High School Physics Tutor

By Joseph Molitoris

Specifically designed to meet the needs of high school students, REA’s High School Physics Tutor presents hundreds of solved problems with step-by-step and detailed solutions. Almost any imaginable problem that might be assigned for homework or given on an exam is covered. Topics include vectors, statics, kinematics, dynamics, energy/power, impulse/momentum, hydrostatics / aerostatics, electric circuits, magnetics, and radiation. Also included are chapter introductions which review major physics principles and their applications to problem-solving. Fully indexed for locating specific problems rapidly.

• Language: English
• Publisher: Research & Education Association
• Release date: Jan 1, 2013
• ISBN: 9780738668413

Book preview

CHAPTER 1

VECTORS

Basic Attacks and Strategies for Solving Problems In this Chapter

Physics deals with many geometric objects such as scalars and vectors. A scalar is a quantity which has only a magnitude such as length, temperature, and speed. A vector (see Figure 1) is a quantity which has both magnitude and direction such as displacement, velocity, and force. The magnitude is given by the length of the vector and a suitable scale, and the direction by the arrow in the figure. Except in relativity, in physics vectors have two (in two dimensions) or three (in three dimensions) components.

Consider two displacement vectors represented in Cartesian coordinates as

The first notation is called unit vector notation. The second is called coordinate, or point notation. The magnitude of either vector (see Figure 2) may be found by the theorem of Pythagoras

and the direction by the tangent rule tan θ = Ay/Ax. Equivalently, one may use sin θ = Ay/A or cos θ = Ax/A.

The two vectors and may be added (See Figure 3) to get the resultant, or sum

Figure 1

Figure 2

The direction is again given by finding θ = Arctan (Ry/ Rx) such that tan θ = Ry/ Rx. A similar rule applies for finding the difference. Note that this is equivalent to walking first along vector A and then along vector to get to the resultant location R This method of adding the components and using the Pythagorean theorem and trigonometry is called the component method. It can easily be generalized to adding more than two vectors or dealing with higher dimensions. For example, in three dimensions, one would get

in unit vector notation.

The other equivalent way to add vectors is by Newton’s parallelogram rule. One connects the vectors head to tail, just as in the component method. One could then find the magnitude of and the direction angle θ graphically using a ruler and protractor as in a force table laboratory exercise. Analytically, one can use geometry and the laws of cosines and sines. From geometry and Figure 3, ∠R = θA + 180°—θB. The law of cosines then gives

The law of sines states that sin ∠R/R = sin ∠B/B. One may thus find ∠B and from it get the direction of the resultant θ = θA+ ∠B.

Vectors cannot be multiplied or divided as scalars can. However, there are two special products: the dot product and the cross product. The dot product, or two vectors and , is a scalar . The dot product may be used to find the work done by a force exerted over a certain distance, for example. The cross product is more complicated and is given by Where

Figure 3

For example, if A points in the x—direction and B in the y—direction, then . More conveniently, the cross product has a magnitude given by C = AB sin θ, θ being the angle from A to B, and a direction given by the right hand rule. Coil the fingers of your right hand from to and stick out the thumb; your thumb then points in the direction of C. For example, if and are in the xy plane (Figure 4), then the cross product points in the z—direction. The cross product is used in physics to find the torque exerted by a force acting at a certain position.

Figure 4

VECTOR FUNDAMENTALS

• PROBLEM 1-1

specified in the figure.

Solution. From the Pythagorean theorem, , and so we get S3 = 5 units. The direction of S3 may be specified by the angle θ which it makes with S1.

Resultant therefore represents a displacement of 5 units from 0 in the direction 37° north of east.

• PROBLEM 1-2

.

