Collisions in two dimensionsEdexcel A-Level Physics: Revision notes

The principle of conservation of linear momentum states that the total momentum of a system of objects is constant, provided no resultant external force acts. Momentum is a vector, so in a two-dimensional collision it is conserved in every direction.

The usual method is to resolve each momentum into components along two perpendicular directions, often parallel and perpendicular to the direction of the initial motion, and apply conservation to each direction separately:

$\sum p_x \text{ (before)} = \sum p_x \text{ (after)}, \qquad \sum p_y \text{ (before)} = \sum p_y \text{ (after)}$

A ball of mass $m$ at speed $u$ strikes an identical stationary ball. Afterwards the first ball moves at $v_1$, angle $\theta_1$ to the original direction, and the second at $v_2$, angle $\theta_2$ on the other side.

Parallel: $mu = mv_1\cos\theta_1 + mv_2\cos\theta_2$

Perpendicular: $0 = mv_1\sin\theta_1 - mv_2\sin\theta_2$

For $u = 2.0$ m s⁻¹, $v_1 = 1.2$ m s⁻¹ at 53°: $1.2\sin 53^\circ = v_2 \sin\theta_2$ and $2.0 = 1.2\cos 53^\circ + v_2\cos\theta_2$. These give $v_2 = 1.6$ m s⁻¹ and $\theta_2 = 37^\circ$.

In an elastic collision the total kinetic energy is the same before and after. In an inelastic collision some kinetic energy is transferred to other forms, such as thermal energy, sound or deformation.

Momentum is conserved in both types, provided no external force acts. If the objects stick together, the collision is perfectly inelastic, with the maximum kinetic energy lost consistent with conserved momentum.

To decide which type, calculate the total kinetic energy before and after. In an elastic collision between a moving object and an identical stationary object, the two move off at right angles to each other.

For a non-relativistic particle, $E_k = \tfrac{1}{2}mv^2$ and $p = mv$. Substituting $v = p/m$ gives:

$E_k = \dfrac{p^2}{2m}$

This is useful when you are given momenta rather than speeds, for example for particles such as neutrons and protons.

Worked example. A neutron has $p = 3.34 \times 10^{-21}$ kg m s⁻¹ and $m = 1.67 \times 10^{-27}$ kg. $E_k = (3.34 \times 10^{-21})^2 \div (2 \times 1.67 \times 10^{-27}) = 3.3 \times 10^{-15}$ J.

Collisions between small spheres can be analysed using a video camera and software:

1. Film the collision on a smooth horizontal surface, with a ruler of known length in the picture to calibrate distances. The frame rate gives the time between frames.
2. Measure the masses of the spheres with a balance.
3. Mark the positions of each sphere in successive frames, and find speeds as distance ÷ time before and after the collision.
4. Resolve the momenta into components, and compare totals before and after.
5. Calculate total kinetic energy before and after to find out whether the collision is elastic.

Uncertainties come from reading positions, the ruler calibration and friction. Repeating for several collisions improves reliability.