Momentum revision notes · AS & A Level Mathematics 9709

1Momentum and the conservation of momentum

Syllabus 4.3

Momentum \(= mv\), measured in \(\text{kg m s}^{-1}\) (or N s), with a sign showing its direction. In a direct collision: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\). If the particles coalesce, \(m_1u_1 + m_2u_2 = (m_1 + m_2)v\).

Fig. 1: two particles moving towards each other coalesce. With right as positive, the 2 kg particle has velocity −2 m s⁻¹, so the combined particle moves right at 2.2 m s⁻¹

Worked example: Change of momentum

A ball of mass 200 g hits a wall at \(15\ \text{m s}^{-1}\) and rebounds along the same line at \(12\ \text{m s}^{-1}\). Find the change in its momentum.

Mass \(= 0.2\) kg. Take the direction towards the wall as positive.
Before: \(0.2 \times 15 = 3\ \text{N s}\). After: \(0.2 \times (-12) = -2.4\ \text{N s}\).
Change \(= -2.4 - 3 = -5.4\ \text{N s}\).

Answer: 5.4 N s, directed away from the wall.

Diagram, signs, one equation.

Diagram: before and after, masses in kg, every velocity marked.
Positive direction: velocities the other way are negative.
Conserve: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\); coalescing uses \((m_1 + m_2)v\).
Interpret: a negative answer means the particle moves in the negative direction.
Chains: one collision at a time, then check whether another collision occurs.

Watch out!

Subtracting the speeds gives \(0.2(15 - 12) = 0.6\) N s, which is wrong. The velocity changed sign, so the sizes ADD: \(0.2(15 + 12) = 5.4\) N s.

Every unknown velocity after the collision should be written as a signed variable, and the sign of the answer tells you the direction. A negative answer is not an error; it means the particle moves in the negative direction.

Do not do both collisions in one equation unless all the particles really do end up moving together. If A and B separate after the first collision, each must be tracked on its own.

Revisit it: Momentum and the conservation of momentum lesson · worked practice