Momentum

Conservation of momentum

Conservation of momentum Supplement 3 min read

Extended (Supplement)

Principle of conservation of momentum

The total momentum of a system of objects is conserved (remains constant) provided no external resultant force acts on the system. \[ \text{total momentum before} = \text{total momentum after} \] \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] where \( u \) = velocity before, \( v \) = velocity after the collision.

This applies to:

All collisions (elastic and inelastic)
Explosions (e.g. a gun firing a bullet, rockets, a compressed spring released between two trolleys)
Any interaction in a closed system with no external force

Always define a positive direction first and keep sign conventions consistent throughout.

Collisions

Type	Momentum conserved?	Kinetic energy conserved?	Example
Elastic	Yes ✓	Yes ✓	Ideal gas molecules, Newton's cradle (approximately)
Inelastic	Yes ✓	No — some KE lost as heat/sound/deformation	Most real collisions (cars, clay, sports)
Perfectly inelastic	Yes ✓	No — maximum KE lost	Objects stick together (railway wagons coupling)

Momentum is always conserved in a collision. Kinetic energy is only conserved in a perfectly elastic collision. For IGCSE, assume collisions are inelastic unless told otherwise — but momentum conservation always holds.

Worked examples

Example 1 — two objects collide and stick together

A 2.0 kg trolley moving at 5.0 m/s collides with a stationary 3.0 kg trolley. They stick together. Find their common velocity after the collision.

Before: \( p = (2.0 \times 5.0) + (3.0 \times 0) = 10\,\text{kg·m/s} \)

After: \( p = (2.0 + 3.0) \times v = 5.0v \)

Conservation: \( 5.0v = 10 \Rightarrow v = 2.0\,\text{m/s} \) (in the original direction)

Example 2 — head-on collision (opposite directions)

A 1.5 kg ball moving at +6.0 m/s collides head-on with a 1.0 kg ball moving at −4.0 m/s. After the collision, the 1.5 kg ball moves at +1.0 m/s. Find the velocity of the 1.0 kg ball.

Positive = rightward.

Total momentum before: \( (1.5 \times 6.0) + (1.0 \times -4.0) = 9.0 - 4.0 = 5.0\,\text{kg·m/s} \)

After: \( (1.5 \times 1.0) + 1.0 v_2 = 5.0 \Rightarrow 1.5 + v_2 = 5.0 \Rightarrow v_2 = 3.5\,\text{m/s (rightward)} \)

Explosions and recoil

In an explosion (or any event where a stationary system splits apart), the total momentum before is zero. After, the two parts move in opposite directions with equal and opposite momenta. \[ 0 = m_1 v_1 + m_2 v_2 \quad \Rightarrow \quad m_1 v_1 = -m_2 v_2 \]

Example 3 — gun recoil

A rifle of mass 4.0 kg fires a bullet of mass 0.010 kg at 400 m/s. Find the recoil velocity of the rifle.

Total momentum before = 0 (both stationary).

After: \( 0 = (0.010 \times 400) + (4.0 \times v_\text{rifle}) \)

\( 4.0 \, v_\text{rifle} = -4.0 \Rightarrow v_\text{rifle} = -1.0\,\text{m/s} \) (opposite to bullet)

Example 4 — astronaut in space pushing off a satellite

An astronaut (80 kg) at rest pushes off a 1000 kg satellite. The satellite recoils at 0.08 m/s. Find the astronaut's velocity.

Before: total \( p = 0 \).

After: \( 0 = (1000 \times -0.08) + 80 v_\text{astro} \)

\( 80\,v_\text{astro} = 80 \Rightarrow v_\text{astro} = +1.0\,\text{m/s} \) (away from satellite)

Show all momentum-conservation working in a clear "before / after" layout. Examiners award marks step by step: (1) correct p before, (2) correct p after expression, (3) correct algebra. State the direction of each velocity in your final answer.