AP Physics C: Mechanics

Magnitude and direction

A scalar is a quantity with a size and nothing else: mass, time, distance, speed, energy and temperature are examples. A vector has a size and a direction, and two vectors are equal only if both agree. Position, displacement, velocity, acceleration, force, momentum and torque are vectors. The size of a vector is its magnitude, and it is never negative.

A vector is drawn as an arrow. The arrow points in the direction of the vector and its length is proportional to the magnitude, so a velocity arrow twice as long stands for twice the speed. Arrows can be moved about the page without changing the vector, provided the length and direction stay the same, which is what makes graphical addition possible.

Distance and speed are scalars; the matching vectors are displacement and velocity. A runner who completes one lap of a track has covered a distance of 400 m and has a displacement of zero, because displacement compares only the start and the finish. The average speed over the lap is positive, and the average velocity is zero.

In a one-dimensional problem a vector has only two possible directions, and these are told apart by sign. Once a positive direction has been chosen, a vector pointing the other way takes a negative sign, so a velocity of $-4\ \mathrm{m/s}$ means 4 m/s in the negative direction. The sign belongs to the component along the chosen axis and says nothing about whether the object is speeding up or slowing down.

Unit vectors and components

A unit vector has magnitude 1 and no units, and is used only to carry a direction. The unit vectors along the $x$, $y$ and $z$ axes are written $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$. Any vector is the sum of its components along these axes:

Each component is a scalar with a sign. Its magnitude and direction are found from the components. In two dimensions, for a vector making an angle $\theta$ with the positive $x$ axis,

and in three dimensions $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$. When $A_x$ is negative the inverse tangent alone gives an angle in the wrong quadrant, so the signs of both components must be checked against a sketch.

The position vector $\vec{r}$ of a point runs from the origin to the point. The unit vector in that direction is $\hat{r}$, found by dividing the vector by its own magnitude:

This unit vector is used whenever a quantity points along the line joining two things, for instance the gravitational force between two masses, which points along $\hat{r}$ or against it.

Adding vectors

The resultant of several vectors is their vector sum. It is found by adding the matching components:

Subtraction works the same way, component by component, because $\vec{A} - \vec{B}$ is $\vec{A}$ added to $-\vec{B}$, a vector of the same length pointing the other way. Graphically the vectors are placed head to tail and the resultant runs from the tail of the first to the head of the last; the order does not matter. Multiplying a vector by a scalar multiplies every component, so the direction is unchanged unless the scalar is negative, which reverses it.

Magnitudes do not add like components. Two vectors of magnitudes 3 and 4 have a resultant anywhere from 1 to 7, and it is 5 only when they are at right angles.

Worked example

Take $\vec{A} = (3\,\hat{\imath} + 4\,\hat{\jmath})\ \mathrm{m}$ and $\vec{B} = (5\,\hat{\imath} - 2\,\hat{\jmath})\ \mathrm{m}$. The resultant is

with magnitude $\sqrt{64 + 4} = 8.25\ \mathrm{m}$ and direction $\tan^{-1}(2/8) = 14.0^\circ$ above the $+x$ axis. The difference is $\vec{A} - \vec{B} = (-2\,\hat{\imath} + 6\,\hat{\jmath})\ \mathrm{m}$, of magnitude $\sqrt{4 + 36} = 6.32\ \mathrm{m}$. Because its $x$ component is negative and its $y$ component positive, its direction lies in the second quadrant, at $180^\circ - \tan^{-1}(6/2) = 108.4^\circ$ from the $+x$ axis.

Products of two vectors

Two ways of multiplying vectors are used in mechanics, and each has its own meaning.

The scalar product (dot product) gives a scalar. For vectors at an angle $\theta$ between their directions,

It measures how much of one vector lies along the other. It is zero for perpendicular vectors, positive when the angle between them is under $90^\circ$ and negative when it is over. Work is the scalar product of force and displacement, and power is the scalar product of force and velocity.

The vector product (cross product) gives a vector. Its magnitude is

and its direction is perpendicular to both vectors, given by the right-hand rule: with the fingers pointing along $\vec{A}$ and curling towards $\vec{B}$ through the smaller angle, the thumb gives the direction. Reversing the order reverses the result, so $\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}$. It is zero for parallel vectors. Torque and angular momentum are vector products. For vectors that lie in the $xy$ plane the product points along $z$, and its component is

With the vectors from the worked example, $\vec{A}\cdot\vec{B} = 3(5) + 4(-2) = 7\ \mathrm{m^2}$ and $(\vec{A}\times\vec{B})_z = 3(-2) - 4(5) = -26\ \mathrm{m^2}$, so $\vec{A}\times\vec{B} = -26\,\hat{k}\ \mathrm{m^2}$. As a check, $A = 5.00\ \mathrm{m}$ and $B = \sqrt{29} = 5.39\ \mathrm{m}$, and the angle from $\cos\theta = 7/26.9$ is $74.9^\circ$, which gives $AB\sin\theta = 26.0\ \mathrm{m^2}$ for the magnitude.

The negative sign shows that turning $\vec{A}$ towards $\vec{B}$ is clockwise when viewed from the positive $z$ axis. That agrees with the picture: $\vec{A}$ points at $53^\circ$ above the $x$ axis and $\vec{B}$ at $22^\circ$ below it, so $\vec{B}$ lies clockwise from $\vec{A}$.