at2Diagnostic Test
1.(A)
Amplitude is volume for sound and brightness for light.
2.(D)
Wave speed is dependent upon the medium in which a wave travels.
3.(B)
Frequency remains constant when a wave enters a new medium. Only wave speed and wavelength are affected by a change in medium.
4.(E)
A uniform magnetic field has the same magnitude and direction at every point within a region of space. The field lines are parallel to each other.
5.(C)
With the thumb of your right hand pointing into the page, your fingers will curl in the clockwise direction and indicate the direction of the magnetic field created by the current. Choice D is the magnetic field created by a current coming out of the page.
6.(B)
Electric field lines point toward negative particles and away from positive particles. Choice A is a positive particle.
7.(A)
This statement describes Albert Einstein and the photoelectric effect.
8.(D)
This statement describes Ernest Rutherford and the gold foil experiment.
9.(C)
The first law of thermodynamics states ΔU = Q + W.
10.(D)
Lenz’s law is a restatement of the law of conservation of energy as it applies to induced electrical currents and their subsequently induced magnetic fields.
11.(E)
The second law of thermodynamics states that the entropy of an isolated system always increases until equilibrium is reached.
12.(B)
The slope of a speed-time graph is acceleration. The slope of the line during interval B is 10 m/s2.
13.(C)
During interval C, the object moved at 30 m/s constantly.
14.(B)
The area under the curve of a speed-time graph is displacement. The greatest distance will occur in the interval that has the greatest area under the graphed interval. The area under B is 40 m.
15.(C)
Acceleration is the rate of change in velocity. It is also a vector quantity consisting of both magnitude and direction. Therefore, only choice C is NOT possible. Choice A is possible because the speed, which is the magnitude of velocity, of a turning object may remain constant even while direction is changing. A constant force, choice E, produces a constant acceleration and is therefore also possible.
16.(E)
The magnitude of velocity is the absolute value of the slope of a position-time graph. All slopes in this graph are constant; therefore, they all represent magnitudes of velocity that remain constant. In intervals A and E, the magnitude of velocity will be negative but the value is not changing (not decreasing).
17.(A)
The slope of a position-time graph is speed. Interval A has the steepest slope and therefore the greatest speed.
18.(D)
No mention was made of mass, so it is assumed to remain constant. Under the conditions described, force, F, and acceleration, a, are directly proportional, F = ma. Doubling the force must then also double the acceleration.
2F = m(2a)
For an object initially at rest, the relationship between displacement and acceleration is:
x = at2
Displacement, x, is directly proportional to acceleration. Doubling acceleration will double the displacement during the same time interval.
2x = (2a)t2
19.(E)
Time is dependent on y-direction variables. For a horizontal launch, the initial velocity in the y-direction is zero. Here, acceleration is g.
The motion in the x-direction has constant velocity. So the horizontal distance traveled by the ball is:
20.(B)
Air resistance is assumed to be negligible unless specifically told otherwise. Under these conditions, the speed (magnitude of true velocity) of a projectile at a specific height will be the same on the way upward and downward at that same height. This projectile lands at the same height as it was launched, and its speed when landing is the same as when it was launched.
21.(B)
Rope 2 and rope 3 are equally sharing the 300 N weight of the 30 kg mass. Therefore, each has a tension of 150 N. The 10 kg mass suspended by rope 1 acts as another “ceiling” for the 30 kg mass. Its mass is not relevant to solving the tension caused by the 30 kg mass suspended beneath it.
22.(D)
The force of friction will always be equal to, but opposite of, the component of force acting in the direction of motion. Since the applied force, F, is at an angle to the direction of motion, the component of force actually acting in the direction of motion will be F cos θ. This is less than the applied force, F.
23.(C)
The masses are connected and act as if they were a single mass. Sum the forces for the entire system.
24.(C)
The masses are connected by a string and act as if they were a single mass of 4m. The 3m mass is being pulled in one direction by the force of gravity. The m mass is being pulled in the other direction, also by the force of gravity. Sum the forces for the entire system all at once.
25.(A)
Since the two forces have equal magnitude but act in opposite directions, their net sum will be zero. The magnitude of acceleration is directly proportional to the magnitude of the net force. If the magnitude of the net force is zero, then acceleration has a magnitude of zero. An acceleration of zero is consistent with constant velocity.
26.(E)
Spring scales measure apparent weight, which is the normal force acting on an object. The normal force acts upward, while the force of gravity acts downward. Sum the forces, and solve for the normal force.
27.(A)
The sum of the force vectors added tip to tail must result in a zero sum for the object to remain stationary or move at a constant, nonzero velocity. Choice A is the only diagram with a zero sum for the vectors. The others all show a resulting force vector and therefore both a net force and a net acceleration.
28.(C)
Determine the period, which is the time to complete one cycle.
Solve for the speed in circular motion.
29.(B)
When a car makes a turn, the net force resulting in the circular motion, FC, is equal to the friction force holding the car in the turn.
30.(E)
Tangential velocity is tangent to the circular path at any given point along the path. Centripetal acceleration always points toward the center of the circle from any given point along a circular path.
31.(D)
In order to complete a loop with a minimum speed at the top, the centripetal force must be a minimum. At the top of the loop gravity and the normal force of the track hold the roller coaster in the loop. Although gravity cannot be decreased, the normal force can be reduced to zero. The normal force is zero for only an instant at the very top of the loop.
32.(C)
Work is equal to the component of force parallel to the motion of an object multiplied by the distance traveled by the object.
33.(A)
There are two ways to arrive at the answer. The first uses force, and the second uses energy. The resisting force of friction acts opposite to the 8 N component of force applied in the direction of motion. This results in a net force of zero. The net work is directly proportional to the net force.
The net work is also equal to the change in kinetic energy (work–kinetic energy theorem). For an object moving at constant velocity, the change in kinetic energy and the net work are both zero.
34.(C)
The area under the curve of a force-displacement graph is the work done in N • m (Joules). This curve is a triangle. As such, the area can be determined by base × height. The magnitude of the mass does not affect the answer.
35.(B)
This question is about conservation of energy. The potential energy at the top of the hill is converted to (and equal to) the kinetic energy at the bottom of the hill.
36.(D)
When an object changes height, its gravitational potential energy changes.
37.(A)
Potential energy stored in a spring can be determined by:
If the potential energy is quadrupled, it would require a doubling of the stretch of the spring, x.
38.(B)
Momentum is conserved in the inelastic collision between the two cars.
39.(E)
Linear momentum and kinetic energy are both conserved in perfectly elastic collisions. However, velocity is not conserved during collisions.
40.(D)
The force of gravity between the small planet and the star is:
The force of gravity between the large planet and the star is:
Dividing the force acting on the smaller planet by the force acting on the larger planet will give the ratio of these forces. This cancels all the variables, represented by letters, and leaves only the numerical coefficients.
41.(B)
The closer a planet is to the central star, the faster its tangential speed and the shorter its period of orbit.
42.(D)
When the two spheres are brought into contact, their charges combine. The negative 4-coulomb charge of the left sphere and the positive 8-coulomb charge of the right sphere combine to make a total of a positive 4-coulomb charge. Upon separation of the spheres, these charges separate equally, leaving both the left and right spheres with a charge of positive 2 coulombs each.
43.(E)
The magnitude of force is determined by the charge multiplied by the magnitude of the electric field, FE = qE. The magnitude of force on both the electron and proton are the same because the particles have equal charge. However, the direction of force is opposite for electrons and protons. Protons move with electric fields, while electrons move against electric fields. Although the magnitude of force for the two particles is equal, the acceleration is not equal. The electron has very little mass compared with protons. The same force will give the electron a much greater acceleration.
44.(D)
In order for the electric field to cancel, the two charges must produce electric fields that are opposite in direction and equal in magnitude.
At positions B, C, and D, the electric field due to point charge +4q is pointed toward the right and the electric field due to point charge +q is pointed toward the left. The magnitude of the electric field is influenced by the size of the charges. The larger charge +4q creates a larger overall electric field. However, the electric field also diminishes with distance. Point D is closer to the small charge +q. So at point D, the smaller charge can generate an electric field that is equal in magnitude to the larger far away charge.
45.(C)
This question involves conservation of energy. The potential energy of the electron, UE = qV, accelerates the electron. When the electron reaches the positive plate, the potential energy has been converted entirely to kinetic energy.
By doubling the voltage to 2V, this will increase the speed to 
46.(E)
Voltage equals the magnitude of the electric field multiplied by the distance between the plates. Convert 10 centimeters to meters.
47.(A)
Capacitance equals the amount of charge divided by the voltage.
48.(D)
Adding resistors in series will increase the resistance to the flow of current, therefore, the current will decrease. With less current flowing, less power is consumed.
49.(B)
The current flows only through series resistors R1 and R2.
Ohm’s law, V = IR, can be used to solve for the current.
50.(D)
When the switch is closed, the 4 Ω resistor is added in parallel into the circuit.
Remember to invert the value above to find the resistance in parallel.
R = 2 Ω
51.(E)
Closing the switch creates a new parallel path through R3 where voltage can be applied and current can flow through. This parallel path does not affect the current flowing through or the voltage across resistors R1 and R2 as they are on a separate parallel path. So the brightness of the lightbulb R1 remains the same.
52.(B)
Resistors in parallel will dissipate the most power.
53.(B)
Resistors in parallel to the battery receive the same voltage. Resistors in circuit A also have the same voltage; however, it is 
V for each of those resistors.
54.(D)
Determine the magnitude of the magnetic force on the moving charge.
The direction of the field can be found with the right-hand rule. However, the moving charge is negative. The right hand can still be used, but then the answer must be reversed. Another way to determine the direction of negative charges is to use the left hand. Point the thumb of the left hand in the direction the charge is moving. Extend the fingers so they point in the direction of the magnetic field (out of the page in this case). The direction in which the palm of the hand pushes is the direction of force (toward the top of the page if using the left hand), which is the +y-direction.
55.(E)
Inducing a current requires a change in flux. Flux is the amount of magnetic field passing through the loop of wire. Only choice E demonstrates no change in the amount of magnetic field passing through the loop. All others show either an increase or a decrease in flux.
56.(A)
Period of oscillation for a spring system is
Doubling the mass will increase the period, T, by 
.
Frequency is the reciprocal of period, f = 1/T. So the inverse of will be applied to the frequency
57.(B)
The potential energy is at its maximum when the spring-mass is at maximum displacement. The potential energy is zero as the spring-mass passes through the equilibrium position, where displacement is zero. Kinetic energy of the system is depicted in choice A, and total energy is depicted in choice E.
58.(A)
Wave speed is affected by the medium in which a wave travels. In a more dense optical medium, light speed decreases. Frequency is unaffected by the medium, and so the wavelength must adjust according to the equation v = f λ.
59.(D)
There is one wavelength and one period in one complete cycle. However, there are four amplitudes. Answer C is a frequently chosen distracter. Amplitude is measured from the equilibrium position to the maximum displacement, and oscillators pass through four amplitudes in one cycle.
60.(D)
According to the Doppler effect, a source moving away has longer wavelengths and lower frequency.
61.(B)
When light enters a more optically dense medium, the speed of light decreases and its wavelength shortens. Snell’s law puts refraction into mathematical terms.
n1 sin θ1 = n2 sin θ2
When light enters glass, the index of refraction in glass, n2, is greater than that in air, n1. In order to maintain the equality in Snell’s law, the angle of refraction in glass, θ2, must be smaller than the angle of incidence in air, θ1.
62.(C)
Light rays passing through the pinhole will converge at the opening and invert as they pass through the opening. This creates a real, inverted image at the back of the camera. Since the image distance is small compared to the object distance, the image will be small.
63.(E)
As the object moves farther away from the focal point, the image continues to get closer to the mirror with the maximum distance of the image being one focal length away from the mirror. The image size, however, continues to decrease as the object moves farther and farther away from the focal point.
64.(E)
Each of the results listed will occur as monochromatic light is projected onto a screen and creates a double-slit interference pattern.
65.(A)
This is the definition of refraction.
66.(C)
Each wavelength of light does have a slightly different index of refraction in an optical medium. Choice A is close but is not the best answer. Color is a function of frequency, and frequency is not affected by the medium.
67.(A)
Process 3 represents the rate of temperature change as the substance is heated from its liquid phase to its boiling point. Latent heats occur only during processes 2 and 4, when there is a phase change and therefore no temperature change.
68.(E)
All of the statements are correct about process 4.
69.(B)
This question is about the first law of thermodynamics. Heat is being added to the system, so Q is positive. If work is being done by the system, the gas must be expanding to move the piston. In doing so, the gas loses energy. So work, W, done by the gas is negative.
70.(B)
When temperatures are given, maximum efficiency is calculated using:
When formulas contain temperature, T, degrees Kelvin are required. Degrees Celsius can only be used when a formula contains a change in temperature, ΔT. When in doubt use degrees Kelvin.
71.(E)
This is the definition of entropy, the measured amount of disorder of isolated systems.
72.(B)
Only discreet states above the ground state, n = 1, exist for a particular atom. Only photons with energies matching the difference between electron energy levels can be absorbed. When the 10 eV photon is added to electrons in the ground state, the electrons acquire an energy of −2 eV and move to the n = 3 energy level.
73.(D)
The electrons will lose energy in order to return to the ground state. In some atoms, the electrons may drop from energy level n = 3 to energy level n = 2, losing 4 eV. Then these electrons will drop from energy level n = 2 to energy level n = 1, losing 6 eV. However, in other atoms the electrons may drop from n = 3 all the way to energy level n = 1, losing 10 eV. Each of these three energy level drops produces a distinct photon with a matching energy.
74.(C)
During one half-life, the sample is reduced to half. In order for only 
of the sample to be remaining, four half-lives must have transpired.
If each half-life lasts 10 days, then four half-lives would last 40 days.
75.(D)
Light is always measured by all observers to be at the constant speed of c.
1: Conventions and Graphing
1.(D)
Key Words:
Derived unit for pressure
Needed for Solution:
P = F/A
Now Solve It:
Formulas are not provided on the examination. The formula was provided in this question for convenience. However, you should memorize formulas for such purposes as derived unit analysis. Replacing the symbols for force, F, and area, A, with their units would produce:
pressure = N/m2
A newton, N, is also a kg • m/s2 so the combined fundamental units for pressure would be kg/ms2.
2.(B)
Key Words:
Derived unit for energy
Needed for Solution:
K = 
mv2
Now Solve It:
Formulas are not provided on the examination. The formula was provided in this question for convenience. However, you should memorize formulas for such purposes as derived unit analysis. Replacing symbols for mass, m, and velocity, v, with their units would produce:
kinetic energy = (kg)(m/s)2
The resulting combination would be kg • m2/s2. The one-half has no effect on the unit combination.
