resolving vectors

Worked example

A ball is thrown at a vertical wall.

The ball is thrown from position S with an initial velocity of 15.0 m s^-1 at 60.0° to the horizontal.

The ball hits the wall at position P and rebounds at a velocity of 4.6 m s^-1

The mass of the ball is 60 × 10^-3 kg.

Calculate the change in momentum of the ball as it bounces off the wall.

Answer:

Step 1: List the known quantities and assign direction

Initial horizontal velocity, v_i = 15.0 m s^-1 at 60.0° to the horizontal
Final horizontal velocity, v_f = −4.6 m s^-1
Mass of ball, m = 60 × 10^-3 kg

Step 2: State the change in momentum equation

$∆ p = m (v_{f} - v_{i})$

Step 3: Calculate the initial velocity

A change in momentum is only due to the horizontal velocity component for projectile motion

$v_{i} = 15.0 \times \cos (60)$

$v_{i} = 7.5 m s^{- 1}$

Step 4: Calculate the change in momentum

$∆ p = m (v_{f} - v_{i})$

$∆ p = (60 \times 10^{- 3}) \times (- 4.6 - 7.5)$

$∆ p = - 0.73 N s$

The answer should be negative because the ball is traveling in the opposite direction after it rebounds

Examiner Tip

For problems in two dimensions, make sure you’re confident resolving vectors.

Here is a little trick to help you remember which component is cosine or sine of the angle for a vector R: Vector components, downloadable AS & A Level Physics revision notes

Resolving vectors with sine and cosine

Conservation of Momentum (CIE A Level Physics)

Revision Note

The principle of conservation of momentum

External and internal forces

Internal and external forces for a mass on a spring

Collisions in one & two dimensions

One-dimensional momentum problems

Worked example

Examiner Tip

Two-dimensional momentum problems

Worked example

Examiner Tip

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Author: Leander