Momentum and Impulse

This lesson explains Momentum and Impulse for Cambridge IGCSE Physics 0625. It covers Supplement content for Extended candidates. Focus on the cause-and-effect explanation and the exact quantities being compared. Work through the example before testing the same skill without notes.

What is momentum and how do I calculate it?

Momentum is mass multiplied by velocity. In symbols, $p = mv$. It is a vector, so direction matters: choose one direction as positive and make the opposite direction negative. The unit is kg m/s.

Quantity	Symbol	Unit
Momentum	$p$	kg m/s
Mass	$m$	kg
Velocity	$v$	m/s
Impulse	$F\Delta t$	N s (= kg m/s)
Resultant force	$F$	N
Time	$t$	s

Impulse is force multiplied by the time for which it acts: $\text{impulse} = F\Delta t = \Delta p = mv - mu$, where $u$ is the initial velocity and $v$ the final velocity. A resultant force equals the change in momentum per unit time: $F = \dfrac{\Delta p}{\Delta t}$. This is the deeper version of $F = ma$, and it explains crumple zones: a longer impact time means a smaller force for the same momentum change.

How does conservation of momentum work in collisions?

In a closed system (no external resultant force), total momentum before an event equals total momentum after. Write it as: total $p$ before = total $p$ after. This holds for collisions and for explosions, where two stationary parts push apart with equal and opposite momentum. Kinetic energy is not usually conserved, but momentum is.

Worked example

A 1200 kg car travelling at 15 m/s collides with a stationary 800 kg van. They lock together and move off as one.

(a) Calculate the velocity of the combined vehicles immediately after the collision. [3] (b) The collision lasts 0.30 s. Calculate the average force on the van. [2]

Solution (a). Equation: total momentum before = total momentum after. Substitute: $(1200 \times 15) + (800 \times 0) = (1200 + 800) \times v$

$18\,000 = 2000v$

Rearrange: $v = 18\,000 \div 2000 = \textbf{9.0 m/s}$ in the car’s original direction.

Solution (b). Equation: $F = \dfrac{\Delta p}{\Delta t}$. Van’s change in momentum: $\Delta p = 800 \times 9.0 - 0 = 7200\ \text{kg m/s}$. Substitute: $F = 7200 \div 0.30 = \textbf{24 000 N}$ ($2.4 \times 10^{4}\ \text{N}$).

Original marking points

• M1: total momentum before $= 1200 \times 15 = 18\,000\ \text{kg m/s}$.
• M1: equating to $(2000)v$ / correct conservation statement applied.
• A1: $9.0\ \text{m/s}$ with unit.
• M1: $\Delta p = 7200\ \text{kg m/s}$ and $F = \dfrac{\Delta p}{\Delta t}$ used.
• A1: 24 000 N. Allow error carried forward from (a).

These marking points belong to this original example. They are not an official Cambridge mark scheme.

Common errors and how to correct them

• Forgetting the combined mass after coupling. Students divide by 1200, not 2000. Fix: write “after: $(m_1 + m_2)v$” before substituting.
• Ignoring direction signs. Head-on collisions need one velocity negative. Fix: draw arrows and label the positive direction first.
• Using kg m/s for force or N for momentum. Fix: momentum is kg m/s; impulse is N s; force is N. Check units in the final line.
• Mixing up $\Delta p$ and $p$. Impulse equals the change in momentum, $mv - mu$, not $mv$ alone. Fix: write both initial and final momentum every time.
• Claiming kinetic energy is conserved. In most 0625 collisions it is not. Fix: only momentum conservation is guaranteed.

How to apply this in an exam

Set out every conservation question as a before/after table: each object’s mass, velocity and momentum before, then after. Total each side and equate. This layout earns the M1 method mark even if your arithmetic slips, and it forces the sign decision before you substitute. Students who adopt this table stop losing the classic minus-sign mark within a couple of practice papers.

Where this skill matters

Momentum appears only on Extended papers. Paper 2 (MCQ) tests $p = mv$, units and simple impulse comparisons. Paper 4 sets multi-mark conservation calculations (coupling trucks, recoiling guns, exploding trolleys), often with a follow-up $F = \dfrac{\Delta p}{\Delta t}$ part like (b) above. Explain-style marks ask why crumple zones or seat belts reduce force: longer time, same $\Delta p$, smaller force. There is no direct Paper 5/6 practical.