3.(D)
Key Words:
Velocity-time graph;
acceleration
Needed for Solution:
Knowledge/definitions
or
Test if units of slope or area match the units of acceleration
Now Solve It:
The units of slope match the units needed in the answer. The slope of velocity versus time is acceleration. The time given, t = 4 s, lies on a constantly sloping line between t = 3 s and t = 5 s.
acceleration = slope = 
= 
= −10 m/s2
The question requires the magnitude (numerical value without direction) of acceleration, which is the absolute value of acceleration.
4.(C)
Key Words:
Velocity-time graph;
displacement
Needed for Solution:
Knowledge/definitions
or
Test if units of slope or area match the units of displacement
Now Solve It:
The units of area match the units needed in the answer. The area of velocity versus time is displacement. Add the areas of the rectangle and triangle, bounded by the function and the x-axis.
displacement = area = height × base
displacement = area of rectangle + area for triangle
(20 m/s × 3 s) + (20 m/s × 2 s)
= 80 m
5.(C)
Key Words:
Which graph;
P = I2R
Needed for Solution:
Are the graphed variables squared, under a square root, or inverted?
Now Solve It:
The independent variable, I, is squared. A squared value indicates a quadratic function, which will graph as a parabola. The independent variable, P, will increase by the square of the dependent variable, I. This creates a parabola consistent with answer C.
6.(E)
Key Words:
Which graph;
p = h/λ
Needed for Solution:
Are the graphed variables squared, under a square root, or inverted?
Now Solve It:
The independent variable, λ, is inverted. The dependent and independent variables are inversely proportional, and the resulting graph is a hyperbola. Only one hyperbola is shown in the answers.
7.(A)
Key Words:
Which graph;
F = kx
Needed for Solution:
Are the graphed variables squared, under a square root, or inverted?
Now Solve It:
The equation contains no squared values, no square roots, and no inverted values. This is a linear equation in the following form:
y = mx + b
There are two possible answers: A and B. Answer B includes a y-intercept. However, the equation F = kx is missing addition of a constant (such as F = kx + b). The y-intercept must therefore be zero. So the answer is A.
2: Vectors
1.(C)
Key Words:
NOT; vector quantity
Needed for Solution:
Knowledge/definitions
Now Solve It:
Answer C reports only the magnitude. There is no mention of a specific direction. This is a scalar and not a vector.
2.(A)
Key Words:
Initial velocity; 30°; y-component; vy
Needed for Solution:
Side relationships for a 30°-60°-90° triangle
Now Solve It:
Visualize or sketch the velocity vector and its components.
vy is opposite the 30° angle, so its magnitude is half of the hypotenuse.
vy = (1/2)(50) = 25 m/s
3.(B)
Key Words:
Two forces; resultant net force
Needed for Solution:
Tip-to-tail vector addition
Now Solve It:
Start at the origin, and add the vectors tip to tail. The resultant is drawn from the origin to the tip of the second vector.
4.(E)
Key Words:
Incline; vertical height; displacement; along the incline
Needed for Solution:
Side relationships for a 30°-60°-90° triangle
Now Solve It:
The displacement along the incline is the hypotenuse of a 30°-60°-90° triangle. The hypotenuse is double the side opposite the 30° angle, which in this case is double the height.
Δd = (2)(2.4 m) = 4.8 m
5.(A)
Key Words:
Three forces;
determine the magnitude and direction of force 
;
resultant force acting on the mass is zero
Needed for Solution:
Side relationships for a 3-4-5 triangle
Now Solve It:
Split the diagonal 5-newton force into components.
The upward 4-newton component vector is canceled by the downward 4-newton vector. The leftward 3-newton component vector must also be canceled.
= 3 newtons in the +x direction.
3: Kinematics in One Dimension
1.(C)
Key Words:
Tossed upward; 24 meters above water; 3 seconds; the surface of the water; initial speed
Needed for Solution:
Involves time, acceleration (gravity), and displacement:
Motion is vertical:
Now Solve It:
Displacement and time are given. Initial velocity is requested. Displacement is negative as it is below the point of origin. Acceleration by gravity is also negative.
(−24 m) = vi(3 s) + (−10 m/s2)(3 s)2
−24 m = vi(3 s) + −45 m
21 m = vi(3 s)
vi = 7 m/s
2.(D)
Key Words:
Moving at 20 meters per second; deceleration of 10 meters per second squared; how far; before it stops
Needed for Solution:
Involves kinematic variables but time is not mentioned:
vf2 = vi2 + 2ax
Hidden variable (stops): vf = 0 m/s
Now Solve It:
Deceleration indicates acceleration in the opposite direction to motion. Initial velocity is positive, so acceleration is negative.
vf2 = vi2 +2ax
(0 m/s)2 = (20 m/s)2 + 2(−10 m/s2)x
−400 m2/s2 = (−20 m/s2)x
x = 20 m
3.(A)
Key Words:
Velocity versus time graph;
acceleration
Needed for Solution:
Knowledge/definitions
or
Test if units of slope or area match the units of acceleration
Now Solve It:
The units of slope match the units needed in the answer. The slope of velocity versus time is acceleration. The slope for this time interval is zero.
4.(E)
Key Words:
Velocity versus time graph; displacement
Needed for Solution:
Knowledge/definitions
or
Test if units of slope or area match the units of displacement
Now Solve It:
The units of area match the units needed in the answer. The area under a velocity versus time graph is displacement. Determine the area between the graph and the x-axis from 3 to 5 seconds. Use the formula for the area of a trapezoid.
displacement = area of trapezoid
= (h1 + h2)(b)
(10 m/s + 5 m/s)(2 s) = 15 m
5.(B)
Key Words:
Velocity versus time graph;
acceleration
Needed for Solution:
Knowledge/definitions
or
Test if units of slope or area match the units of acceleration
Now Solve It:
The units of slope match the units needed in the answer. The slope of velocity versus time is acceleration. The slope between t = 3 s and t = 5 s is:
Magnitude is the absolute value, 2.5 m/s2.
6.(B)
Key Words:
Vertical displacement versus time
Needed for Solution:
Knowledge/definitions
and graphing
Now Solve It:
The phrase “vertical displacement versus time” indicates displacement is on the y-axis and time on the x-axis. The ball starts at the origin and initially moves upward (+y). After reaching its maximum height, it moves downward. The ball passes its starting point (y = 0) and finishes below its starting point (−y). The graph of displacement appears nearly identical to the diagram of the throw shown in the problem.
7.(B)
Key Words:
How high is the ceiling; thrown upward from the floor at 10.0 meters per second; barely touches the ceiling
Needed for Solution:
Involves kinematic variables, but time is not mentioned:
vf2 = vi2 + 2ax
Hidden variable (barely touches ceiling):
Now Solve It:
vf = 0 m/s
vf2 = vi2 + 2ax
(0 m/s)2 = (10 m/s)2 + 2(−10 m/s2)x
−100 m2/s2 = (−20 m/s2)x
x = 5 m
8.(C)
Key Words:
Constant acceleration; from rest; 4.0 meters per second squared; what is the displacement;10 seconds
Needed for Solution:
Involves time, acceleration, and displacement:
x = vit + at2
Hidden variable (from rest):
vi = 0 m/s
Now Solve It:
Initially at rest (vi = 0 m/s) simplifies the equation.
x = at2
x = (4 m/s2)(10 s)2
x = 200 m
9.(E)
Key Words:
Uniform acceleration; time doubles
Needed for Solution:
Involves time, acceleration, and displacement:
x = vit + at2
Now Solve It:
Initially at rest (vi = 0 m/s) simplifies the equation.
x = at2
Time is squared. Doubling time quadruples the right side of the equation. The left side of the equation must also quadruple in order to maintain the equality.
(4)x = a(2t)2
10.(A)
Key Words:
Oval-shaped track; constant speed; returning to the starting point; velocity
Needed for Solution:
Knowledge/definitions
Now Solve It:
Constant speed does not necessarily mean constant velocity. Velocity is a vector relying on displacement. Displacement is the distance between the initial and final position of an object. When an object returns to its original position, displacement is zero.
4: Kinematics in Two Dimensions
1.(A)
Key Words:
Compared with the observers
Needed for Solution:
Understanding of relative velocity
Now Solve It:
The observers in car X measure the motion of objects relative to themselves. When doing this, they see themselves as stationary. To have a speed of 0 m/s, the actual 20 m/s must be subtracted. Subtracting 20 m/s from every car will show their speeds relative to car X. When this is done, car A has a velocity of −10 m/s.
2.(A)
Key Words:
Crossing a 40-meter-wide river
Needed for Solution:
Kinematic equations solve independently in the x- and y-directions.
y = vyit
Now Solve It:
The y-displacement is given, and the x- and y-variables must be used in separate equations. Use the y-velocity with the y-displacement to solve for the crossing time.
3.(D)
Key Words:
How far downstream;
as a result of the river current
Needed for Solution:
Although the x- and y-directions solve independently, time is the key variable they share.
x = vxit
Now Solve It:
Motion is simultaneous in the x- and y-directions, and time is the key variable shared by both equations. The time was determined in question 2. The current and downstream distance are both x-direction variables.
x = vxit = (3)(10) = 30 m
4.(C)
Key Words:
Cross a flowing river;
minimizing the time
Needed for Solution:
Crossing the river involves motion in the y-direction.
y = vyit
Now Solve It:
Rearrange the equation to solve for cross-river time.
The cross-river distance, y = 40 m, is fixed. To minimize time, the velocity must be maximized. Therefore, try to swim directly across the river. If you try to swim either upriver or downriver, the cross-river component of velocity will be less than 4 m/s, causing you to take longer to cross the river.
5.(D)
Key Words:
Variable(s) remain(s) constant;
projectile
Needed for Solution:
Projectile motion definition
Now Solve It:
Projectiles are thrown, or shot, in the presence of gravity alone. Gravity is the only acceleration affecting a projectile; therefore, the acceleration of gravity remains constant during a projectile’s flight. Answer C is correct. However, the x- and y-directions are independent, and gravity acts in only the y-direction. This implies that there is no acceleration in the x-direction, so the horizontal component of velocity must also remain constant. Answer A is also correct. Therefore, answer D is the most complete response.
6.(C)
Key Words:
Thrown horizontally; 20-meter-tall structure;
lands 30 meters from the base of the structure
Needed for Solution:
Horizontally launched projectile
Now Solve It:
Time is usually the key variable, and time is dependent on y-direction variables.
Once you have determined the time, solve the remaining equation.
For horizontally launched projectiles, 
i = vix.
7.(B)
Key Words:
Car . . . constant velocity;
ball is thrown straight upward;
no air resistance
Needed for Solution:
Projectile motion knowledge, specifically the independence and types of x- and y-direction motions.
Now Solve It:
Before the ball is thrown, it is moving at constant velocity in the x-direction. The throw gives the ball an additional velocity in the y-direction where gravity acts. However, gravity in the y-direction has no effect on the constant velocity in the x-direction. Therefore, the ball continues to move forward in the x-direction at the same speed as the car. The ball will land in the car as long as there is no air resistance.
8.(D)
Key Words:
Speed; projectile; maximum height
Needed for Solution:
Projectile motion knowledge.
Components of initial launch velocity.
vx = v cos θ
vy = v sin θ
Now Solve It:
The true velocity at any point during the flight of a projectile will be tangent to the parabolic path at that point. At maximum height, the vertical component of velocity will be zero, vy = 0, and the true velocity will be completely horizontal, = vx. The horizontal component of velocity is constant during the flight of a projectile, and the magnitude of velocity is speed, v.
v = vx = v cos θ
9.(B)
Key Words:
Speed; projectile; at the same height
Needed for Solution:
Projectile motion knowledge.
Now Solve It:
At any two points having the same height, a projectile will have the same speed. Point C has the same height as point A.
vC = vA = v
If the problem had requested the velocity at C, the answer would have the same magnitude as the velocity at A. However, the direction would be negative θ since the projectile would strike the ground at a fourth-quadrant angle.
10.(E)
Key Words:
Acceleration,
projectile
Needed for Solution:
Definition of a projectile’s motion
Now Solve It:
The only acceleration in projectile motion problems is gravity. Gravity has a constant magnitude, 10 m/s2, and a constant downward direction at all times. Examining the other answers reveals that they seem to address the changing velocity of a projectile. If you read the question too quickly, you may be distracted into thinking about velocity instead of acceleration.
5: Dynamics
1.(E)
Key Words:
Removes his foot from the accelerator; air resistance is negligible
Needed for Solution:
Newton’s first law
Now Solve It:
When the driver stops pressing the accelerator, the forward force and acceleration of the engine become zero. If air resistance is negligible, then it is so small that it is treated as zero. With no horizontal forces acting on the car, the automobile will follow Newton’s first law. The car will continue moving forward at its current speed.
2.(D)
Key Words:
A small object is attracted to Earth; the object pulls back on Earth with a force of B
Needed for Solution:
Newton’s third law
Now Solve It:
Whenever an object and an agent interact, there is an equal and opposite force between them. Two objects of different size are given in an attempt to trick students into matching the size of the force with the size of the object.
3.(B)
Key Words:
Thrown upward in the absence of air resistance; net force
Needed for Solution:
Force diagram:
Now Solve It:
Drawing a free-body diagram reveals only one force: the force of gravity acting in the downward direction. The force of gravity is also the weight of an object. By stating that the object stops instantaneously at the top of its flight, this problem distracts the reader with the velocity of the object. Read all questions carefully to ensure your answers always match the questions.
4.(D)
Key Words:
Two strings support; the tension in the string is 1
Needed for Solution:
Draw a force diagram, and split the diagonal tension vectors into components:
Now Solve It:
Orient the problem: y-direction. Gravity pulls the object downward, and the strings are pulling upward.
Determine the type of motion: Stationary, ΣF = 0.
Sum the forces in the relevant direction: Two tensions are pulling upward, and the diagonal tension vectors must be split into component vectors.
ΣF = 2Ty − Fg
Substitute and solve: The weight of the object is 10 newtons, Fg = 10 N.
0 = 2Ty − 10
Ty = 5 N
The actual tension is directed diagonally, and this vector is longer than the vertical 5-newton component. Therefore, the force of tension in each rope must be greater than 5 newtons. However, it cannot be greater than the 10-newton total weight of the object. As a result, the tension in each string is between 5 and 10 newtons.
5.(C)
Key Words:
70-kilogram person; accelerating upward at 2 meters per second squared;
normal force
Needed for Solution:
Draw a force diagram, and then sum the forces acting in the vertical direction:
Orient the problem: y-direction. The floor of the elevator pushes the person upward with a normal force, and gravity acts downward.
Determine the type of motion: Acceleration, ΣF = ma.
Sum the forces in the relevant direction:
ΣF = N − Fg
Substitute and solve:
ma = N − mg
N = ma + mg
N = (70)(2) + (70)(10)
N = 840 N
6.(A)
Key Words:
Remains at rest on an incline; free-body diagram
Needed for Solution:
Create a force diagram:
Now Solve It:
The force of gravity acts downward, while the normal force acts perpendicular to the incline (Figure 1 below). These two forces are out of balance. They add together to create a net force acting parallel to the incline (Figure 2). This net force is not drawn in free-body diagrams since it is the sum of two acting forces. Essentially, the force of gravity and the normal force acting together pull objects down the incline. In this problem, the object remains at rest. This means a third force has to act upward along the incline to cancel the effect of gravity and the normal force (Figure 3).
