CK-12 Physics Concepts - Intermediate
Answer Key
Chapter 4: Vectors
4.1 Graphical Methods of Vector Addition
Practice
Questions
1. What is a resultant?
2. What are the steps necessary to add vectors in two dimensions?
Answers
1. A resultant is the sum of two (or more) vectors.
2. To add vectors, place the vectors such that the tail of the second vector is placed
at the head of the first vector. Draw a resultant vector from the tail of the first
vector to the head of the second vector. Using the Pythagorean theorem,
calculate the magnitude of the resultant vector. Use geometry or a protractor to
determine the angle of direction of the resultant vector.
Review
Questions
1. On the following number line, add the vector 7.5 m/s and the vector -2.0 m/s.
2. On a sheet of graph paper, add a vector that is 4.0 km due east and a vector that is
3.0 km due north.
Answers
1. The resultant vector is 5.5 m/s.
2. The resultant vector is 5.0 km long and 37 north of east. (This is a 3, 4, 5 right
triangle)
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4.2 Resolving Vectors into Axial Components
Practice
Questions
1. What does SohCahToa mean?
2. Why is SohCahToa relevant to resolving a vector into components?
3. Why is the sum of the components larger than the resultant vector?
Answers
1. SohCahToa is an abbreviation and a way to remember geometry relationships.
Soh: Sine = Opposite over Hypotenuse; Cah: Cosine = Adjacent over
Hypotenuse; Toa: Tangent = Opposite over Adjacent.
2. SohCahToa is relevant to resolving a vector because it is a way to determine the
relationship between the length of the two relevant sides and the angle.
3. The sum of the components is larger than the resultant because the resultant is
calculated using the Pythagorean theorem a2+b2=c2.
Review
Questions
1. A force of 150. N is exerted 22° north of east. Find the northward and eastward
components of this force.
2. An automobile travels a displacement of 75 km 45° north of east. How far east
does it travel and how far north does it travel?
Answers
1. The northward component is calculated as 150 (sin(22)) = 56.19, indicating 56.2
N north, and the eastward component is calculated as 150 (cos(22)) = 139.07,
indicating 139 N east.
2. The automobile travels 53 km both north and east, as calculated by 75 (cos (45))
= 53.03, and 75 (sin (45)) = 53.03.
4.3 Vector Addition
Practice
Questions
1. What are the steps the teacher undertakes in order to calculate the resultant
vector in this problem?
2. How does she find the components of the individual vectors?
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3. How does she use the individual vector’s components to find the components of
the resultant vector?
4. Once the components are determined, how does she find the overall resultant
vector?
Answers
1. First, she sketches all the vectors and the resultant vector. Then, she draws each
vector and determines its components. Using a table, she adds the components
to determine the components of the resultant. Finally, using Pythagorean
Theorem, she calculates the resultant’s length and direction.
2. She finds the components using trig functions. She uses SohCahToa.
3. She identifies the individual components of each vector and sums them together
using the table to help her.
4. She draws a final sketch and uses Pythagorean Theorem and trig functions to
determine the resultant direction and magnitude.
Review
Questions
1. A hiker walks 11 km due north from camp and then turns and walks 11 km due
east.
a. What is the total distance walked by the hiker?
b. What is the displacement (on a straight line) of the hiker from the camp?
2. While flying due east at 33 m/s, an airplane is also being carried due north at 12
m/s by the wind. What is the plane’s resultant velocity?
3. Two students push a heavy crate across the floor. John pushes with a force of
185 N due east and Joan pushes with a force of 165 N at 30° north of
east. What is the resultant force on the crate?
4. An airplane flying due north at 90. km/h is being blown due west at 50.
km/h. What is the resultant velocity of the plane?
Answers
1.
a. The hiker walks 22 km (11 km N + 11km E)
b. The hiker’s displacement is 11(√2) = 15.56 indicating an approximate
displacement of 16 km.
2. The plane is traveling at √(33)2 + 122 = 35.11 or 35 m/s at 20 north of east.
3. Resolve the vectors: Force A = 0 N north + 185 N east, Force B = 82.5 N north +
143 N east. The resultant is 82.5 + 0 = 82.5N north and 185N + 143N = 328N
east. The force on the crate is 338N at 14 north of east
4. The airplane’s resultant velocity is 103 km/hr at 29 west of north.
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4.4 Gravitational Force and Inclined Plane
Practice
Questions
1. What force is broken down into components in this video?
2. What two components are formed from this original force?
Answers
1. The gravitational force acting on the box, or the weight, is broken down into
components.
2. The weight is broken down into a W || (parallel force) and a W  (normal force).
Review
Questions
1. A car weighing 12,000 N is parked on a 36° slope.
a. Find the force tending to cause the car to roll down the hill.
b. Find the acceleration of the car rolling down the hill.
Answers
1.
a. The force tending to cause the car to roll is the parallel force:
12000N(sin(36))=7053.42. The force is 7,053N.
b. Using 𝐹 = 𝑚𝑎: 7,503𝑁 = (
12,000𝑁
9.8𝑚
𝑠2
) (𝑎) =
car is 5.8 m/s2, down the inclined plane.
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7503𝑁
1224𝑚
𝑠2
=
5.76𝑚
𝑠2
The acceleration of the