Fig. A

Fig. B

Fig. C

Solution: When vectors are added (or subtracted), their components in the directions of the unit vectors add (or subtract) algebraically. Thus since

then it follows that

Similarly,

units = 5.66 units, and lies in the y- z plane, making an angle θ with the y - axis, as shown in figure (a), where tan θ = 4/4 = 1. Thus θ = 45°.

units = 2.82 units, and lies in the x - y plane, making an angle φ with the x-axis, as shown in figure (b), where tan φ = + 2/-2 = - 1. Thus φ = 315°.

units = 6.32 units, and lies in the x - z plane at an angle χ to the x-axis, as shown in figure (c), where tan χ = 6/2 = 3. Thus χ = 71°34’.

• PROBLEM 1-3

We consider the vector

Find the length of A.

What is the length of the projection of A on the xy plane?

Construct a vector in the xy plane and perpendicular to A.

Construct the unit vector B.

of the vector C = 2x.

in a reference frame obtained from the old reference frame by a rotation of π/2 clockwise looking along the positive z axis.

in the primed coordinate system.

.

.

Fig. 1

Fig. 3

Fig. 2

Fig. 4

The primed reference frame x’, y’, z’, is generated from the unprimed system x, y, z, by a rotation of +π/2 about the z axis.

.

.

(c) Construct a vector in the xy plane and perpendicular to A. We want a vector of the form

). Hence

On taking the scalar product we find

or

The length of the vector B is not determined by the specification of the problem. We have therefore determined just the slope of vector B, not its magnitude. See diagram 2.

must satisfy the following two equations:

or

The vector B is then:

(e) Converting the vectors into coordinate form and computing the dot product (scalar product):

by (see fig. 3)

; where y appeared, we now have x , so that

and C into coordinate form in the primed coordinate system, giving us the following dot product:

This is exactly the result obtained in the unprimed system.

is defined as

We have

DISPLACEMENT VECTORS

• PROBLEM 1-4

Two hikers set off in an eastward direction. Hiker 1 travels 3 km while hiker 2 travels 6 times the distance covered by hiker 1. What is the displacement of hiker 2?

Solution: From the information given the displacement vector is directed east. The magnitude of the displacement vector for hiker 2 is 6 times the magnitude of the displacement vector for hiker 1. Therefore, its magnitude is 6 x (3 km) = 18 km

• PROBLEM 1-5

Two wires are attached to a corner fence post with the wires making an angle of 90° with each other. If each wire pulls on the post with a force of 50 pounds, what is the resultant force acting on the post? See Figure.

Solution: As shown in the figure, we complete the parallelogram. If we measure R and scale it, we find it is equal to about 71 pounds. The angle of the resultant is 45° from either of the component vectors.

If we use the fact that the component vectors are at right angles to each other, we can write

whence

R = 71 pounds approximately at 45° to each wire.

• PROBLEM 1-6

If a person walks 1 km north, 5 km west, 3 km south, and 7 km east, find the resultant displacement vector.

Fig. A

Fig. B

. The magnitude of this vector is 2.8 km. The direction, as measured with a protractor, is 45° south of east, or the tangent may be used to find the direction, since a right triangle is formed.

We shall also compute the solution analytically.

is shown. We can see from the graph that side A and side B each equal 2 km. Thus, by the Pythagorean theorem:

• PROBLEM 1-7

? How far will he be from where he started?

is as shown in the diagram. Thus, the sum of the components is:

The recruit’s final distance from the starting point will be the magnitude of R:

• PROBLEM 1-8

A ship leaving its port sails due north for 30 miles and then 50 miles in a direction 60° east of north. See the Figure. At the end of this time where is the ship relative to its port?

Solution by Parallelogram Method:

which is found to represent about 70 miles. Angle r is found to be about 38.2° east of north.

Solution by Component Method:

lie along the same direction in this problem, we add them directly to get 30 miles + 25 miles, or 55 miles. We then have a right triangle with one side equal to 55 miles and the other side equal to 43 miles. From these data we find the resultant R according to the equation :

R² = 55² + 43²

whence R = about 70 miles

Solution by the Cosine Law:

In solving this problem by means of the cosine law, we write

whence the magnitude of R is

whence

VELOCITY VECTORS

• PROBLEM 1-9

An aircraft is climbing with a steady speed of 200 m/sec at an angle of 20° to the horizontal (see figure). What are the horizontal and vertical components of its velocity?