7.(C)
Key Words:
Pulled along a rough horizontal surface;
accelerating;
friction force
Needed for Solution:
Use the force problem-solving method, including a force diagram.
Orient the problem: Friction, f, acts in the x-direction and opposes the forward applied force, F. However, friction also depends on the normal force, N, which acts in the y-direction.
Determine the type of motion: Stationary in the y-direction, ΣFy = 0, while accelerating in the x-direction, ΣFx = ma.
Sum the force vectors in the relevant direction: Forces must be summed in the y-direction to find the normal force. However, it is not necessary to sum the forces in the x-direction. Simply use the friction formula.
ΣFy = N − Fg and f = μN
Substitute and solve:
0 = N − mg
N = mg
f = μN
f = μ(mg)
8.(D)
Key Words:
Pulled along a horizontal surface; constant velocity; the frictional force
Needed for Solution:
Use the force problem-solving method, including a force diagram.
Now Solve It:
Orient the problem: The friction force, f, acts in the x-direction. The applied force, F, acts at an angle and must be resolved into its components. The x-component of the applied force, Fx, pulls the object in the positive direction. Friction slows the object and is negative.
Determine the type of motion: Constant velocity, ΣFx = 0.
Sum the force vectors in the relevant direction:
ΣFx = Fx − f
0 = F cos θ − f
f = F cos θ
9.(C)
Key Words:
Three masses are connected with a string; acceleration of the 2-kilogram mass
Needed for Solution:
This is a compound body problem.
To solve for acceleration, treat all the masses as one large mass (one system).
Now Solve It:
Orient the problem: Visualize the masses as a single system.
Determine the type of motion: acceleration
ΣFsys = (m1 + m2 + m3)a
Sum the force vectors in the relevant direction: Only the 12-newton force creates acceleration.
ΣFsys = 12 N
(m1 + m2 + m3)a = 12
(1 + 2 + 3)a = 12
a = 2 m/s2
10.(E)
Key Words:
2-kilogram mass lies on a horizontal rough surface; 1-kilogram mass hangs vertically;
minimum coefficient of friction; rest
Needed for Solution:
This is a compound body problem involving a pulley. Treat the two masses as a single system. Sum the forces acting parallel to the string.
Now Solve It:
Orient the problem: The force of gravity acting on the 1-kg mass pulls it downward while simultaneously dragging the 2-kg mass along the rough horizontal surface. Friction acts on the 2-kg mass, opposing the effect of gravity.
Determine the type of motion: Friction is strong enough to keep the system at rest, ΣFsys = 0.
Sum the force vectors in the relevant direction: The tension vectors cancel. The force of gravity acts on the 1-kg mass, while friction acts on the 2-kg mass.
Fsys = Fgm1 − fm2
Substitute and solve:
ΣFsys = m1g − μm2g
0 = (1)(10) − μ(2)(10)
μ = 0.5
6: Circular Motion
1.(D)
Key Words:
Tangential velocity;
centripetal acceleration
Needed for Solution:
Knowledge/definitions
Now Solve It:
Tangential velocity is in the direction of motion and is tangent to the curving motion. Centripetal acceleration is directed toward the center of a circular path.
2.(E)
Key Words:
Tangential velocity
Needed for Solution:
Use either kinematics:
or force:
Now Solve It:
Given the available variables, using kinematics is the only possible option. However, this requires you to find the period but the problem has given the frequency:
3.(D)
Key Words:
Centripetal acceleration
Now Solve It:
Use either kinematics
or force:
Fc = mac
Now Solve It:
Given the available variables, using kinematics is the only possible option. This also requires you have the answer to the previous problem.
4.(B)
Key Words:
Coefficients of static and kinetic friction; maximum speed; before the child begins to slip.
Needed for Solution:
Sum the forces for circular motion:
Fc = Fto center − Faway from center
Now Solve It:
Orient the problem: The only force acting in the plane of the circle is friction. The force of gravity and the normal force are equal, oppose each other, and act perpendicular to the circling child:
N = Fg = mg
Determine the type of motion: This is circular motion.
Sum the force vectors in the relevant direction:
Fc = f
Substitute and solve:
Use the coefficient of static friction. While the merry-go-round is moving, the child is not moving relative to its surface:
5.(C)
Key Words:
Vertical loop; minimum speed; top of the loop
Needed for Solution:
Sum the forces for circular motion:
Fc = Fto center − Faway from center
Now Solve It:
Orient the problem: Both the force of gravity and the normal force (acting from the track on the roller coaster) point down toward the center of the circle.
Determine the type of motion: This is circular motion.
Sum the force vectors in the relevant direction:
Fc = N + Fg
To solve for the minimum speed, Fc must be as small as possible. The force of gravity cannot change. However, the normal force can be reduced to zero: N = 0.
Substitute and solve:
7: Energy, Work, and Power
1.(D)
Key Words:
Potential energy
capable of doing work
Needed for Solution:
Knowledge/definitions
Now Solve It:
Potential energy is related to an object’s location or position. To be capable of doing work, an object must be able to move through a distance and be able to generate kinetic energy by losing potential energy.
Choice B is almost correct, but the object is not necessarily in a position that is capable of transforming potential energy to kinetic energy.
2.(C)
Key Words:
Spring;
distance of x;
force of F;
potential energy, U
Needed for Solution:
Hooke’s law
Fs = kx
Elastic potential energy
Us = kx2
Now Solve It:
Spring force follows Hooke’s law, where force is directly proportional to spring displacement.
(2Fs) = k(2x)
Spring potential energy is directly proportional to the square of displacement.
(4Us) = k(2x2)
Doubling the displacement doubles the restoring force and quadruples the spring’s elastic potential energy.
3.(E)
Key Words:
Total mechanical energy
Needed for Solution:
Knowledge/definitions
Total energy
E = K + U
Now Solve It:
The total mechanical energy of a system is the sum of the kinetic and potential energies at a specific instant in time. Why are the other answers wrong?
(A)The total energy does not always remain constant. Nonconservative forces, such as friction, can change the energy of the system.
(B)The net work equals the change in kinetic energy only. This answer overlooks potential energy.
(C)Stored energy is potential energy. This answer is missing kinetic energy.
(D)This answer does not include potential energy.
4.(A)
Key Words:
Uniform circular orbit;
work done by gravity
Needed for Solution:
Parallel force and distance
W = Favgdparallel
Changing height
Wg = mgΔh
Now Solve It:
You can arrive at the answer in two ways.
■Force/displacement method: In uniform circular motion, such as circular orbits, force and displacement are perpendicular. Perpendicular forces do no work.
■Energy/work method: The work of gravity requires a change in height, Δh. In a uniform circular orbit, the height above Earth’s surface is constant.
5.(B)
Key Words:
Force-displacement graph;
initial speed;
final speed
Needed for Solution:
Graphs often involve slope or area. The key is
W = area
Work–kinetic Energy theorem
Wnet = ΔK
Now Solve It:
The work done by force F must be responsible for the change in speed of the mass. Set the two work equations in the middle column equal to each other.
6.(A)
Key Words:
Sliding down a frictionless 30° incline;
work done by gravity
Needed for Solution:
Work of gravity
Wg = mgΔh
Weight/force of gravity
Fg = mg
Now Solve It:
The work done by gravity depends on only the object’s change in height, Δh. Be careful. In this problem, the mass is not given. Instead, you are given the weight of the object in newtons.
Wg = FgΔh
Wg = (3.0 N)(5.0 m) = 15 J
7.(D)
Key Words:
Sliding down a frictionless 30° incline; initially at rest; final speed
Now Solve It:
Conservation of energy
Ki + Ui = Kf + Uf
Now Solve It:
Initially, the mass starts at rest and possesses only gravitational potential energy due to its initial height. At the end of the motion, all of the potential energy is gone and the block now possesses only kinetic energy.
8.(C)
Key Words:
Lift an object 1 meter; incline takes twice as long as lifting it straight up; compared to
Needed for Solution:
Work
Wg = mgΔh
Power
Now Solve It:
The work against gravity is associated with only the change in vertical height, Wg = mgΔh. Therefore, the work done is the same whether the object is lifted straight up or follows a diagonal, inclined, path.
Power is inversely proportional to time .
Pushing the object up the incline uses twice the time and requires only half the power.
9.(D)
Key Words:
10-meter-tall ledge; kinetic energy; halfway point
Needed for Solution:
Conservation of energy
Ki + Ui = Kf + Uf
Now Solve It:
Initially, the rock is at rest and possesses only potential energy. At the half way point it has half its initial height and now has kinetic energy.
0 + mghi = Kf + mghf
(5.0 kg)(10 m/s2)(10 m)
= Kf + (5.0 kg)(10 m/s2)(5 m)
Kf = 250 J
An alternate solution comes from realizing that the rock has lost half of its potential energy when it is halfway to the ground. The lost potential energy transforms into kinetic energy.
Kf = (5.0 kg)(10 m/s2)(5 m) = 250 J
10.(B)
Key Words:
Pendulum; NOT correct
Needed for Solution:
Total Energy
ΣE = K + U
Conservation of energy
Ki + Ui = Kf + Uf
Now Solve It:
A pendulum illustrates the transformation of energy from potential to kinetic, and back again, in a repeating process. During the swing, energy is changing forms between potential and kinetic. However, total mechanical energy is conserved and remains constant. Answer B is incorrect because it states that the total energy is changing from A to C and C to E. This does not occur.
8: Momentum and Impulse
1.(C)
Key Words:
Conserved; perfectly elastic collision
Needed for Solution:
Knowledge/definitions
Now Solve It:
As long as no external forces act, total linear momentum is conserved in all collisions and explosions. Velocity is not a conserved quantity, eliminating answer A. The individual momentums of objects may change during collisions, which eliminates answer B. In any type of inelastic collision, kinetic energy is not conserved, eliminating answers D and E.
2.(D)
Key Words:
Inelastic collision; quantities decrease
Needed for Solution:
Knowledge/definitions
Now Solve It:
Linear momentum, total energy, and mass are always conserved in collisions. However, kinetic energy is not conserved in inelastic collisions. During these collisions, some kinetic energy is lost. This energy becomes thermal energy. This conversion to another form of energy keeps the total energy constant.
3.(D)
Key Words:
Impulse is equal to
Needed for Solution:
Knowledge/definitions
Now Solve It:
Answer I is the expression for work, not impulse. Answer III is not correct because impulse does not equal momentum. Impulse is equal to both 
Δt and a change in momentum, mΔ.
4.(C)
Key Words:
Force versus time graph; initial speed
Needed for Solution:
Graphing knowledge and the impulse–momentum theorem
Now Solve It:
Use the two expressions on the right.
Note: the conversion factors for kilonewtons and milliseconds cancel each other. These units are frequently encountered in force versus time graphs.
5.(D)
Key Words:
Ice skaters; initially stationary; push off of each other
Needed for Solution:
Several possible solutions exist involving Newton’s third law, the impulse–momentum theorem, and the conservation of momentum in an explosion
Now Solve It:
The easiest solution involves seeing the separating skaters as an explosion starting from rest.
The skater with less mass must have a greater velocity in order to have equal and opposite momentum. Answers A, B, and C cannot be correct. Newton’s third law dictates that the force on each skater will be identical. If the force is the same, then the impulse and the change in momentum of each skater is the same.
6.(B)
Key Words:
Person running; stationary cart; resulting speed of combined cart and person
Needed for Solution:
Conservation of momentum in an inelastic collision
Now Solve It:
7.(E)
Key Words:
Striking a wall; some of them are stopped; some of them bounce back; greatest change in momentum
Needed for Solution:
Change in momentum
Now Solve It:
Objects that bounce have a greater change in momentum. Essentially, they are changing momentum twice. First they slow down, then they speed back up. The object with the largest combination of mass and change in velocity will have the greatest change in momentum.
Remember that when an object bounces, it reverses direction and its final velocity is now negative. This results in a negative change in momentum, which is consistent with the force acting opposite the initial velocity of the ball. Although negative, it still has the largest magnitude and creates the greatest change.
9: Gravity
1.(B)
Key Words:
Twice the mass; a radius three times; gravity
Needed for Solution:
Gravity formula
Now Solve It:
The gravity formula would solve for the gravity of Earth if the mass and radius of Earth were used.
Doubling the mass and the tripling radius will solve for gravity on a planet that is larger than Earth by those factors.
The left side of the equation must be multiplied by 
in order to maintain the equality.
2.(A)
Key Words:
Mass; speed; height
Needed for Solution:
Orbital speed formula
Now Solve It:
The mass of an orbiting body is not part of the orbital speed formula. This means that the masses m and 5m are distracters and should be ignored. The formula indicates that orbital speed depends on the central mass M, in this case Earth, and the distance from it. Both satellites are orbiting the same mass M, and both are at the same height, h. This means both satellites will have the same speed, v.
3.(D)
Key Words:
Height; period
Needed for Solution:
Kepler’s third law
T2 ∝ r3
Now Solve It:
Although the formula is incomplete, any changes to one side must be matched on the other side to maintain proportionality.
(?T)2 ∝ (2r)3
The right side has to be multiplied by 23 or 8.
(?T)2 ∝ 8r3
A number that when squared is equal to 8 is needed, which is 
.
4.(C)
Key Words:
Mass; height; speed
Needed for Solution:
Orbital speed formula
Now Solve It:
If r is multiplied by , then the left side of the equation must be multiplied by 
.
5.(B)
Key Words:
Height of 2rEarth
Needed for Solution:
Inverse square law and/or gravity formula
Now Solve It:
The inverse square of 2 is 
. The gravity at twice the orbital distance is of its value on Earth.
Analyzing the equation yields the same answer.
6.(A)
Key Words:
Gravitational force; 16 newtons;
distance between the masses is quadrupled
Needed for Solution:
Inverse square law and/or Newton’s law of gravity
Now Solve It:
The inverse square of 4 is 
, and of 16 newtons is 1 newton. Analyzing the equation yields the same answer.
7.(A)
Key Words:
Kepler’s laws
Needed for Solution:
Knowledge/definitions
Now Solve It:
Kepler’s first law describes the orbital path of planets as ellipses.
8.(B)
Key Words:
Planets; acceleration on their surfaces
Needed for Solution:
Gravity formula
Now Solve It:
Both planet II and planet III have a gplanet of 2g.
9.(D)
Key Words:
Kepler’s second law; swept; equal interval of time
Needed for Solution:
Kepler’s laws
Now Solve It:
Kepler’s second law states that a line connecting the Sun and a planet will sweep out equal areas in equal intervals of time.
