## AP Physics C Exam

# Part IV

# Content Review for the AP Physics C Exam

3 Vectors

4 Kinematics

5 Newton’s Laws

6 Work, Energy, and Power

7 Linear Momentum

8 Rotational Motion

9 Laws of Gravitation

10 Oscillations

11 Electric Forces and Fields

12 Electric Potential and Capacitance

13 Direct Current Circuits

14 Magnetic Forces and Fields

15 Electromagnetic Induction

16 Solutions to the Chapter Review Questions

# Chapter 3

# Vectors

## INTRODUCTION

Vectors will show up all over the place in our study of physics. Some physical quantities that are represented as vectors are: displacement, velocity, acceleration, force, momentum, and electric and magnetic fields. Since vectors play such a recurring role, it’s important to become comfortable working with them; the purpose of this chapter is to provide you with a mastery of the fundamental vector algebra we’ll use in subsequent chapters. For now, we’ll restrict our study to two-dimensional vectors (that is, ones that lie flat in a plane).

## DEFINITION

A **vector** is a quantity that involves both magnitude and direction and obeys the **commutative law for addition**, which we’ll explain in a moment. A quantity that does not involve direction is a **scalar**. For example, the quantity *55 miles per hour* is a scalar, while the quantity *55 miles per hour to the north* is a vector. Other examples of scalars include: mass, work, energy, power, temperature, and electric charge.

Vectors can be denoted in several ways, including:

In textbooks, you’ll usually see one of the first two, but when it’s handwritten, you’ll see the last one.

Displacement (which is the difference between the final and initial positions) is the prototypical example of a vector:

When we say that vectors obey the commutative law for addition, we mean that if we have two vectors of the same type, for example another displacement,

then **A + B** must equal **B + A**. The vector sum **A + B** means *the vector* **A***followed by* * B*, while the vector sum

**B + A**means

*the vector*

**B***followed by*

*. That these two sums are indeed identical is shown in the following figure:*

**A**Two vectors are equal

if they have the same

magnitude and the same

direction.

## VECTOR ADDITION (GEOMETRIC)

The figure above illustrates how vectors are added to each other geometrically. Place the tail (the initial point) of one vector at the tip of the other vector, then connect the exposed tail to the exposed tip. The vector formed is the sum of the first two. This is called the “tip-to-tail” method of vector addition.

**Example 1** Add **A** + **B** using the two following vectors:

**Solution**. Place the tail of **B** at the tip of **A** and connect them:

## SCALAR MULTIPLICATION

A vector can be multiplied by a scalar (that is, by a number), and the result is a vector. If the original vector is **A** and the scalar is *k*, then the scalar multiple *k***A** is as follows:

magnitude of *k***A** = |*k*| × (magnitude of **A**)

**Example 2** Sketch the scalar multiples 2**A**, **A**, –**A**, and –3**A** of the vector **A**:

**Solution.**

Keep in mind that when you multiply a vector times a scalar of *k,* the vector becomes *k* times longer.

## VECTOR SUBTRACTION (GEOMETRIC)

To subtract one vector from another, for example, to get **A** – **B**, simply form the vector –**B**, which is the scalar multiple (–1)**B**, and add it to **A**:

**A** – **B** = **A +** (–**B**)

**Example 3** For the two vectors **A** and **B**, find the vector **A** – **B**.

**Solution.** Flip **B** around—thereby forming –**B**—and add that vector to **A**:

It is important to know that vector subtraction is not commutative: you must perform the subtraction in the order stated in the problem.

## STANDARD BASIS VECTORS

**Two-dimensional vectors,** that is, vectors that lie flat in a plane, can be written as the sum of a horizontal vector and a vertical vector. For example, in the following diagram, the vector **A** is equal to the horizontal vector **B** plus the vertical vector **C**:

The horizontal vector is always considered a scalar multiple of what’s called the **horizontal basis vector**, **i**, and the vertical vector is a scalar multiple of the **vertical basis vector**, **j**. Both of these special vectors have a magnitude of 1, and for this reason, they’re called **unit vectors.** Unit vectors are often represented by placing a hat (caret) over the vector; for example, the unit vectors **i** and **j** are sometimes denoted and , or and .

For instance, the vector **A** in the figure below is the sum of the horizontal vector **B** = 3 and the vertical vector **C** = 4.

The vectors **B** and **C** are called the **vector components** of **A**, and the scalar multiples of and which give **A**—in this case, 3 and 4—are called the **scalar components** of **A**. So vector **A** can be written as the sum *A _{x}* +

*A*, where

_{y}*A*and

_{x}*A*are the scalar components of

_{y}**A**. The component

*A*is called the

_{x}**horizontal**scalar component of

**A**, and

*A*is called the

_{y}**vertical**scalar component of

**A**.

## VECTOR OPERATIONS USING COMPONENTS

The use of components makes the vector operations of addition, subtraction, and scalar multiplication pretty straightforward.

