Momentum and impulse in vector formEdexcel International A Level Further Maths: Revision notes

The momentum of a particle of mass $m$ and velocity $\mathbf{v}$ is $m\mathbf{v}$, a vector in the same direction as the velocity, measured in kg m s$^{-1}$ (or N s). Mass is a scalar, so multiplying scales each component: a $2$ kg particle with velocity $4\mathbf{i}-\mathbf{j}$ has momentum $8\mathbf{i}-2\mathbf{j}$. The speed is the magnitude of the velocity, $\sqrt{v_x^2+v_y^2}$, and the direction is found with $\tan\theta=\frac{v_y}{v_x}$ (check the quadrant). Momentum and velocity are vectors, so they add as vectors; speeds do not.

The impulse of a constant force $\mathbf{F}$ acting for a time $t$ is $\mathbf{I}=\mathbf{F}t$, measured in N s. The impulse-momentum principle says that the impulse equals the change in momentum: $\mathbf{I}=m\mathbf{v}-m\mathbf{u}=m(\mathbf{v}-\mathbf{u}).$ In vector form this applies to each component separately. Example: a $0.4$ kg ball changes velocity from $6\mathbf{i}+8\mathbf{j}$ to $-2\mathbf{i}+5\mathbf{j}$: $\mathbf{I}=0.4(-8\mathbf{i}-3\mathbf{j})=-3.2\mathbf{i}-1.2\mathbf{j}$ N s. If the impulse is given, rearrange: $\mathbf{v}=\mathbf{u}+\frac{\mathbf{I}}{m}$.

The magnitude of $\mathbf{I}=a\mathbf{i}+b\mathbf{j}$ is $\sqrt{a^2+b^2}$ and its direction is the direction of the force that acted on the particle. For $-3.2\mathbf{i}-1.2\mathbf{j}$ the magnitude is $\sqrt{11.68}=3.42$ N s, at $\tan^{-1}\frac{1.2}{3.2}=20.6^\circ$ below the direction of $-\mathbf{i}$. To find the angle through which a particle is turned, find the direction of $\mathbf{u}$ and of $\mathbf{v}$ (each from $\tan^{-1}$, checking the quadrant) and subtract, or use the scalar product.

When two particles collide and there is no external impulse (for example on a smooth horizontal plane), the total momentum is unchanged: $m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2.$ This vector equation gives two scalar equations, one for $\mathbf{i}$ and one for $\mathbf{j}$. If the particles coalesce they move together, so $m_1\mathbf{u}_1+m_2\mathbf{u}_2=(m_1+m_2)\mathbf{v}$. Newton's third law means the impulse on each particle is equal in size and opposite in direction, which is why total momentum is conserved.

$A$ ($2$ kg) has velocity $4\mathbf{i}-\mathbf{j}$ and $B$ ($3$ kg) has velocity $-2\mathbf{i}+3\mathbf{j}$. After the collision $A$ has velocity $-\mathbf{i}+2\mathbf{j}$. Total momentum: $2(4\mathbf{i}-\mathbf{j})+3(-2\mathbf{i}+3\mathbf{j})=2\mathbf{i}+7\mathbf{j}$. So $2\mathbf{i}+7\mathbf{j}=2(-\mathbf{i}+2\mathbf{j})+3\mathbf{v}_B$, giving $\mathbf{v}_B=\frac43\mathbf{i}+\mathbf{j}$ and speed $\frac53$ m s$^{-1}$. Method: (1) write each momentum as a vector; (2) apply the principle; (3) solve component by component; (4) find magnitudes and angles last. The impulse on one particle is $m(\mathbf{v}-\mathbf{u})$; on the other it is minus that.