10.(D)
Key Words:
Planet X has twice the mass of Earth; the orbital radius of the moon orbiting Planet X is the same as the orbital radius of Earth’s moon
Needed for Solution:
Orbital speed formula
The orbital speed is proportional to the square root of the central body’s mass
Now Solve It:
Orbital speed depends on the central mass, M (Planet X), and the orbital radius, r. Planet X has twice the mass of Earth, but the orbital radius of its moon is the same as Earth’s moon.
If mass M is doubled, then the left side of the equation must be multiplied by to maintain equality.
10: Electric Field
1.(C)
Key Words:
Adding extra electrons; atom
Needed for Solution:
Knowledge/definitions
Now Solve It:
Adding negatively charged electrons to an atom results in the creation of a negatively charged ion.
2.(B)
Key Words:
Identical conducting spheres; charge; touch; separated
Needed for Solution:
Conservation of charge
Now Solve It:
Total charge is conserved and remains constant.
Total charge = (−3 C) + (+2 C) = −1 C
When the spheres touch, their excess charges neutralize, leaving only a −1 C charge on both spheres together. Since the spheres are equal in size, each sphere must contain half of this charge, −1/2 C.
3.(A)
Key Words:
Electron; between two charged plates
Needed for Solution:
Knowledge/definitions
Now Solve It:
An electric field always points away from a positive charge and toward a negative charge. The vector field arrows of the charged plates point to the left.
4.(E)
Key Words:
Proton; electron; uniform electric field; acceleration
Needed for Solution:
FE = qE
The sum of forces yields acceleration
Now Solve It:
The charge on a proton and an electron is the same, but their masses are very different. The larger mass of the proton results in a smaller acceleration in the same electric field.
5.(C)
Key Words:
Electric force; charge Q; compare; charge 2Q
Needed for Solution:
Newton’s third law
Now Solve It:
Newton’s laws of motion apply to all forces. Anytime two objects interact, the third law requires consideration. There is always an equal and opposite force between interacting objects.
6.(D)
Key Words:
Charge; magnitude of electric field; force
Needed for Solution:
FE = qE
Now Solve It:
Rearrange the equation to solve for the electric field:
The mass was given as a distractor.
7.(E)
Key Words:
Charged sphere; electric field; 1/2d
Needed for Solution:
Inverse square law
Now Solve It:
For spherical charges, any change in distance causes the electric field to change by the inverse square. The inverse of 1/2 is 2, and its square is 4.
8.(A)
Key Words:
Two point charges; where the electric field is zero
Needed for Solution:
Superposition
Now Solve It:
For the total field to be zero, the individual electric field vectors of each charge must be equal in magnitude but opposite in direction. The zero point must be closer to the smaller charge so it can equal the field of the larger, more distant charge. This narrows the possibilities to answers A and B only. The answer must be A since the vectors of the electric field of each charge are opposite at point A. At point B, they point in the same direction and cannot cancel.
9.(C)
Key Words:
Two negative charges; electric force
Needed for Solution:
Coulomb’s law
Now Solve It:
Substitute the values into Coulomb’s law:
10.(D)
Key Words:
Two negative charges; released; force, acceleration, and velocity vectors
Needed for Solution:
Attraction/repulsion
Coulomb’s law
Sum of forces
ΣF = FE
ma = FE
Definition of acceleration and its relationship to the velocity final
Now Solve It:
Like charges repel. The charges will separate, increasing the distance r. According to Coulomb’s law, the force of electricity is inversely proportional to the square of separation. As r increases, FE decreases. Acceleration is directly proportional to the net force acting on an object, in this case the electric force. Since force decreases, acceleration also decreases. Many students will be fooled by answer C, believing a decreasing acceleration will slow an object. However, to slow an object, the force and resulting acceleration must be opposite the velocity of the object. However, as each charge moves, the force and acceleration are acting in the direction of motion. The charges are speeding up. Acceleration is the rate of change in velocity (change in velocity per second). When acceleration decreases, the change in velocity in the next second is smaller. However, the acceleration is still speeding up the object.
11: Electric Potential
1.(E)
Key Words:
Charged plates; distance; potential; electric field
Needed for Solution:
V = Ed
Rearranged to solve for electric field
Now Solve It:
Convert 10 centimeters to 0.1 meters.
For this question, the charge of 3.0 C and the plate area are not needed. Note that both N/C and V/m are valid electric field units.
2.(A)
Key Words:
Charged plates; charge; potential; capacitance
Needed for Solution:
Q = CV
Rearranged to solve for capacitance
Now Solve It:
For this question, distance between the plates is not needed.
3.(D)
Key Words:
Charged plates; charge; potential; energy
Needed for Solution:
Either equation works since the variables for both are now known.
Now Solve It:
4.(D)
Key Words:
Potential; electric field; direction
Needed for Solution:
Knowledge/definitions
Now Solve It:
The electric field is a vector that is perpendicular to the lines of equal potential and that points from high potential (12 V) to low potential (−12 V). Therefore, the electric field direction is −y.
5.(C)
Key Words:
Charge; at rest; released; travels 2.0 m; uniform 30 V/m electric field; work
Needed for Solution:
Work–kinetic energy theorem
W = ΔK = –ΔUE
Potential difference in uniform electric fields
V = E Δd
Now Solve It:
Work is related to changes in kinetic energy, which are equal in value to changes in potential energy. Changes in potential energy are related to changes in electric potential
W = ΔK = –ΔUE = –q ΔV
W = –qE Δd
The variables given require using potential energy to solve for work, but there was no indication of the correct signs on the variables. Find the absolute value for work. Then assess the resulting motion of the object to confirm the correct sign on work.
|W| = (1.5 C)(3.0 V/m)(2.0 m) = 9.0 J
Since the object started at rest, it must be speeding up. The change in kinetic energy is positive, and work is positive. In addition, force and displacement can be compared. When the direction of the force and the displacement vectors are the same, work is positive. When these vectors oppose each other, work is negative.
6.(B)
Key Words:
Charge; mass; initially at rest; travels 2.0 m; uniform 3.0 volt per meter electric field; speed
Needed for Solution:
Conservation of energy starting from rest
mv2 = q ΔV
Potential difference in uniform electric fields
V = E Δd
Now Solve It:
The change in distance is related to a change in electric potential and in potential energy. The change in potential energy equals the change in kinetic energy.
7.(D)
Key Words:
Four charges; point P
Needed for Solution:
Superposition of both electric field (see Chapter 10) and electric potential
Now Solve It:
Each charge has the same magnitude, |q|, and is the same distance, r, from point P. Therefore, the magnitude of the four resulting field vectors will all be the same. Vector direction is the critical factor in many problems. Drawing the four electric field vectors, due to each charge, at point P reveals that all the vectors cancel each other, E = 0. However, electric potential is not a vector. The sign of the charge is included in calculations, but the sign on distance is not. The potential of the four charges is summed.
8.(B)
Key Words:
Mass; charge; initially at rest; speed; potential difference
Needed for Solution:
Conservation of energy
Now Solve It:
9.(C)
Key Words:
Capacitor; fully charged; isolating; plates are pulled apart
Needed for Solution:
Now Solve It:
First equation: Pulling the plates apart increases the separation distance, d, which is inversely proportional to capacitance, C. As a result, capacitance decreases.
Second equation: When a capacitor is isolated, charge remains constant since the charges cannot move. Capacitance is inversely proportional to electric potential. So the decrease in capacitance, C, causes an increase in electric potential, V.
10.(B)
Key Words:
Distance; doubled; energy stored; capacitor
Needed for Solution:
Now Solve It:
Doubling distance, d, cuts capacitance, C, in half.
Since the battery is still connected, the voltage remains constant. Using Uc = CV2 yields the quickest answer.
12: Circuit Elements and DC Circuits
1.(B)
Key Words:
Current will flow through the 4 Ω resistor
Needed for Solution:
Voltage remains the same in parallel
Vtotal = V1 = V2 = · · ·
V = IR
Now Solve It:
Parallel resistors receive the same voltage.
Vtotal = V1 = V2 = 4 V
Use Ohm’s law to solve for current.
2.(C)
Key Words:
Total equivalent resistance of the circuit
Needed for Solution:
Resistance in parallel
Now Solve It:
Invert to solve for total resistance.
RT = 4/3 Ω
3.(D)
Key Words:
Total power dissipated in the circuit
Needed for Solution:
Three possible power equations
P = IV = V2/R = I2R
Now Solve It:
Select the equation for power that matches the given variables. The total voltage at the battery is given, and the previous problem solved for total resistance.
Using alternate variations of the power formula will produce the same result, but they require you to solve for total current first.
4.(C)
Key Words:
Current will flow through the 4 Ω resistor
Needed for Solution:
Resistors in parallel
Resistors in series
Rs = R1 + R2
Ohm’s law
V = IR
Current remains the same in series
Itotal = I1 = I2 = · · ·
Now Solve It:
Add resistors 1 and 2 in parallel.
Add resistors R12 and R3 in series to find the total resistance
RT = R12 + R3 = 1 + 4 = 5 Ω
Use Ohm’s law to find the total current in the circuit.
The 4 Ω resistor is in series, and the same 2 A current passes through it.
5.(D)
Key Words:
Voltage drop around the 4 Ω resistor
Needed for Solution:
Ohm’s law
V = IR
Now Solve It:
Use the current found in the previous question.
V = IR = (2 A)(4 Ω) = 8 V
6.(E)
Key Words:
Heat is generated; 2 seconds
Needed for Solution:
Joule’s law
Q = I2Rt
Now Solve It:
Q = I2Rt
Q = (2 A)2(4 Ω)(2 s) = 32 J
7.(C)
Key Words:
Power is dissipated; each second
Needed for Solution:
P = I2R
Now Solve It:
Power is the rate of energy use (joules per second).
P = I2R = (2 A)2(4 Ω) = 16 W
8.(A)
Key Words:
Voltage drop; R1
Needed for Solution:
Voltage adds in series
Vs = V1 + V2 + · · ·
Voltage is the same in parallel
Vp = V1 = V2 = · · ·
Now Solve It:
The battery has a voltage of 10 V. The 4-ohm resistor is in series with the battery and with resistors 1 and 2 combined. In question 5, the voltage of the 4-ohm resistor was found to be 8 volts.
VT = V12 + V3
(10 V) = V12 + (8 V)
V12 = 2 V
Resistors 1 and 2 are in parallel.
V12 = V1 = V2 = 2 V
9.(B)
Key Words:
Dissipate the most power
Needed for Solution:
Knowledge/definitions
P = I2R
Now Solve It:
The more resistors in parallel, the lower the overall resistance is. Lower overall resistance leads to more current flow.
10.(D)
Key Words:
Same voltage, V, across each one of its resistors
Needed for Solution:
Knowledge/definitions
Vp = V1 = V2 = · · ·
Now Solve It:
Resistors in parallel receive the same voltage.
13: Magnetism
1.(D)
Key Words:
Wire; current out of page: magnetic field
Needed for Solution:
Knowledge/definitions
Right-hand rule (curled fingers version)
Now Solve It:
The magnetic field forms circles around current-carrying wires, which means C and D are possible. The right-hand rule indicates that the current is counterclockwise, which is answer D.
2.(C)
Key Words:
Wire; current; distance; magnitude of the magnetic field
Needed for Solution:
Now Solve It:
The magnetic field is directly proportional to current and inversely proportional to distance.
Doubling both current and distance results in the same strength in the magnetic field: a factor of 1.
3.(C)
Key Words:
Fields; cannot change the speed
Needed for Solution:
Knowledge/definitions
Now Solve It:
Gravity and electric fields can speed up objects and slow them down when objects move parallel to these fields. However, uniform magnetic fields cause objects to circle at constant speed.
4.(B)
Key Words:
Negative charge; enters a uniform +z magnetic field; direction of force
Needed for Solution:
This charge is negative
Left-hand rule
Now Solve It:
Point the thumb of the left hand in the direction of the velocity vector. While keeping the fingers straight, point them out of the page. The palm is now oriented so that it pushes up the page in the +y-direction.
5.(D)
Key Words:
Force of magnetism; doubling the magnetic field
Needed for Solution:
FB = qvB
Now Solve It:
The force of magnetism is directly proportional to the strength of the magnetic field.
(2FB) = qv(2B)
The magnitude of force is doubled.
6.(D)
Key Words:
Acceleration; doubling its initial velocity
Needed for Solution:
Knowledge/definitions
FC = FB
Now Solve It:
Charges moving in a uniform magnetic field experience uniform circular motion with centripetal acceleration.
Doubling the velocity doubles the acceleration.
7.(B)
Key Words:
Diagram; current runs clockwise through the loop; magnetic field; force
Needed for Solution:
Right-hand rule (straight finger version)
Now Solve It:
The right hand is used for currents, which are considered positive. When force is involved, the fingers are kept straight and point into the page, −z-direction. The thumb points in the direction of the current, which is circling. However, when the hand is placed at any point on the loop, the palm always points outward away from the center of the loop.
8.(E)
Key Words:
Parallel wires; currents in the same direction; 2-ampere current; 1-ampere current; force
Needed for Solution:
Knowledge/definitions
Right-hand rule
Now Solve It:
Two parallel wires with current in the same direction will attract one another. However, the key to this problem is actually Newton’s third law. When two objects interact, there is an equal and opposite force between them. The only answer involving equal force is E. The two different currents act as distracters.
9.(A)
Key Words:
Loop; entering a uniform magnetic field
Needed for Solution:
Knowledge/definitions
Now Solve It:
Magnetic flux is the intersection of the magnetic field and the area of the loop. Initially, the magnetic flux is zero. In the end, it is at a maximum.
10.(B)
Key Words:
Loop; entering a uniform magnetic field: current is induced
Needed for Solution:
Now Solve It:
Rearrange Ohm’s law, adapted for emf, to solve for current.
Substitute BLv from the motional emf formula.
14: Simple Harmonic Motion 14
1.(B)
Key Words:
20 cycles . . . 50 seconds . . . determine the period
Needed for Solution:
Now Solve It:
Note: A cycle does not have any units. It is merely the count of a specific event.
2.(A)
Key Words:
20 cycles . . . 50 seconds . . . determine the frequency
Needed for Solution:
Either
Now Solve It:
Either
Or use the answer from question 1.
3.(C)
Key Words:
Spring . . . lowered to equilibrium . . . stretches 2.0 meters
Needed for Solution:
At equilibrium, forces are equal and opposite
Now Solve It:
Finding the spring constant is often a critical first step.
4.(E)
Key Words:
Factors affects the period and frequency of an oscillating spring
Needed for Solution:
Period of a spring
Now Solve It:
Only the mass and spring constant are contained in the formula for the period of a spring, and only these factors affect its period.
5.(A)
Key Words:
Factors affects the period and frequency of a pendulum
Needed for Solution:
Now Solve It:
Only length and gravity are contained in the formula for the period of a pendulum, and only these factors affect its period. Gravity is most often Earth’s gravity, which remains constant. However, if the pendulum is moved into a new gravity, then gravity will matter.
