Impulse-momentum principle in vector form?

\mathbf{I}=m\mathbf{v}-m\mathbf{u}=m(\mathbf{v}-\mathbf{u})

Conservation of momentum in vector form?

m1\mathbf{u}1+m2\mathbf{u}2=m1\mathbf{v}1+m2\mathbf{v}2

Relationship between the impulses two colliding particles exert on each other?

Equal in magnitude and opposite in direction (Newton's third law).

Angle between a\mathbf{i}+b\mathbf{j} and the \mathbf{i} direction?

\tan\theta=\frac{b}{a}, taking care with the quadrant.

Average force from an impulse?

\mathbf{F}=\frac{\mathbf{I}}{t}

m=0.5, \mathbf{u}=4\mathbf{i}-3\mathbf{j}, \mathbf{v}=-2\mathbf{i}+5\mathbf{j}. Impulse?

0.5(-6\mathbf{i}+8\mathbf{j})=-3\mathbf{i}+4\mathbf{j} N s

How do you find a velocity from total momentum and a known particle?

Subtract the known particle's momentum, then divide by the mass of the unknown one.

Why do the impulses on the two particles sum to zero?

Because each is the negative of the other, so total momentum is unchanged.

Momentum and impulse as vectorsEdexcel A-Level Further Maths: Flashcards

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Momentum as a vector?

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Momentum as a vector?: $\mathbf{p}=m\mathbf{v}$ , in the direction of the velocity.
Impulse-momentum principle in vector form?: $\mathbf{I}=m\mathbf{v}-m\mathbf{u}=m(\mathbf{v}-\mathbf{u})$
Impulse of a constant force $\mathbf{F}$ acting for time $t$ ?: $\mathbf{I}=\mathbf{F}t$
Magnitude of $a\mathbf{i}+b\mathbf{j}$ ?: $\sqrt{a^2+b^2}$
Conservation of momentum in vector form?: $m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2$
Relationship between the impulses two colliding particles exert on each other?: Equal in magnitude and opposite in direction (Newton's third law).
Angle between $a\mathbf{i}+b\mathbf{j}$ and the $\mathbf{i}$ direction?: $\tan\theta=\frac{b}{a}$ , taking care with the quadrant.
Direction of the impulse from a smooth wall?: Perpendicular to the wall.
Average force from an impulse?: $\mathbf{F}=\frac{\mathbf{I}}{t}$
$m=0.5$ , $\mathbf{u}=4\mathbf{i}-3\mathbf{j}$ , $\mathbf{v}=-2\mathbf{i}+5\mathbf{j}$ . Impulse?: $0.5(-6\mathbf{i}+8\mathbf{j})=-3\mathbf{i}+4\mathbf{j}$ N s
How do you find a velocity from total momentum and a known particle?: Subtract the known particle's momentum, then divide by the mass of the unknown one.
Why do the impulses on the two particles sum to zero?: Because each is the negative of the other, so total momentum is unchanged.

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).