Conservation of momentumAQA A-Level Further Maths: Revision notes

The momentum of a particle of mass $m$ moving with velocity $\mathbf{v}$ is $\mathbf{p}=m\mathbf{v}$, measured in kg m s⁻¹ (or N s). It is a vector, in the same direction as the velocity. In one dimension, choose a positive direction and give velocities opposite to it a negative sign. In two dimensions with perpendicular unit vectors $\mathbf{i}$ and $\mathbf{j}$, the $\mathbf{i}$ and $\mathbf{j}$ components of momentum are treated separately. Example: $3$ kg at $4$ m s⁻¹ and $2$ kg at $1$ m s⁻¹ in the same direction have total momentum $12+2=14$ kg m s⁻¹.

When two particles collide, they exert equal and opposite forces on each other for the same time (Newton's third law), so the impulses are equal and opposite. If no external horizontal force acts, the total momentum is unchanged by the collision: $m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2.$ Method: draw a before-and-after diagram, mark a positive direction, write one momentum equation. Momentum is conserved in every collision of an isolated system, whether or not kinetic energy is conserved. On smooth horizontal ground, friction and the table's reaction do not change horizontal momentum.

• Coalescence: the particles join and move together, so $m_1u_1+m_2u_2=(m_1+m_2)v$.
• Rebound or separation: the particles move apart with different velocities. Conservation of momentum alone gives one equation; a second piece of information (such as one final velocity given, or Newton's experimental law in a later topic) is needed to find two unknown velocities.
• Explosion or recoil: particles start together, often at rest, so the total momentum is zero. A gun of mass $M$ firing a shell of mass $m$ at speed $v$ recoils with speed $\frac{mv}{M}$ in the opposite direction. In recoil questions, check whether a speed is given relative to the ground or relative to the gun. If it is relative to the gun, add or subtract the recoil speed.

When velocities are given as $a\mathbf{i}+b\mathbf{j}$, conserve momentum as a vector equation. This means the $\mathbf{i}$ components balance and the $\mathbf{j}$ components balance separately. At AS level you are not required to resolve velocities into components yourself. Example: $A$ (mass $2$) with velocity $4\mathbf{i}+\mathbf{j}$ and $B$ (mass $3$) with velocity $-2\mathbf{i}+3\mathbf{j}$ have total momentum $2\mathbf{i}+11\mathbf{j}$. If $A$ then moves with velocity $\mathbf{i}+\mathbf{j}$, then $3\mathbf{v}_B=2\mathbf{i}+11\mathbf{j}-2(\mathbf{i}+\mathbf{j})=9\mathbf{j}$, so $\mathbf{v}_B=3\mathbf{j}$. The speed is the magnitude: $|a\mathbf{i}+b\mathbf{j}|=\sqrt{a^2+b^2}$.

Kinetic energy $\frac12mv^2$ is not the same as momentum and is not conserved in most collisions. Some is transformed to sound, heat and deformation. The kinetic energy after a collision is never greater than before, unless energy is added (for example an explosion). Kinetic energy lost $=$ KE before $-$ KE after. For vector velocities use $v^2=a^2+b^2$. Example: $3$ kg at $4$ m s⁻¹ coalescing with $2$ kg at $1$ m s⁻¹: before $25$ J, after $\frac12(5)(2.8)^2=19.6$ J, so $5.4$ J is lost.