6.(B)
Key Words:
Pendulum . . . length of the string were quadrupled . . . resulting period
Needed for Solution:
Now Solve It:
The quadrupled length is under a square root.
This will increase the period to 2Tp.
7.(D)
Key Words:
Hooke’s law
Needed for Solution:
F = −kx
Now Solve It:
You should memorize equations and relevant names associated with key laws and major principles. The minus sign in the equation indicates that the vector representing the restoring force, F, is opposite the direction of x, the vector representing the spring stretch or compression. When solving for the magnitude of the restoring force, the minus sign is ignored, F = kx.
8.(B)
Key Words:
Diagram . . . spring-mass systems in oscillation . . . kinetic energy greatest
Needed for Solution:
Knowledge/definitions
Now Solve It:
The kinetic energy is greatest when speed is greatest. This occurs at equilibrium.
9.(E)
Key Words:
Diagram . . . spring-mass systems in oscillation . . . acceleration the greatest
Needed for Solution:
Knowledge/definitions
Now Solve It:
This occurs at maximum displacement when the spring is stretched the farthest from equilibrium. At this position, the sum of force and acceleration are the most extreme. Strangely, this is the position where motion stops instantaneously. Therefore, students are tricked into thinking acceleration is zero when it is really at its maximum. At equilibrium, acceleration is zero, but this is where the speed is greatest. Watch out for this!
10.(E)
Key Words:
Graph . . . total energy
Needed for Solution:
Knowledge/definitions
Now Solve It:
Total energy (potential plus kinetic) is conserved during an oscillation and must remain constant. Choice A is the graph of kinetic energy, and choice B is the graph of potential energy. Add these two graphs together. The result is a straight horizontal line at the maximum (total) energy during the oscillation.
15: Waves
1.(A)
Key Words:
Longitudinal waves
Needed for Solution:
Knowledge/definitions
Now Solve It:
Sound waves are always longitudinal. The other choices form transverse waves.
2.(E)
Key Words:
Frequency were doubled . . . how would the speed and wavelength be affected
Needed for Solution:
Knowledge/definitions
v = f λ
Now Solve It:
Wave speed in a given medium is constant, so speed is unaffected. As a result, frequency and wavelength are inversely proportional.
Wavelength is halved when frequency doubles.
3.(C)
Key Words:
Light moving in air . . . enters a block of glass . . . property(ies) do NOT change
Needed for Solution:
Knowledge/definitions
v = f λ
Now Solve It:
When a wave changes mediums, the frequency remains the same.
4.(D)
Key Words:
Siren rapidly approaches an observer . . . perceive
Needed for Solution:
Knowledge/definitions
Doppler Effect
Now Solve It:
When a sound source approaches an observer, the sound’s wavelength appears to be less than it really is. This causes the perceived frequency to increase.
5.(A)
Key Words:
Superposition
Needed for Solution:
Knowledge/definitions
Superposition
Now Solve It:
When two waves superimpose, their amplitudes are added together.
6.(C)
Key Words:
Wave pulses travel toward each other . . . superposition
Needed for Solution:
Superposition
Now Solve It:
Sketch the waves as they meet each other:
The area below the equilibrium line cancels out the rectangular area above the equilibrium line. This leaves the small square on the top left as the sum of the two pulses.
7.(B)
Key Words:
Node
Needed for Solution:
Knowledge/definitions
Now Solve It:
Destructive superposition of two waves causes a node.
8.(A)
Key Words:
Brightness . . .volume
Needed for Solution:
Knowledge/definitions
Now Solve It:
The amplitude of light waves is brightness. The amplitude of sound waves is volume.
9.(B)
Key Words:
Pitch . . . color
Needed for Solution:
Knowledge/definitions
Now Solve It:
The frequency of sound waves is pitch. The frequency of light waves is color.
10.(D)
Key Words:
Shortest wavelength to longest wavelength
Needed for Solution:
Knowledge/definitions
Electromagnetic spectrum
Now Solve It:
The shortest wavelengths are the electromagnetic waves with the highest energy and frequency. X-rays is first and then visible light. Radio waves are the longest wavelengths with the lowest frequencies.
16: Geometric Optics
1.(E)
Needed for Solution:
Plane mirror . . . image
Needed for Solution:
Knowledge/definitions
Reflection of a plane mirror
Now Solve It:
The image created by a plane mirror is nearly identical to the object. The image is upright, which means that it is a virtual image. The image is also the same size (M = 1) as the object. The image forms behind the mirror at the same distance as the object (di = do). Answers A through D are all incorrect.
2.(C)
Key Words:
Light ray . . . from point 1 in air to point 2 in water
Needed for Solution:
Knowledge/definitions
Refraction
Now Solve It:
Light moving in a denser medium has a lower speed and a shorter wavelength. This causes the angle of its path, measured from the normal, to be smaller than the angle measured in a less dense medium. Medium 2 is denser. So the angle of refraction is smaller than the incident angle in medium 1.
3.(D)
Key Words:
Light enters a denser medium
Needed for Solution:
Knowledge/definitions
Refraction
Now Solve It:
Denser mediums have higher indexes of refraction (answers A, C, and D). Light moves slower in denser mediums (answers C, D, and E). When light changes mediums, its frequency remains constant. As a result, wavelength is directly proportional to speed. When light slows, its wavelength decreases (answers A, B, D, and E). Only answer D satisfies all these conditions.
4.(B)
Key Words:
Light ray refracts . . . three medium . . . rank the indexes of refraction; greatest to least
Needed for Solution:
Knowledge/definitions
Snell’s law
n1 sin θ1 = n2 sin θ2
Now Solve It:
Snell’s law indicates that the index of refraction, n, and the angle of refraction, θ, are inversely proportional. Although denser mediums have higher indexes of refraction, they have smaller angles of refraction. The smallest angle is seen in medium 2, which has the greatest index of refraction. Medium 1 has the next smallest angle, followed by medium 3.
5.(B)
Key Words:
Critical angle
Needed for Solution:
Knowledge/definitions
Critical angles
Total internal reflection
Now Solve It:
If the angle of the incident ray of light is increased, the angle of refraction increases. When the incident angle becomes critical, the refracted angle is 90°. At the critical angle, light enters the second medium and all light is reflected back into the first medium.
6.(D)
Key Words:
Pinhole camera
Needed for Solution:
Knowledge/definitions
Pinhole cameras
Now Solve It:
Pinhole cameras focus a small, inverted image on the film at the rear of the camera. Inverted images are real images that are projected onto a screen (the film).
7.(B)
Key Words:
Type of lens . . . refracts light . . . to the far focal point
Needed for Solution:
Knowledge/definitions
Lenses
Now Solve It:
Converging instruments refract parallel light to the focal point. The converging lens is convex.
8.(B)
Key Words:
Screen
Needed for Solution:
Knowledge/definitions
Now Solve It:
Images projected onto a screen are real. Real images are always inverted.
9.(C)
Key Words:
Convex mirror
Needed for Solution:
Knowledge/definitions
Convex mirror
Now Solve It:
Convex mirrors are divergent instruments. Divergent optical instruments can create only small, virtual, upright images.
10.(A)
Key Words:
Diagram of a concave mirror . . . object is initially at 2f . . . how will the image change as the object moves toward . . . f
Needed for Solution:
Knowledge/definitions
Concave mirror
Now Solve It:
Concave mirrors are convergent instruments. When the object is outside the focal point, the image is real, is inverted, and appears on the near side of the mirror. When the object is at 2f, the image will be the same size and will also be at 2f. According to
object distance, do, and image distance, di, vary inversely. If the object moves toward the focus, the image moves away from the focus and away from the mirror. If image distance is increasing while object distance is decreasing, then according to
both magnification and image height increase.
17: Physical Optics
1.(C)
Key Words:
Bending of light around obstacles
Needed for Solution:
Knowledge/definitions
Diffraction
Now Solve It:
Diffraction is the bending of light around obstacles or through openings. Refraction is the bending of light due to the change in speed when light enters a new medium.
2.(D)
Key Words:
Very small opening
Needed for Solution:
Knowledge/definitions
Diffraction
Now Solve It:
The smaller an opening becomes, the more circular the wave fronts leaving the opening become.
3.(D)
Key Words:
Shadow region . . . more pronounced
Needed for Solution:
Knowledge/definitions
Diffraction and their shadow regions
Now Solve It:
The shadow region occurs where there is little or no diffraction. The question is the complete opposite of the previous question. Wider slits cause less diffraction, resulting in more distinct shadow regions.
4.(E)
Key Words:
Double-slit experiment . . . evidence that light
Needed for Solution:
Knowledge/definitions
Double-slit experiment
Now Solve It:
The double-slit interference pattern is a characteristic of a wave phenomenon. Young experimentally showed that light acts like a wave.
5.(A)
Key Words:
Interference pattern . . .distance between the bright . . .maximums . . . can be increased
Needed for Solution:
Now Solve It:
In the equation, the spacing between the maximums, xm, is inversely proportional to the slit separation, d. The slit spacing must be decreased in order to increase the space between maximums. Be careful to distinguish between the space between slits, d, and the width of the slits. Answers C and D are distractors.
6.(B)
Key Words:
Diagram above shows the paths of light . . . second dark fringe . . . path difference
Needed for Solution:
Path difference = mλ
Now Solve It:
Students should know that the dark fringes are assigned numbers ending in one-half. There is a tendency to number the second dark fringe incorrectly as m = 2.5. The second dark fringe is actually numbered as m = 1.5. Therefore, the path difference is 1.5 λ.
7.(E)
Key Words:
Polarizing filter . . . light . . . affected . . . ; filter rotated about the x-axis
Needed for Solution:
Knowledge/definitions
Polarization of light
Polarizing filters
Now Solve It:
Answers A and B are true when two filters are used and only one of them is rotated. However, a single polarizing filter rotated as shown in the diagram cannot change the amount of light passing through it. Rotating the filter will change which light waves pass through the filter. At every instant, only the correctly aligned light waves will pass through. If the filter’s orientation is changing, then so is the orientation of the light moving through it. C and D are both correct.
8.(C)
Key Words:
Prism disperses . . . as each color of light moves through the prism . . . different
Needed for Solution:
Knowledge/definitions
Dispersion
Now Solve It:
Prisms disperse light because each color of light has a different wavelength and, as a result, a slightly different index of refraction in the prism. A different index of refraction leads to a different angle of refraction for each wavelength of light.
9.(E)
Key Words:
Light striking . . . surface
Needed for Solution:
Knowledge/definitions
Reflection, scattering, polarization, and absorption
Now Solve It:
Choices A to D can all occur when light strikes the surface of an object. Some may be more prominent in certain cases, but they are all a possibility.
18: Thermal Properties
1.(E)
Key Words:
Effect of heating
Needed for Solution:
Knowledge/definitions
Thermal expansion
Now Solve It:
When an object is heated, the entire object expands proportionally. Both diameters A and B will increase. Students are often fooled by answer D. An analogy may help. What would happen if a picture of a person were resized? Would the head expand while the mouth became smaller? No. Both would expand.
2.(E)
Key Words:
Rate of heat transfer
Needed for Solution:
Knowledge/definitions
Thermal conductivity
Now Solve It:
Thermal conductivity is a physical property indicating how well heat transfers. Note that this question can be easily reworded into four additional questions testing all of the given terms. It is important to know the definitions and the effects of each of the physical properties listed as choices in this question.
3.(C)
Key Words:
Average kinetic energy . . . from 50°C to 100°C
Needed for Solution:
Now Solve It:
Although the temperature appears to be doubling, this is actually a trick question. Formulas containing T require the Kelvin temperature scale (ΔT uses either scale).
50°C = 323 Kand100°C = 373 K
Increasing temperature from 323 K to 373 K is a 15.5% increase. Average kinetic energy is directly proportional to temperature, an increase by a factor of 0.155. Answer C is the only possible answer. When in doubt, or to be safe, use Kelvin temperatures.
4.(B)
Key Words:
Temperature of a trapped gas is tripled . . . volume of the gas doubles . . . new pressure
Needed for Solution:
Ideal gas law
PV = nRT
Now Solve It:
Modify the variables that change, and determine the coefficient for P that will maintain the equality.
Pressure is 
times its original value: P.
5.(C)
Key Words:
Does not contact any surfaces . . . vacuum . . . method of heat transfer
Needed for Solution:
Knowledge/definitions
Conduction
Convection
Radiation
Now Solve It:
If the container does not touch any surfaces, then heat transfer by conduction cannot take place. Heat transfer by convection requires fluids. Although a fluid is inside the container, the vacuum surrounding the container has no fluids to transfer heat into the container. This leaves radiation. There are many forms of electromagnetic radiation. All can move through a vacuum and some can penetrate walls and containers.
6.(D)
Key Words:
Specific heat . . . how much heat
Needed for Solution:
Q = mc ΔT
Now Solve It:
Q = (5.0 kg)(3000 J/kg • K)(20°C)
Q = 300,000 J
7.(A)
Key Words:
Determine the change in temperature . . . melts
Needed for Solution:
Knowledge/definitions
Heating and cooling
Now Solve It:
The temperature does not change during a phase change. The values for the specific heat, heat of fusion, and heat of vaporization are all distracters inviting unnecessary calculations.
8.(D)
Key Words:
Thermal equilibrium
Needed for Solution:
Knowledge/definitions
Now Solve It:
When two systems are in thermal equilibrium, they are at the same temperature and no net flow occurs between them.
9.(B)
Key Words:
Gas . . . sealed in a container . . . heated . . . temperature to increase at a constant rate
Needed for Solution:
Knowledge/definitions
Heat
Temperature
Thermal conductivity
Average speed
Average kinetic energy
Pressure
Now Solve It:
When heat is added to a gas, the temperature of the system increases. As a result, the gas particles move faster and have a higher average kinetic energy. However, thermal conductivity is a property of a substance and remains constant for that substance. Thermal conductivity influences the rate at which heat transfers.
10.(E)
Key Words:
Thermal conductivities . . . insulate
Needed for Solution:
Knowledge/definitions
Heat transfer
Now Solve It:
Air has a very low thermal conductivity. As a result, it transfers heat very slowly. This makes air an ideal insulator. Insulation materials are often designed to trap air pockets in order to slow heat transfer.
19: Thermodynamics
1.(E)
Key Words:
Isothermal
Needed for Solution:
Knowledge/definitions
Now Solve It:
In an isothermal process, the temperature remains constant and the change in internal energy is zero. Answer A may catch the eye, but it is wrong. It states that the temperature is zero, not that the change is zero.
2.(E)
Key Words:
Adiabatic
Needed for Solution:
Knowledge/definitions
Now Solve It:
In an adiabatic process, no heat is added to or removed from a system.
3.(C)
Key Words:
First law
Needed for Solution:
Knowledge/definitions
Now Solve It:
The first law of thermodynamics is a statement of conservation of energy for systems involving thermal energy.