Vector addition: *Add the respective components*

**A** + **B** = (*A _{x}* +

*B*) + (

_{x}*A*+

_{y}*B*)

_{y}Vector subtraction: *Subtract the respective components*

**A** – **B** = (*A _{x}* –

*B*) + (

_{x}*A*–

_{y}*B*)

_{y}Scalar multiplication: *Multiply each component by k*

*k***A** = (*kA _{x}*) + (

*kA*)

_{y}**Example 4** If **A** = 2 – 3 and **B** = –4 + 2, compute each of the following vectors: **A** + **B**, **A** – **B**, 2**A**, and **A** + 3**B**.

**Solution.** It’s very helpful that the given vectors **A** and **B** are written explicitly in terms of the standard basis vectors and :

**A** + **B** = (2 – 4) + (–3 + 2) = –2 –

**A** – **B** = [2 – (–4)] + (–3 –2) = 6 – 5

2**A**= 2(2) + 2(–3) = 4 – 6

**A** + 3**B** = [2 + 3(–4)] + [–3 + 3(2)] = –10 + 3

## MAGNITUDE OF A VECTOR

The magnitude of a vector can be computed with the Pythagorean theorem. The magnitude of vector **A** can be denoted in several ways: *A* or |A| or ||A||. In terms of its components, the magnitude of **A** = *A _{x}* +

*A*is given by the equation

_{y}*A* =

which is the formula for the length of the hypotenuse of a right triangle with sides of lengths *A _{x}* and

*A*.

_{y}## DIRECTION OF A VECTOR

The direction of a vector can be specified by the angle it makes with the positive *x*-axis. You can sketch the vector and use its components (and an inverse trig function) to determine the angle. For example, *θ* if denotes the angle that the vector **A** = 3 + 4 makes with the +*x* axis, then tan*θ* = 4/3, so *θ* = tan^{–1}(4/3) = 53.1°. Make sure your calculator is in the correct mode, either radian or degree, when using trig functions.

If **A** makes the angle *θ* with the +*x* axis, then its *x*- and *y*-components are *A* cos *θ* and *A* sin *θ*, respectively (where *A* is the magnitude of **A**).

In general, any vector in the plane can be written in terms of two perpendicular component vectors. For example, vector **W** (shown below) is the sum of two component vectors whose magnitudes are *W* cos *θ* and *W* sin *θ*:

## THE DOT PRODUCT

A vector can be multiplied by a scalar to yield another vector, as you saw in Example 2. For instance, we can multiply the vector **A** by the scalar 2 to get the scalar multiple 2**A**. The product is a vector that has twice the magnitude and, since the scalar is positive, the same direction as **A**.

Additionally, we can form a scalar by multiplying two vectors. The product of the vectors in this case is called the **dot product** or the **scalar product**, since the result is a scalar. Several physical concepts (work, electric and magnetic flux) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that’s parallel to the first. The dot product was invented specifically for this purpose.

Consider these two vectors, **A** and **B**:

In order to find the component of **B** that’s parallel to **A**, we do the following:

As this figure shows, the component of **B** that’s parallel to **A** has the magnitude *B* cos *θ*. So if we multiply the magnitude of **A** by the magnitude of the component of **B** that’s parallel to **A**, we would form the product *A*(*B* cos *θ*). This is the definition of the dot product of the vectors **A** and**B**

**A** • **B** = *AB* cos *θ*

where *θ* is the angle between **A** and **B**. Notice that the dot product of two vectors is a scalar.

The angle between the vectors is crucial to the value of the dot product. If *θ* = 0, then the vectors are already parallel to each other, so we can simply multiply their magnitudes: **A** • **B** = *AB*. If **A** and **B** are perpendicular, then there is no component of **B** that’s parallel to **A** (or vice versa), so the dot product should be zero. And if *θ* is greater than 90°,

then the component of one that’s *parallel* to the other is actually *antiparallel* (backwards), and this will give the dot product a negative value (because cos *θ* < 0 if 90° < *θ* < 180°).

The value of the dot product can, of course, be figured out using the definition above (*AB* cos *θ*) if *θ* is known. If *θ* is not known, the dot product can be calculated from the components of **A** and **B** in this way:

**A** • **B** =(*A _{x}* +

*A*) • (

_{y}*B*+

_{x}*B*) =

_{y}*A*+

_{x}B_{x}*A*

_{y}B_{y}This means that, to form the dot product, simply add the product of the scalar *x* components and the product of the scalar *y* components.

**Example 5**

(a) What is the dot product of the unit vectors and ?

(b) What is the product of the unit vectors and ?

**Solution.**

(a) Since and are perpendicular to each other (*θ* = 90°), their dot product must be zero, because cos 90° = 0. (In fact, unless **A** or **B** already has magnitude zero, it’s also true that two vectors are perpendicular to each other when their dot product is zero.)

(b) Because and are parallel to each other, their dot product is just the product of the magnitudes, which is 1 • 1 = 1.

