Momentum and impulse as vectorsEdexcel A-Level Further Maths: Revision notes

Section 1

Momentum as a vector

Momentum is the product of a scalar and a vector, so it is a vector in the direction of the velocity: $\mathbf{p}=m\mathbf{v}$ . In two dimensions with horizontal unit vectors $\mathbf{i}$ and $\mathbf{j}$ , a particle of mass $2$ kg with velocity $(3\mathbf{i}+\mathbf{j})$ m s $^{-1}$ has momentum $(6\mathbf{i}+2\mathbf{j})$ N s. Add and subtract momenta component by component.

Key termsmomentum vector

Section 2

Impulse in vector form

The impulse-momentum principle in vector form is $\mathbf{I}=m\mathbf{v}-m\mathbf{u}=m(\mathbf{v}-\mathbf{u}),$ and for a constant force $\mathbf{I}=\mathbf{F}t$ . The impulse is a vector: its direction is that of the change in velocity, not of the velocity itself. Example: $m=0.5$ , $\mathbf{u}=4\mathbf{i}-3\mathbf{j}$ , $\mathbf{v}=-2\mathbf{i}+5\mathbf{j}$ gives $\mathbf{I}=0.5(-6\mathbf{i}+8\mathbf{j})=-3\mathbf{i}+4\mathbf{j}$ N s.

Key termsimpulse vector

Common mistake

Taking $\mathbf{u}-\mathbf{v}$ instead of $\mathbf{v}-\mathbf{u}$ . That gives the impulse on the other body, with the opposite sign.

Section 3

Magnitude and direction

For $\mathbf{I}=a\mathbf{i}+b\mathbf{j}$ , the magnitude is $|\mathbf{I}|=\sqrt{a^2+b^2}$ and the angle with the $\mathbf{i}$ direction satisfies $\tan\theta=\frac{b}{a}$ (take care with the quadrant). Given $\mathbf{I}=-3\mathbf{i}+4\mathbf{j}$ , $|\mathbf{I}|=5$ N s and the average force over $0.1$ s is $\mathbf{F}=\frac{\mathbf{I}}{t}=(-30\mathbf{i}+40\mathbf{j})$ N with magnitude $50$ N. Divide a vector by the scalar $t$ before finding its magnitude, or divide the magnitude: both give the same result.

Key termsmagnitude

Exam tip

If a question asks for a magnitude, give a scalar with units. If it asks for the impulse, give a vector.

Section 4

Conservation of momentum in vector form

For a collision between two particles on a smooth surface, total momentum is conserved as a vector: $m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2.$ Equate the $\mathbf{i}$ components and the $\mathbf{j}$ components separately if needed. Example: $A$ ( $2$ kg, $3\mathbf{i}+\mathbf{j}$ ) and $B$ ( $3$ kg, $-2\mathbf{i}+4\mathbf{j}$ ) have total momentum $14\mathbf{j}$ . If $A$ then moves at $-3\mathbf{i}+4\mathbf{j}$ , then $3\mathbf{v}_B=14\mathbf{j}-(-6\mathbf{i}+8\mathbf{j})=6\mathbf{i}+6\mathbf{j}$ , so $\mathbf{v}_B=2\mathbf{i}+2\mathbf{j}$ .

Key termsconservation of momentum

Common mistake

Forgetting to divide by the mass when finding a velocity from a momentum. Always finish with $\mathbf{v}=\frac{\mathbf{p}}{m}$ using the right particle's mass.

Section 5

Impulses between colliding particles

The impulse that $A$ exerts on $B$ is equal and opposite to the impulse $B$ exerts on $A$ . Find either one from a change of momentum and write down the other as its negative. For the example above, the impulse on $B$ is $3[(2\mathbf{i}+2\mathbf{j})-(-2\mathbf{i}+4\mathbf{j})]=12\mathbf{i}-6\mathbf{j}$ , and on $A$ it is $2[(-3\mathbf{i}+4\mathbf{j})-(3\mathbf{i}+\mathbf{j})]=-12\mathbf{i}+6\mathbf{j}$ . For a smooth fixed wall, the impulse is perpendicular to the wall, so only that component of the velocity changes.

Key termsNewton's third law

Exam tip

Use the impulse found from both particles as a check: the two answers must be exact negatives of each other.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Momentum and impulse as vectors

A ball of mass $0.5$ kg moves on a smooth horizontal surface with velocity $(4\mathbf{i}-3\mathbf{j})$ m s $^{-1}$ when it is struck by a club. Immediately afterwards its velocity is $(-2\mathbf{i}+5\mathbf{j})$ m s $^{-1}$ . The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and perpendicular. The club is in contact with the ball for $0.1$ s.
Find the magnitude of the average force exerted by the club on the ball.2 marks
Particles $A$ and $B$ , of masses $2$ kg and $3$ kg, move on a smooth horizontal surface and collide. Before the collision $A$ has velocity $(3\mathbf{i}+\mathbf{j})$ m s $^{-1}$ and $B$ has velocity $(-2\mathbf{i}+4\mathbf{j})$ m s $^{-1}$ . After the collision $A$ has velocity $(-3\mathbf{i}+4\mathbf{j})$ m s $^{-1}$ .
Find the impulse exerted by $A$ on $B$ .2 marks
A ball of mass $0.4$ kg moves on a smooth horizontal surface and strikes a smooth fixed vertical wall that is parallel to the unit vector $\mathbf{i}$ . Just before the impact its velocity is $(6\mathbf{i}-8\mathbf{j})$ m s $^{-1}$ and just after it is $(6\mathbf{i}+4\mathbf{j})$ m s $^{-1}$ . The ball is in contact with the wall for $0.02$ s.
Find the impulse exerted by the wall on the ball, and state the direction of this impulse relative to the wall.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).