4.(B)
Key Words:
Heat removed . . . work done on the gas . . . change in internal energy
Needed for Solution:
1st law of thermodynamics
ΔU = Q + W
Now Solve It:
Heat removed is negative. Work done on the gas is positive, and the gas is compressed.
5.(A)
Key Words:
Adiabatic . . . internal energy
Needed for Solution:
Knowledge/definitions
1st law of thermodynamics
ΔU = Q + W
Now Solve It:
In an adiabatic process, no heat is added or removed.
Positive work is done on the gas. Since internal energy increased, the temperature increased.
6.(D)
Key Words:
Isometric . . . heat removed
Needed for Solution:
Knowledge/definitions
1st law of thermodynamics
ΔU = Q + W
Now Solve It:
In an isometric process, volume is constant and work is zero.
When internal energy decreases, temperature decreases.
7.(B)
Key Words:
25°C and 100°C . . . efficiency
Needed for Solution:
Now Solve It:
This requires Kelvin temperatures.
Note that answer E uses the wrong temperature scale.
8.(A)
Key Words:
Heat engine absorbs . . . exhausts . . . efficiency
Needed for Solution:
Now Solve It:
Net work is the difference between heat absorbed and heat exhausted.
9.(E)
Key Words:
Entropy
Needed for Solution:
Knowledge/definitions
Now Solve It:
Isolated systems follow natural trends. The environment cannot interfere with them. Entropy for an isolated system always increases.
10.(E)
Key Words:
Second Law
Needed for Solution:
Knowledge/definitions
Now Solve It:
All of the answers are consequences of the second law of thermodynamics.
20: Atomic and Quantum Phenomena
1.(C)
Key Words:
Alpha particle
Needed for Solution:
Knowledge/definitions
Now Solve It:
An alpha particle is a helium nucleus consisting of 2 protons and 2 electrons. It is an ion with a +2 charge.
2.(B)
Key Words:
Rutherford
Needed for Solution:
Knowledge/definitions
Now Solve It:
The atom is mostly empty space. II is true.
Answer I is Thomson’s atomic model, and answer III is a characteristic of Bohr’s atomic model.
3.(D)
Key Words:
Bohr
Needed for Solution:
Knowledge/definitions
Now Solve It:
Answers I and III are elements of the Bohr model. The Bohr model was concerned with the interior structure of the atom, not electrons emitted by photoelectric effect.
4.(A)
Key Words:
Emission . . . longest wavelength
Needed for Solution:
Photon energy is the difference between energy levels:
Now Solve It:
Be careful. Students see long wavelength and they think of big energy drops. However, energy is inversely proportional to wavelength. Long wavelengths result from small energy changes. The smallest energy change occurs when electrons move from n = 4 to n = 3, a drop of only 1 eV.
5.(D)
Key Words:
Frequency
Needed for Solution:
Photon energy is the difference between energy levels:
E = hf
Now Solve It:
When the electron transitions from n = 4 to n = 2, it loses 4 eV of energy, which is released as a photon.
Ephoton = E4 − E1 = (−5 eV)− (−1 eV)= 4 eV
The frequency of the photon can be determined by rearranging E = hf.
6.(E)
Key Words:
Photoelectric effect . . .experimental evidence
Needed for Solution:
Knowledge/definitions
Now Solve It:
The photoelectric effect is one of several experiments indicating that light has a particle characteristic. (Note that Young’s double-slit experiment is evidence of light’s wave characteristic.)
7.(E)
Key Words:
Photoelectric experiment
Needed for Solution:
Distinguishing between the effects of increasing photon frequency and increasing intensity
Now Solve It:
Only answers A–D are consistent with increasing the frequency of light. Answer E has to do with intensity (see the next question).
8.(E)
Key Words:
Photoelectric experiment
Needed for Solution:
Distinguishing between the effects of increasing photon frequency and increasing intensity
Now Solve It:
Increasing the intensity of light increases the rate that photons strike the photocell. This does not affect the photocell’s threshold frequency. Increasing the intensity does not change the energy of each photon and therefore does not change the energy of the individual electrons emitted. Increasing the rate of photons emitted does increase the number of electrons emitted.
9.(C)
Key Words:
Graph . . .maximum kinetic energy . . . frequency
Needed for Solution:
Kmax = hf − ϕ
y = mx + b
Now Solve It:
The slope of this function is Planck’s constant, h.
10.(B)
Key Words:
Work function
Needed for Solution:
Kmax = hf − ϕ
Now Solve It:
The answers to this question all contain the threshold frequency. At the threshold frequency, the maximum kinetic energy is zero, Kmax = 0.
Kmax = hf − ϕ
(0)= hf0 − ϕ
ϕ = hf0
21: Nuclear Reactions
1.(D)
Key Words:
Ranks the fundamental particles . . . massive
Needed for Solution:
Knowledge of fundamental particles 
and 
Now Solve It:
Compare the mass numbers. Although the proton and neutron appear to have the same mass, the neutron is slightly more massive. Even though the beta particle originates from the nucleus, it is an electron and has the same mass as an electron.
2.(E)
Key Words:
Alpha particle . . . change in the nucleus
Needed for Solution:
Alpha decay 
Now Solve It:
The alpha particle is ejected and leaves the nucleus. Subtract the atomic and mass numbers of an alpha particle from the original nucleus. This reduces the atomic number by 2 and the mass number by 4.
3.(B)
Key Words:
Transmutation . . . nuclear reaction
Needed for Solution:
Assess how the mass number and atomic number have changed.
Now Solve It:
The mass number remained constant, while the atomic number increased by 1. This is consistent with the removal of a beta particle from the nucleus. Removing (subtracting) the atomic number (−1) of the beta particle is the same as adding 1 to the atomic number.
4.(E)
Key Words:
Product of a nuclear reaction . . . diagram
Needed for Solution:
Knowledge/definitions
Now Solve It:
Only an uncharged entity would follow a straight path when passing through charged plates. Gamma rays are not particles. They are photons of light energy and have neither mass nor charge.
5.(A)
Key Words:
Half-life of 3 days . . . remaining . . . after 15 days
Needed for Solution:
During each half-life, the remaining sample of radioactive material is halved.
Now Solve It:
To find the number of half-lives that have taken place, divide the length of time that has passed by the length of one half-life.
Halve the original sample 5 times. However, this problem is asking for a fraction of the original sample. Treat the original sample as having a value of 1.
6.(D)
Key Words:
Half-life of 4 days . . . 
of its radioactive material remaining
Needed for Solution:
Determine how many times needs to be multiplied by itself to arrive at .
Now Solve It:
raised to the 3rd power is .
Three half-lives have taken place. Each half-life is 4 days long. As a result, it takes 12 days for the sample to decay to of its original size.
7.(C)
Key Words:
Isotope
Needed for Solution:
Knowledge/definitions
Now Solve It:
An isotope has the same number of protons but a different number of neutrons than the original atom. The name (symbol) of the atom remains unchanged.
8.(D)
Key Words:
Forces in nucleus . . . NOT correct
Needed for Solution:
Strong force versus the electrostatic force.
Now Solve It:
Neutrons and protons attract each other with the strong force. However, neutrons have no charge, and they do not add to the electrostatic force that repels the protons in the nucleus.
9.(D)
Key Words:
How many neutrons . . . following nuclear reaction
Needed for Solution:
Balance the mass numbers and atomic numbers.
Now Solve It:
Mass numbers:
235 + 1 = 90 + 142 + ?
Atomic numbers:
92 + 0 = 38 + 54 + ?
The missing mass number is 4, and the missing atomic number is 0. This reaction produces 4 additional neutrons: 
.
10.(E)
Key Words:
Categorize . . . nuclear reaction
Needed for Solution:
Knowledge/definitions
Now Solve It:
The largest nucleus is on the product (right) side of the reaction equation. Fusion is the process where smaller nuclei are combined to create a larger nucleus.
22: Relativity
1.(D)
Key Words:
Flashes a laser beam . . . observer . . . register . . . speed
Needed for Solution:
Second postulate of special relativity
Now Solve It:
The second postulate of special relativity states that the speed of light is not dependent upon the motion of the light source. The speed of light will be measured as c = 3 × 108 m/s in inertial reference frames.
2.(B)
Key Words:
Speed of a clock . . . length . . . mass
Needed for Solution:
Time dilation, length contraction, and mass effect
Now Solve It:
A stationary observer will perceive the clock on a moving object to be slower, its length (in the direction of motion) to be shorter, and its mass to be larger.
3.(C)
Key Words:
Traveling at 0.8c . . . rod of length L . . .astronaut inside the spaceship
Needed for Solution:
Length contraction
Now Solve It:
The astronaut inside of the spaceship will not see any change in the length of the rod. The astronaut is in the same inertial reference frame as the rod.
4.(D)
Key Words:
Speed of 0.85c . . . moving right; appearance of the box . . . stationary observer
Needed for Solution:
Length contraction
Now Solve It:
Length contraction affects only the direction of motion. The spacecraft is moving in the x-direction. The box will appear shorter in the x-direction. The height and depth will remain unchanged.
23: Historical Figures and Contemporary Physics
1.(D)
Key Words:
Potential, current, and resistance
Needed for Solution:
Knowledge Problems in this chapter deal with specific facts that are either memorized or recognized.
Now Solve It:
The relationship among these variable was described by Georg Simon Ohm and is known as Ohm’s law, V = IR.
2.(A)
Key Words:
Light . . . packets . . . E = hf
Needed for Solution:
Knowledge Problems in this chapter deal with specific facts that are either memorized or recognized.
Now Solve It:
This was suggested by Albert Einstein in his work on the photoelectric effect.
3.(B)
Key Words:
Link . . . work . . . heat
Needed for Solution:
Knowledge Problems in this chapter deal with specific facts that are either memorized or recognized.
Now Solve It:
James Joule conducted the famous experiment demonstrating heat-work equivalence. Energy can be added to a system as either mechanical work or heat.
4.(B)
Key Words:
Complex . . . random systems
Needed for Solution:
Knowledge Problems in this chapter deal with specific facts that are either memorized or recognized.
Now Solve It:
This is chaos theory.
5.(B)
Key Words:
Properties of both a conductor and an insulator
Needed for Solution:
Knowledge Problems in this chapter deal with specific facts that are either memorized or recognized.
Now Solve It:
Semiconductors have the properties of both a conductor and an insulator. Semiconductors are widely used in all modern electronic devices.
Practice Test 1
1.(B)
The atomic number (number of protons) increases from 83 to 84. For this to happen, a neutron must eject a beta particle (an electron) and become a proton. The overall mass of the atom remains virtually unchanged because the mass of the ejected electron is very small. As a result, the atomic mass remains unchanged.
2.(D)
Fission is the breaking apart of an atom with a large number of protons and neutrons into two smaller atoms. Free neutrons are often released along with a tremendous amount of energy.
3.(A)
An alpha particle is the nucleus of a helium atom, 
. When two protons and two neutrons are released from thorium-227, the resulting isotope is radium-223. Adding together the atomic numbers of helium and radium-223 as well as adding together their atomic masses will reveal thorium-227.
4.(D)
The dot representing the sound source is located at the extreme boundary of the sound wave it has created. The sound waves are superimposing on each other at this point, creating the sound barrier.
5.(C)
The dot representing the sound source is located behind the boundary of the sound wave it has created.
6.(E)
The dot representing the sound source is located in front of the sound wave it has created.
7.(B)
Current is a measure of the amount of charge, Q, flowing per time, t. It is measured in units of amps (amperes).
8.(C)
Power is the rate of energy dissipation in a circuit. It is measured in units of watts.
9.(E)
Voltage is synonymous with potential difference. It is measured in joules per coulomb.
10.(A)
The rays will converge beyond the focal point on the right side of the lens, forming an inverted and real image.
11.(E)
The rays will diverge. The negative back ray traces will form an upright, virtual image.
12.(A)
The rays will reflect off the mirror beyond the focal point on the left side of the mirror, creating an inverted, real image.
13.(B)
The area under a speed-time graph is the displacement. Finding the area under interval A requires solving for the area of a triangle.
Displacement = Area = (30 m/s)(1 s) = 15 m
14.(C)
During interval C, the object is moving at a constant speed of 20 m/s.
15.(D)
The slope of a speed-time graph is acceleration. At t = 6 seconds, there is a negative, nonzero slope, indicating an acceleration. At t = 6 seconds, the instantaneous speed is also zero. In order to return to the origin, the displacement must be zero. Displacement is the area under a speed-time graph. For the first 6 seconds, the accumulated area is positive, indicating that the object has moved away from the origin in the positive direction. The zero value at t = 6 seconds is the object’s instantaneous speed, not its position.
16.(D)
Try 
rearranged to solve for height.
At max height, the final velocity is zero.
Doubling the initial speed of the ball will increase the maximum height by 4.
17.(A)
The initial velocity is zero.
18.(C)
Velocity is a vector quantity and is calculated using the displacement vector. Displacement is the final position minus the initial position. It is the shortest straight-line distance from the initial position to the final position. Although the object moved along a circular path, its displacement is the diameter of the circle and is equal to 20 m. So the velocity equals 4 m/s.
19.(C)
The time for the boat to cross the river is independent of the motion of the flowing river. The boat takes 10 seconds to cross the 50-meter river at a speed of 5 meters per second. During these same 10 seconds, the boat drifts downstream because of the river flowing at 2 meters per second. In 10 seconds, the boat will have drifted 20 meters downstream.
20.(E)
Horizontal motion does not affect vertical motion. The amount of time the ball takes to hit the ground can be determined as follows:
The horizontal motion of a projectile is constant velocity.
21.(C)
Before throwing the ball vertically, both the person and the ball have a horizontal velocity of 5 meters per second. Even though the ball is thrown straight up at 20 meters per second in the vertical direction, it continues to move independently at 5 meters per second horizontally. The cart also maintains the same horizontal speed. As a result, the ball stays above the cart during its motion and lands where it started, in the thrower’s hand.
22.(D)
The horizontal velocity remains constant during flight at v cos θ. At the top of the flight, the vertical velocity becomes zero. However, the horizontal velocity remains unchanged.
23.(E)
24.(B)
The component of gravity acting parallel to an incline is mg sin θ.
If the object is stationary, the forces acting along the incline must have equal magnitudes in opposite directions.
T = mg sin θ
The sine of 30° is equal to .
T = mg
25.(E)
The direction of motion is parallel to the string. The sum of the forces acting on the system (both bodies simultaneously) will be the forces that are parallel to the string. The two equal and opposite tensions cancel.
26.(D)
Newton’s third law states that each force is balanced by an equal but opposite reactionary force. If Earth pulls on a person with a force of 600 N, then the person will also pull on Earth with a force of 600 N. The effect on the acceleration of the person, however, is much greater than the effect on the acceleration of the Earth. Given the same force, the more massive Earth does not accelerate as much as the less massive person.