Fig. A

Fig. B

Solution: Using trigonometric relations for right triangles, the velocity can be broken down into two components perpendicular to each other.

Horizontal component = 200 cos 20°

Vertical component = 200 sin 20°.

Trigonometric tables tell us that cos 20° = 0.9397 and sin 20° = 0.3420

Therefore, horizontal component = 200 x 0.9397

= 187.94 m/sec

Vertical component = 200 x 0.3420

= 68.40 m/sec.

as the hypotenuse,

• PROBLEM 1-10

An automobile driver, A, traveling relative to the earth at 65 mi/hr on a straight, level road, is ahead of motorcycle officer B, traveling in the same direction at 80 mi/hr. What is the velocity of B relative to A? Find the same quantity if B is ahead of A.

solution: The velocity of B relative to A is equal to the velocity of B relative to the earth minus the velocity of A relative to the earth, or

If B is ahead of A, the velocity of B relative to A is still the velocity of B relative to the earth minus the velocity of A relative to the earth or 15 mi/hr.

In the first case, B is overtaking A, and, in the second, B is pulling ahead of A.

• PROBLEM 1-11

City A is 100 miles north and 200 miles west of city B. An airplane flies in a direct line between the cities in a time of one hour. What are the vectors that describe the distance of A from B, and the velocity of the airplane?

Solution: We will define first a coordinate system with B at the origin (see the figure below). The x-direction is east and the y-direction is north. The vector BA is specified by its coordinates

x = -200 mi

y = 100 mi

or by its magnitude and direction

The velocities are given in a similar way. Since they are constant

• PROBLEM 1-12

A certain boat can move at a speed of 10 mi/hr in still water. The helmsman steers straight across a river in which the current is 4 mi/hr. What is the velocity of the boat?

Solution: The boat has a speed of vb = 10 mi/hr perpendicular to the river due to the power of the boat. The current gives it a speed of vc = 4 mi/hr in the direction of flow of the river. The boat’s resultant velocity (having both magnitude and direction) can be found through vector addition.

The magnitude of the velocity which is the speed of the boat is found using the Pythagorean theorem (see figure).

The angle θ, which determines the direction of the velocity is,

• PROBLEM 1-13

A boy can throw a baseball horizontally with a speed of 20 m/sec. If he performs this feat in a convertible that is moving at 30 m/sec in a direction perpendicular to the direction in which he is throwing (see figure), what will be the actual speed and direction of motion of the baseball?

Solution; Since the baseball is originally travelling with the convertible, it has the speed of 30 m/sec in the direction the car is travelling. When the boy throws the ball perpendicular to the car’s path, he imparts an additional velocity of 20 m/sec in that direction. The ball’s velocity is then 30 m/sec in the direction the convertible is moving and 20 m/sec perpendicular to this movement. Its resultant velocity can be found through adding vectors as shown in the diagram.

If the resultant velocity is R m/sec at an angle θ to the direction in which the convertible is moving, then

R² = (20)² + (30)² = 1300

From tables of tangents, θ = 33.69°. Therefore, the ball has a speed of 36.06 m/sec in a direction at an angle of 33.69° to the direction in which the convertible is travelling.

• PROBLEM 1-14

An airplane, whose ground speed in still air is 200 mi/hr, is flying with its nose pointed due north. If there is a cross wind of 50 mi/hr in an easterly direction, what is the ground speed of the airplane?

Solution: The cross wind causes the plane to travel 50 mi/hr to the east in addition to its speed of 200 mi/hr to the north. To find its speed with respect to ground, use vector addition. Vectors are quantities that have both magnitude and direction; and velocity fits this specification. Using the Pythagorean theorem, we can find the magnitude of the resultant velocity v. This magnitude is the plane’s speed. Speed does not have direction (note that speed is not a vector).