**Example 6** If **A**= –2 + 4 and **B**= 6 + *B _{y}*, find the value of

*B*such that the vectors

_{y}**A**and

**B**will be perpendicular to each other.

**Solution.** Two vectors are perpendicular to each other if their dot product is zero. The dot product of **A** and **B** can be determined as follows: Multiply the scalar *x* components, (–2)(6) = –12, multiply the scalar *y* components, (4)(*B _{y}*), and add them: –12 + 4

*B*. Setting this equal to 0, we see that

_{y}*B*must equal 3.

_{y}## THE CROSS PRODUCT

Some physical concepts (torque, angular momentum, magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that’s *perpendicular* to the first. The cross product was invented for this specific purpose.

Consider these two vectors **A** and **B**:

In order to find the component of one of these vectors that’s perpendicular to the other one, we do the following:

As this figure shows, the component of **B** that’s perpendicular to **A** has magnitude *B* sin *θ*. Therefore, if we multiply the magnitude of **A** by the magnitude of the component of **B** that’s perpendicular to **A**, we form the product *A*(*B* sin *θ*). This is the magnitude of what’s called the **cross product** of the vectors **A** and **B**, denoted **A** × **B**:

|**A** × **B**| = *AB* sin *θ*

where *θ* is the angle between **A** and **B** (such that 0° ≤ *θ* ≤ 180°).

The equation above gives the magnitude of the cross product. The cross product of two vectors is another vector that’s always perpendicular to both **A** and **B**, with its direction determined by a procedure known as the *right-hand rule*. The direction of **A** × **B** is perpendicular to the plane that contains **A** and **B**, but this leads to an ambiguity, since there are two directions perpendicular to a plane (one on either side; they point in opposite directions). The following description resolves this ambiguity.

Make sure you are using your right hand. Point your index finger in the direction of the first vector, **A**, then point your middle finger in the direction of the second vector, **B**, and your thumb now points in the direction of the cross product, **A** × **B**.

This is called the **right-hand rule**.

You know that the standard basis vectors and can be used to write any two-dimensional vector, but now that we have introduced the cross product we need a third basis unit vector to write three-dimensional vectors. This third unit vector is denoted and, like and , it points along the direction of a coordinate axis. The vector is the unit vector that points along the *x*-axis, along the *y*, and now along the *z*. The three basis vectors are mutually perpendicular:

The coordinate axes shown define a *right-handed* coordinate system, because the directions of the *x*, *y*, and *z* axes obey the right-hand rule. That is, × points in the direction of . In fact, × actually equals , since the magnitude of × is 1. As a *right-handed* coordinate system, × = and × = .

Unlike the dot product, the cross product is not commutative; **A** × **B** is not equal to **B** × **A**. This is because applying the right-hand rule to determine the direction of **B** × **A** would give a vector that points in the direction opposite to that of **A** × **B**. Therefore, **B** × **A** = –(**A** × **B**).

The cross product can be computed directly from the scalar components of **A** and **B** without first determining the angle *θ*, as follows: If **A** = *A _{x}* +

*A*+

_{y}*A*and

_{z}**B**=

*B*+

_{x}*B*+

_{y}*B*, then the cross product (magnitude and direction) of

_{z}**A**and

**B**is

**A** × **B** = *(A _{y}B_{z}*

*– A*+ (

_{z}B_{y})*A*

_{z}B_{x}*– A*) + (

_{x}B_{z}*A*

_{x}B_{y}*– A*)

_{y}B_{x}This formula requires a lot of memorization. Another method for determining the cross product is to realize that the cross product is the determinant of the following 3 × 3 matrix.

**A×B** = = − +

By taking the determinant of each 2 × 2 matrix you will realize that you get the same formula as shown above. This may appear to be equally difficult to memorize; however, you can just memorize two facts. Each 2 × 2 matrix is missing the column associated with the basis vector that multiplies it and the terms alternate in sign.

**Example 7** Calculate the cross product of the vectors **A** = 2 + 3 and **B** = – + + 4, and verify that it’s perpendicular to both **A** and **B**.

**Solution.** First we figure out the cross product:

*A* × **B** = [(3)(4) – (0)(1)] – [(2)(4) – (0)(–1)] + [(2)(1)– (3(–1)] = 12 – 8 + 5

Using the determinant method

**A×B** = = − +

**A×B** = [(3)(4)−(0)(1)]−[(2)(4)−(0)(1)]+[(2)(1)−(3)(−1)]

**A×B** = 12−8+5

Now, to verify that this vector is perpendicular to both **A** and **B**, we use the property of the dot product: Two vectors are perpendicular to each other if their dot product is zero. Extending the computation of the dot product in terms of the scalar components to three dimensions, we get:

(**A** × **B**) · **A**= (12 – 8 + 5) · (2 + 3) = (12)(2) + (–8)(3) + (5)(0) = 0

(**A** × **B**) · **B**= (12 – 8 + 5) · (– + + 4) = (12)(–1) + (–8)(1) + (5)(4) = 0