27.(C)
In order for the mass that is acted upon by these forces to remain at rest, the sum of the forces must be zero. Force F must cancel the combined pull of the 8 N and 6 N forces. When the 8 N and 6 N forces are added using vector addition, they form the sides of a 3-4-5 triangle. Their vector sum is 10 N directed in the third quadrant. In order to cancel this 10 N force, force F must pull with 10 N in the opposite direction.
28.(E)
Humans sense weight by interacting with surfaces. When asked for the apparent (feeling of) weight of a person, solve for the normal force.
Essentially, the acceleration of the spacecraft is added to the force of gravity.
29.(A)
All three blocks are stationary. There is no acceleration. Therefore, there is no net force.
30.(B)
The centripetal force, Fc, is the sum of the forces acting on the mass and is directed toward the center of the circular motion. When the mass reaches the bottom of the circle, tension is directed toward the center of the circle and is positive. Gravity acts downward, is directed away from the center of the circle, and is negative.
31.(C)
At the top of the circle, both tension and gravity act downward toward the center of the circle and both are positive.
The mass, radius, and gravity are all fixed variables that cannot be changed. However, as the speed of the circling mass is reduced, the tension in the string will decrease. The minimum speed occurs when the tension is zero at the exact instant that the mass is at the top of the loop.
32.(E)
The mass does not affect the answer. Any object traveling under these conditions will have the same velocity.
33.(D)
34.(C)
The centripetal force, Fc, is the sum of the forces acting on the mass and directed toward the center of the circular motion. For a car making a turn, the force acting toward the center of the turn is friction.
While the car is moving in the horizontal plane, it is not moving vertically. Therefore, the normal and gravity are canceling in the y-direction. They must have equal magnitudes: N = mg.
Mass does not affect the outcome. It is not part of the equation.
35.(B)
As the speed is constant, the sum of the forces acting on the block is zero. The force acting on the block down the ramp is given by mg sin θ. The force pulling the block uphill, F, must therefore be equal but opposite to mg sin θ. The sine of 30° is equal to .
36.(E)
As there is no friction on the hill, the amount of work, W, done by force F pulling the block uphill is equal to the change in potential energy, ΔUg, of the block regardless of the path taken by the block.
37.(C)
Power, P, is the rate of work: work divided by time. However, the time needed for the block to move up the hill has not been given. An alternate solution is needed. Work, W, is equal to the force, F, parallel to the motion of an object multiplied by the displacement, Δx, of the object.
The displacement divided by the elapsed time is velocity, v.
The force and velocity must be parallel to use this formula. If they are not parallel, use the component of force that is parallel to the velocity.
38.(B)
This involves conservation of energy. The potential energy of the spring is converted into the kinetic energy of the block:
39.(D)
Impulse, J, is defined as a change in momentum, Δp. Impulse can be calculated by multiplying the force, F, acting on an object by the elapsed time, t, that the force acts. The change in momentum is also equal to the mass, m, of an object multiplied by the change in velocity, Δv.
40.(E)
Linear momentum and total energy are always conserved, regardless if the collision is inelastic or elastic. However, in an inelastic collision, some of the kinetic energy is transformed into thermal energy that is then lost as heat to the environment.
41.(E)
This is conservation of momentum. The total momentum of both objects added together must be the same before and after the collision. This is an explosion. Although the momentums of the two objects are in opposite directions, their magnitudes must be equal to conserve momentum.
42.(B)
Orbits involve circular motion. The centripetal force, Fc, is the sum of the forces acting on the mass and directed toward the center of the circular motion. For planets, this is the force of gravity.
Fc = Fg
Two force of gravity equations can be substituted into this equation. Both are shown in the next step. Either one can be used to answer this question.
In both of these equations, the orbiting mass m cancels. The mass M in the right equation is the mass of the large central star. Since the mass of the orbiting body cancels, all planets at the same radius will have the same orbital speeds.
43.(A)
The measurement given is from the surface of Earth. The value needed for gravity calculations is the distance from the center of a planet. The point in space is located 3rE from the center of Earth. The acceleration of gravity, g, at a point in space can be found using the gravity equation.
Tripling the distance from the center of Earth will cause the gravity to be 
the value it is on the surface of Earth.
44.(B)
Coulomb’s law can be expressed as:
45.(E)
By definition, the direction of an electric field is in the same direction as the force on a positive charge.
46.(A)
Electric potential, also called voltage, for a point charge can be determined by the following formula:
When there is more than one point charge, the electric potential is the sum of the individual potentials.
The sign on the charge is included. However, the distance from each charge to the point in question is an absolute value.
47.(E)
This is conservation of energy. The electric potential energy is converted into kinetic energy.
48.(C)
Add resistors R1 and R2 in parallel:
Add R12 and R3 in series.
Use Ohm’s law, V = IR, to determine the total current:
Resistor R3 is in series with the battery. In series, the current is constant.
I3 = Itotal = 2.0 A
49.(A)
The current flowing through R3 has been determined to be 2.0 A. Use Ohm’s law, V3 = I3R3, to determine the voltage drop across R3:
V = (2.0 A)(1 Ω) = 2 V
50.(A)
Any of these three power formulas can be used:
51.(B)
Resistors in parallel receive the most current, dissipate the most power, and will also glow the brightest when they are lightbulbs.
52.(C)
The right-hand rule states that the thumb of the right hand points in the direction of motion of a charged particle, the extended fingers point in the direction of the magnetic field (into the page in this case), and the palm of the hand points in the direction of force (upward in this case). The right-hand rule applies to positively charged particles. Since this is a negatively charged particle, it will do the opposite and be forced downward in the –y direction. This matches the answer if the left hand had been used instead of the right hand.
53.(B)
The induced emf, ε, is caused by a change in flux, ϕ. Flux can be determined by the area of the loop multiplied by the magnetic field passing through the loop. The amount of flux will change as the loop is moved with velocity v through the magnetic field, B. When a linear length of wire, L, enters a magnetic field, the emf generated in the wire can be determined with the following formula:
ε = BLv
54.(C)
The period of a pendulum can be described as:
The period is affected only by changes in the length, L, of a pendulum’s string and the gravity field, g, it is in. Changes in the displacement, x, and mass, m, do not affect the period of a pendulum.
55.(A)
The kinetic energy is at maximum when the mass is passing through the position of zero amplitude. The kinetic energy is zero as the mass reaches the maximum and minimum displacements of +A and –A and velocity becomes zero. The potential energy of the system is depicted in B. Total mechanical energy is shown in E.
56.(E)
Wave speed is determined by the medium in which it travels. Since the medium has not changed the wave speed must remain a constant v. The product of frequency and wavelength is the wave speed. Therefore, the wavelength must be reduced to λ when the frequency is doubled in order for wave speed to remain constant in the same medium.
57.(A)
When the two waves superimpose, they will add destructively since they are each on opposite sides of the axis of propagation. The wave pulse on the left is a larger and inverted version of the pulse on the right.
58.(A)
For strings and open tubes, the wavelengths of the harmonics can be found using the following formula where n is the number assigned to the harmonic:
The second harmonic is n = 2.
59.(D)
For objects placed between the focus and a convex lens, the image will be magnified, virtual, and upright.
60.(C)
Light will refract toward a normal line drawn perpendicularly to the surface of the medium it is entering if that medium has a higher index of refraction than the one it is exiting. Glass has a higher optical density and a higher index of refraction than air.
61.(C)
A convex mirror is a divergent optical instrument. It can form only virtual images that are upright.
62.(E)
In the above formula, xm is a measure between the bright regions seen on the screen in Young’s double-slit experiment. If the space between the bright regions decreases, then xm will decrease and vice versa. Decreasing the wavelength (λ), decreasing the distance from the slits to the screen (L), and increasing the slit spacing (d), will all decrease the xm and the space between bright regions on the screen. If the experiment is performed under water, then the wave speed will decrease, causing a corresponding decrease in wavelength. All of these alterations will result in a closer spacing pattern.
63.(B)
This is the definition of diffraction.
64.(C)
When polarizing filters are oriented perpendicularly to each other, no light can pass through. When oriented in a parallel position, the maximum amount of light can pass through.
65.(C)
The ideal gas law describes the relationship among the pressure, volume, and temperature of a gas.
There is no change in the temperature.
66.(E)
Heating a metal will cause the metal to expand. The entire ring, outer diameter, and inner diameter will all expand proportionally.
67.(E)
68.(D)
During an isothermal process, the temperature remains constant and the change in internal energy, ΔU, is zero. When heat is removed, the engine loses energy and heat is negative, Q = −600 J.
69.(D)
During an isometric process, the volume of gas remains unchanged. No work is done since there is no change in volume.
70.(C)
The temperatures must be in kelvins.
71.(B)
9 eV is enough energy to raise the electron from the ground state, n = 1, to an excited state at n = 4. The possible photon’s energies emitted can be determined by taking the difference between any two energy levels on the electron’s way back to the ground state. All of those energies are possible except for 2 eV.
72.(E)
Once the threshold frequency has been reached, the current begins. The current cannot increase with increasing the frequency. To increase the current, the intensity of the light must be increased. Increasing the intensity of the light increases the number of photons. This would cause more electrons to be ejected.
73.(B)
If each half-life lasts for 25 days, then 100 days is four half-lives.
During each half-life the sample is reduced by half.
74.(A)
As objects approach the speed of light, their length in the direction of motion will appear to decrease to an outside observer. Since this spacecraft is moving in the x-direction, only this dimension appears shorter.
75.(B)
Kepler’s three laws of planetary motion are based upon observations of the orbit of Mars.
Practice Test 2
1.(D)
To identify if an object is speeding up or slowing down, assess the trend in the speed (absolute value of velocity) of each object. Object A is speeding up from 0 to 30 m/s. Object B has a constant speed of 15 m/s. Object C is slowing during the first 3 seconds and then speeding up during the next three seconds. Object D is slowing during the entire interval from 15 m/s to zero and is therefore the correct answer. Object E does have negative velocity. However, its speed (absolute value of velocity) is increasing from 20 m/s to 30 m/s.
2.(A)
Acceleration is the slope of the velocity-time graph. The magnitude of acceleration is the absolute value of the slope. The greatest acceleration will have the steepest slope. Object A has the steepest slope and therefore the greatest acceleration, 5 m/s2.
3.(E)
The area between the velocity-time graph and the x-axis is the displacement of an object. The area between the graph of object E and the x-axis represents a displacement of 135 m from the origin in the negative direction. Although negative, it is still the largest displacement.
4.(E)
Work is a change in energy. Power is the rate of work, which is also the rate of change in energy.
5.(C)
Total mechanical energy is the sum of the kinetic and potential energies of a system.
6.(B)
Potential energy is the energy associated with the instantaneous position of an object.
7.(B)
The direction of the electric field is the same as the direction of the force on a positive test charge. This means the electric field lines are drawn away from positive plates, or charges, and toward negative plates or charges.
8.(A)
Negative charges move in the direction opposite that of electric field lines.
9.(D)
The electric force described in answer 8 will begin to accelerate the electron to the right. Use the right-hand rule to find the force of magnetism on the moving charge. The thumb of the right hand points in the direction of motion of a charged particle, the extended fingers point in the direction of the magnetic field (into the page in this case), and the palm of the hand points in the direction of force (upward in this case). The right-hand rule applies to positively charged particles. Since this is a negatively charged particle, it will do the opposite and be forced downward in the −y-direction. As an alternative, you can use the left hand to solve for negative charges moving in magnetic fields.
10.(D)
Refraction is the bending of light, or any wave, as it moves from one medium into another.
11.(A)
Diffraction is the spreading of waves that results when waves pass through an opening or encounter an obstacle.
12.(B)
Wave superposition causes interference patterns in waves.
13.(D)
The photoelectric effect experiment provided evidence that light behaves as a particle made up of discrete packets of energy known as quanta.
14.(E)
The interference patterns observed by passing light through two small slits in Young’s experiment imitated the behavior of water waves passing through two small slits.
15.(C)
While trying to prove the existence of the ether, the medium in space in which light waves supposedly travel, the Michelson-Morley experiment failed to find the ether but did determine an accurate measurement for the speed of light.
16.(C)
An object moving at a constant speed may be turning. When an object turns, it is changing direction and is therefore changing it velocity. If the velocity is changing, the object is accelerating. Uniform circular motion is an example. The magnitude of velocity is constant (A), the object is changing direction (B), and centripetal force is the net force acting on the object and is greater than zero (D). The work–kinetic energy theorem states that the net work is equal to a change in kinetic energy. If an object is moving at constant speed, then the change in kinetic energy and net work are both zero (E).
17.(A)
18.(B)
The equation above simplifies to the following when vi = 0 m/s:
Velocity is proportional to the square root of acceleration.
Doubling acceleration increases the final velocity by .
19.(B)
The horizontal and vertical motions solve independently. The y-direction accelerates under the influence of gravity. In a horizontal launch, the initial velocity in the y-direction is zero, viy = 0.
The x-direction has constant velocity.
20.(E)
Acceleration due to gravity is a constant 10 meters per second squared during the entire flight of the object. Even at the top of its arc, when the instantaneous velocity becomes 0, the acceleration remains 10 meters per second squared.
21.(B)
The friction force must be equal but opposite to the horizontal component of force, Tx, in order to maintain a constant velocity.
22.(D)
The string is lifting the object by an amount equal to T sin θ. The sum of the normal force, N, and T sin θ is equal to the weight of the object. Therefore, the normal force must be less than the weight of the object.
23.(D)
Treat the blocks as a single system with a combined mass.
24.(E)
The string between the two blocks provides the force accelerating the 2.0 kg block at the same rate as the acceleration of the system.
25.(E)
Apparent weight is equal to the normal force acting on the person.
26.(B)
In order for the system to remain at rest, the opposing forces acting on the two masses must be equal and opposite. The tension in the string cancels. Set the magnitude of the friction force acting on mass 1 equal to the force of gravity acting on mass 2.
27.(B)
The friction force, f, holds the car in the turn and creates the net centripetal force, Fc, acting toward the center of the circular motion.
28.(B)
Centripetal acceleration is equal to the square of the velocity divided by the radius. If the radius were doubled, the acceleration would be halved.
29.(D)
At the top of the loop, both the normal force to the track and the force of gravity act downward toward the center of the circle. Both forces are positive.
The mass, radius, and gravity are all fixed variables that cannot be reduced. However, as the speed of the roller coaster is reduced, the normal force of the track decreases. The minimum speed occurs when the normal force reaches zero at the top of the loop.
30.(D)
The spring constant is the slope of a force-displacement graph.
31.(B)
The work is the area underneath a force-displacement graph.
32.(B)
This is conservation of energy. The potential energy at the top of the curved section of track is transformed into kinetic energy. The initial height is equal to the radius of the curved section of track, h = r.
33.(C)
Even though the question asks for the total energy at point P, you do not need to solve for projectile motion and then determine the landing velocity. Total mechanical energy is conserved and is the same at the beginning and at the end of the problem. It is easier to solve for the total mechanical energy at the beginning where it consists entirely of kinetic energy.
