Momentum and conservation of momentumEdexcel International A Level Physics: Revision notes

The momentum of a body is the product of its mass and velocity: $p = mv$. The unit is kg m s⁻¹ (equivalently N s). Because velocity is a vector, momentum is a vector with the same direction as the velocity.

In one dimension, choose a positive direction and give momentum in the opposite direction a negative sign. A 0.80 kg trolley moving at $1.5\ \text{m s}^{-1}$ has $p = 0.80 \times 1.5 = 1.2\ \text{kg m s}^{-1}$.

Newton's second law states that the resultant force on a body is equal to its rate of change of momentum, in the direction of the force:

$F = \frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$

For constant mass this becomes $F = m\frac{\Delta v}{\Delta t} = ma$. The change in momentum is $\Delta p = m(v - u)$, and the quantity $F\Delta t$ is called the impulse.

For a fixed change in momentum, a longer contact time means a smaller average force. This is why crumple zones, seat belts and padded surfaces reduce injury.

When body A exerts a force on body B, body B exerts a force on body A that is equal in magnitude, opposite in direction and of the same type. The two forces act on different bodies, so they never cancel each other out on one body.

The forces act for exactly the same time. Using $F = \Delta p/\Delta t$, A and B therefore receive equal and opposite changes of momentum. This is why momentum is conserved in an interaction.

The principle of conservation of linear momentum states that, for a system of interacting bodies, the total momentum before an interaction equals the total momentum after it, provided no external resultant force acts on the system.

For two bodies in one dimension: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.

In a collision or explosion the interaction time is short, so resistive forces are usually negligible. Examples are trolleys colliding on a track, a rifle recoiling as a bullet is fired, or two skaters pushing apart. When bodies start at rest, the total momentum is zero both before and after, so the momenta of the parts are equal and opposite.

A $0.80\ \text{kg}$ trolley at $1.5\ \text{m s}^{-1}$ hits a stationary $0.40\ \text{kg}$ trolley and they stick together.

Momentum before: $0.80 \times 1.5 + 0.40 \times 0 = 1.2\ \text{kg m s}^{-1}$.

Momentum after: $(0.80 + 0.40)v = 1.2v$.

So $v = 1.0\ \text{m s}^{-1}$ in the original direction.

Check using Newton's third law: the stationary trolley gains $0.40 \times 1.0 = 0.40$ kg m s⁻¹, and the first trolley loses $1.2 - 0.80 \times 1.0 = 0.40$ kg m s⁻¹.