In one line
If no net external impulse acts on a system, its total linear momentum stays the same before and after an interaction.
Explained simply
When you jump off a small boat onto the jetty, the boat moves backward. Before the jump, you and the boat had zero total momentum, so after the jump your forward momentum must be balanced by the boat's backward momentum.
Technically, for a system of particles, if \( \Sigma \vec{F}_{ext} = 0 \) (or its impulse is negligible over the short interaction), then \( \Sigma m\vec{v}_{before} = \Sigma m\vec{v}_{after} \). Internal forces (contact forces in a collision, explosion forces in a gun) cancel in pairs by Newton's third law.
How it works
- Define the system so that the large interaction forces are internal.
- Check that external impulses are zero or negligible during the short event (weight over a few milliseconds is usually negligible).
- Choose a positive direction and write total momentum before = total momentum after.
- Apply along each direction separately; momentum may be conserved in one direction only.
- Kinetic energy is generally not conserved in such events.
Key formula
\( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \); masses in kg, velocities in m/s with signs.
Quick example
A 4 kg gun fires a 20 g bullet at 400 m/s. Initially both at rest: \( 0 = 0.02(400) + 4V \), so the gun recoils at \( V = -2 \) m/s. The bullet gets \( \tfrac12(0.02)(400)^2 = 1600 \) J and the gun only \( \tfrac12(4)(2)^2 = 8 \) J, so energy is clearly not shared equally even though momentum is.
Common mistakes
- Applying conservation when a significant external force acts (e.g. friction over a long time).
- Forgetting signs for opposite directions.
- Assuming kinetic energy is also conserved.
How MU asks it
Theory 3 marks: 'State the law of conservation of momentum'. Numericals 5 to 6 marks: gun recoil, a man jumping off a boat, a bullet embedding in a block. Write the system, the condition, the equation and the result, and add a line on KE loss if asked.