34.(E)
Linear momentum is conserved during both elastic and inelastic collisions. In perfectly elastic collisions, kinetic energy is conserved as well.
35.(D)
Change in momentum is equal to the area underneath a force-time graph.
36.(B)
This is conservation of momentum. This collision is perfectly inelastic. The mass sticks together to form one large, combined mass.
37.(A)
Impulse is equal to the change in momentum. This question is asking for the change in momentum of only the 3.0 kg mass.
38.(D)
Regardless of the mass of an object in orbit around a planet or a star, the velocity at a specific distance from the center of the planet or star is constant for that specific distance. All of the other statements are true.
39.(A)
The universal law of gravitation which can be expressed as:
The radial distance, r, between the two masses is a line drawn from the center of one mass to the center of the other mass. The actual radius of the masses is not a factor in the equation. Since the star did not lose any mass as it collapsed, the force between the star and planet remains unchanged.
40.(D)
The negative rod near sphere 1 causes negative charges to be repelled from sphere 1 onto sphere 2. As a result, sphere 1 has a greater positive charge while sphere 2 has a greater negative charge. If the spheres are separated, the spheres become charged. Since the rod did not touch either sphere, this was accomplished by induction. If sphere 1 receives a charge of Q, then due to conservation of charge sphere 2 must have an equal and opposite charge of −Q. If the spheres are again touched, they will neutralize and neither sphere will then have a charge.
41.(E)
For the total electric field to equal zero, the individual electric field vectors of the two charges must be opposite and equal to one another. The electric field vector due to q points toward q and the electric field vector due to Q points away from Q. The only locations where these two field vectors are opposite each other is to the left of charge q and to the right of charge Q. However, in order for the magnitudes of the vectors to be equal, the zero point has to be closer to the smaller charge q. Therefore, the only location where the electric fields can cancel is to the left of charge q. This narrows the choices to A, C, or E. The magnitude of the electric field of each point charge can be found using the following equations:
r1 and r2 are the distances from each charge to the zero point.
In order to cancel each other, E1 and E2 must be equal. So the location of the zero point can be found by setting the two equations equal to each other where r2 = r1 + d.
Since this is a multiple-choice question, a guess-and-check approach may be more efficient than solving the equations. Substitute the value for r1, and set r2 equal to r1 + d.
42.(A)
Doubling the values for q, Q, and r would yield:
This is equivalent to the original expression.
43.(D)
44.(B)
45.(B)
The electric potential energy, UE, of charge, q, located a distance, d, from the charged plate that it is attracted to can be determined as follows.
UE = qEd = (0.40 C)(30 N/C) (0.10 m)= 1.2 J
The charge is located at the midpoint between the plates. Its energy is related to its position, not the distance between the plates. If the positive charge had been located initially on the positive plate, then the distance between the plates would have been used.
46.(D)
First add the two 8 Ω resistors in parallel.
Add this resistor to the 2 Ω in series.
47.(A)
Brightness is determined by the power dissipated in the bulb. Power can be determined by using the following formula:
P = I2R
All of the bulbs have the same resistance, R, so their brightness is determined by the amount of current flowing through them. All of the current must pass through bulb 1. After flowing past bulb 1, the current reaches a junction. Some current must go down the path leading to bulb 2, and the rest of the current must go down the path leading to bulbs 3 and 4. Current will take the path of least resistance, so more current will flow down the path leading to bulb 2 than the path leading to bulbs 3 and 4. However, the exact same amount of current will pass through both bulbs 3 and 4 because they are in series and receive the same current. As a result, bulbs 3 and 4 will have the same brightness.
48.(C)
The voltage of R1 is shown to be 4 V. The current flowing through R2 is shown to be 2 A. This same current must also flow through R1 because they are in series with each other. Using Ohm’s law, V = IR, the resistance of R1 can be determined:
49.(B)
The components in any loop of the circuit must use the voltage supplied by the battery. A loop exists containing R1, R2, and R3. These must add up to the voltage of the battery.
50.(D)
Power can be determined using the following formula:
P = IV
The voltage drop across R3 is shown to be 6 V. The current across R3 is equal to the total current flowing in the circuit minus the current flowing through the parallel resistor, R4. The total current of 2 A is flowing through R2 and splits up between R3 and R4. Since the current flowing through R4 is 0.5 A, the current flowing in R3 must be 2.0 A − 0.5 A = 1.5 A.
P = IV = (1.5 A)(6 V) = 9 W
51.(D)
The right-hand rule states that the thumb of the right hand points in the direction of motion of a charged particle, the extended fingers point in the direction of the magnetic field (into the page in this case), and the palm of the hand points in the direction of the force (upward in this case). The right-hand rule applies to positively charged particles. The charge will circle counterclockwise. The magnitude of its radius can be determined by setting the centripetal force, Fc, equal to the force of the magnetic field on a moving charge.
52.(C)
Magnetic fields apply a force in a direction that is perpendicular to the motion of the object. As a result, the forward speed is not changed, only the velocity is changed. Velocity changes because there is a change in direction but not in magnitude.
53.(B)
The induced emf, ε, is caused by a change in flux, ϕ. Flux can be determined by the area of the loop multiplied by the magnetic field passing through the loop. The amount of flux will change as the loop is moved with velocity v through the magnetic field, B. The equation to describe this is ε = BLv. The ε is essentially an induced voltage. The current can be found by applying Ohm’s law: ε = V = IR.
54.(E)
In order to double the period T, the length must be quadrupled.
55.(E)
During the oscillation, the total amount of energy will not change. The kinetic energy and potential energy will transform between each other, but their sum will remain constant.
56.(C)
This is the order of the electromagnetic spectrum listed from shortest to longest wavelength. It is also the electromagnetic spectrum listed from highest to lowest frequency.
57.(B)
Dispersion is caused by the slightly different indexes of refraction within a particular medium depending upon the wavelength. White light is made up of multiple wavelengths that experience dispersion in a prism.
58.(D)
There are 2.5 wavelengths visible in the drawing: 2.5λ = 5 m.
59.(E)
An object placed between the focal point, f, and twice the focal point, 2f, will form a real and inverted image beyond the 2f point on the opposite side of a convex lens.
60.(A)
Snell’s law can be expressed as:
n1 sin θ1 = n2 sin θ2
When total internal reflection occurs, θ2 will equal 90° and the sine of 90° is 1. Air, n2, has an index of refraction of 1.
61.(D)
Different wavelengths of light have slightly different indexes of refraction in glass. The shorter the wavelength, the greater the refraction resulting in the dispersion of the colors of light.
62.(C)
This pattern is caused by the interference of the light upon itself as it passes through the single slit and is evidence of the wave nature of light.
63.(C)
When light passes from one medium into another, the frequency does not change. Light moving from air, which has nearly the same index of refraction as a vacuum, will slow down.
v = fλ
When the frequency is constant, wavelength is directly proportional to wave speed. If the speed decreases, then the wavelength also decreases.
64.(B)
Heat, Q, is equal to the mass, m, multiplied by the specific heat, c, and the change in temperature, ΔT:
The coefficient of linear expansion is not needed for this part. When in doubt, use kelvins. However, if the formula involves a change in temperature, ΔT, then either the Celsius or the Kelvin scale can be used; 1° is equal to 1 K.
65.(C)
The change in length, ΔL, is equal to the coefficient of linear expansion, α, multiplied by the original length, L0, and the change in temperature, ΔT:
66.(D)
Use the ideal gas law. The change in both pressure and volume must offset the increase, by a factor of 6 in temperature.
Find the combination that does not work.
Increasing the pressure by a factor of 15 and decreasing the volume to will only raise the temperature to 5 times its initial value.
67.(E)
During an adiabatic process, the pressure and volume change so rapidly that no heat is exchanged with the surroundings, which is consistent with the correct answer, E. Why are the other answers wrong? Examine the first law of thermodynamics: ΔU = Q + W. In an adiabatic process, heat exchange is zero (Q = 0). This modifies the first law to ΔU = W. In this process, the 1,200 joules of work are done by gas. When work is done by gas, the gas loses energy, ΔU = W = −1,200 J. This means the internal energy decreases, invalidating answers A and B. If the internal energy decreases, then the temperature decreases, invalidating answers C and D.
68.(B)
The net work, W, can be determined by subtracting the heat exhausted to the cold reservoir, QC, from the heat absorbed by the engine, QH.
W = QH – QC = 2,500 J – 1,500 J = 1,000 J
Efficiency, e, can be determined by dividing the net work done by the engine by the amount of heat added, QH.
69.(C)
Michelson and Morely’s experiment provided evidence that light travels through space without the need of a medium. It became evidence of the particle nature of light in addition to confirming the speed of light.
70.(A)
When the threshold frequency is reached, photoelectrons will be emitted with a kinetic energy above 0. According to the graph, that occurs when the frequency is equal to 1.0 × 1015 Hz. That narrows the choices to A and C. The work function is the amount of energy that must be added to reach the threshold frequency. The y-intercept of the graph, −4 eV, indicates the energy of the electrons occupying the lowest energy level inside the atom. To reach the threshold frequency, +4 eV of energy must be added to these electrons. Therefore, the work function is +4 eV, which is also consistent with answers A and C. Planck’s constant is the slope of the function.
This narrows the answer to A.
71.(E)
The slope of the line is Planck’s constant. Constants do not change, so the slope of the line cannot change.
72.(D)
Beta decay occurs when a neutron releases an electron and becomes a proton. The result will be an increase in the atomic number by 1 but no change to the atomic mass because the mass of a neutron is essentially the same as that of a proton. The loss of mass from the release of an electron is insignificant.
73.(C)
In a balanced nuclear equation, the sum of the mass numbers on the left side of the equation must equal the sum of the mass numbers on the right side of the equation. Similarly, the sum of the atomic numbers on both sides of the equation must be equal. The sum of the mass numbers on the left side is 236. A liberation of 3 neutrons would result in a total mass number of 236 on the right side of the equation.
74.(C)
The result of an object reaching the speed of light is that its mass will increase and its length will shorten in the direction of travel.
75.(B)
The motion of stars and galaxies is not consistent with the mass that can be seen with telescopes. Mathematically, a large portion of the mass of the universe is missing. Since the missing mass cannot be observed directly, it has been called dark matter. A search is underway to identify the validity, composition, and properties of dark matter.
Practice Test 3
1.(D)
Slope is the rise, Δy, divided by the run, Δx. The rise of a position-time graph is position, in meters. The run is time in seconds. The slope has the units of meters per second. Slope can be positive or negative, indicating direction. Speed is not correct because it is the absolute value of velocity and consists of magnitude without a specific direction.
2.(E)
The rise of a velocity-time graph is velocity in meters per second. The run is time in seconds. The units of slope, meters per second per second, match the units of acceleration.
3.(B)
The area bounded by a function can be determined by multiplying the rise by the run. The product of the rise, in meters per second, and the run, in seconds, is meters. This is the unit of displacement.
4.(C)
Also known as voltage, electric potential is the potential ability to generate electric energy should a charged particle occupy a particular point in space. The charged objects surrounding the point in space generate this potential. When a new charged object is positioned at the point in space, it acquires potential energy due to the surrounding charges.
5.(E)
Capacitance is the amount of charge, Q, stored per voltage, V, measured in units of farads.
6.(A)
Charges are surrounded by an electric field. This is similar to the electric potential in question 4. The electric field is the capability of the charge to create an electric force on any other charge that occupies a particular location within the electric field.
7.(C)
The concave lens is a diverging lens.
8.(A)
A converging lens will form a real and inverted image on the far side of the lens from the object if the object is placed outside of the focal length.
9.(B)
A converging lens will form a virtual and upright image on the near side of the lens if the object is placed inside of the focal length. This is the effect produced by a magnifying lens.
10.(E)
An alpha particle is the nucleus of a helium atom. It has a charge of +2.
11.(A)
A beta particle is an electron ejected from a neutron. As a result, the neutron becomes a positive proton. Beta particles have a charge of –1.
12.(C)
Neutrons are generally released as a by-product of fission. Neutrons have a neutral charge.
13.(D)
At the top of its flight, the ball will have an instantaneous velocity equal to zero. As a result, the initial and final velocities are both known. The vertical displacement is needed.
14.(D)
The final speed can be determined with the following kinematic equation:
vf = vi + at
When the initial speed equals zero, the equation simplifies to vf = at and the final speed is directly proportional to time. Doubling the time will double the final speed.
(2vf) = a (2t)
Displacement can be described with the following kinematic equation:
x = vit + at2
When the initial speed equals zero, the equation simplifies to x = at2 and displacement is proportional to the square of time. Doubling the time will quadruple the displacement.
(4x) = a (2t)2
15.(C)
Acceleration is the slope of a velocity-time graph. It is not the slope of a position-time graph. Choice A is good distracter. However, this scenario can be true for objects that are changing direction at constant speed, such as objects in uniform circular motion.
16.(B)
To travel north, the plane must have a component of velocity directed north. To overcome the eastward wind, the plane must also have a component of velocity that opposes the wind and is therefore directed to the west. Resolving the components results in a true velocity directed northwest.
17.(B)
Since the ball is thrown horizontally, the time, t for the ball to hit the ground is the same as if it were simply dropped from a height of y.
Time is proportional to the square root of vertical displacement. Doubling the vertical displacement to 2y increases the time to t.
The horizontal motion solves independently, and it has constant velocity.
x = vxt
Horizontal displacement is directly proportional to time. If the time is t, then the horizontal displacement is x.
18.(E)
Acceleration due to gravity acts on only the motion in the vertical direction. The acceleration remains a constant 10 meters per second squared downward. This action continually changes the vertical component of velocity. Horizontal velocity is unaffected by acceleration due to gravity and continues unabated in the horizontal direction.
19.(D)
At first glance, answer B may appear to be the obvious choice. However, the normal force is not always equal to an object’s weight. The y-component of force F must be included.
N = mg – F sin θ
f = μ(mg – F sin θ)
This is not one of the available answers. The key to this problem is the motion experienced by the object. In order for the velocity to be constant, the frictional force must be opposite in direction but equal in magnitude to the component of force acting in the direction of motion.
f = Fx = F cos θ
20.(E)
The component of force F pulling in the vertical direction can be described as F sin θ. The normal force, N, is reduced by the amount of force pulling in the vertical direction.
21.(A)
The masses are connected by a string and therefore act as if they were a single system with a mass of 8.0 kg. Mass m1 is being pulled in one direction by the force of gravity. Mass m2 is being pulled in the other direction by the force of gravity.
22.(C)
Rockets work in the vacuum of space as well as on Earth. Therefore, their motion cannot be the result of pushing against either Earth or air molecules. The expulsion of gas is the action force, and the reaction force is the motion of the rocket in the opposite direction.
23.(B)
The force acting on the object remains mg, its weight, throughout its flight. As the object rises, the force acts to slow it down. Upon reaching its highest point, the force acts to change the object’s direction and bring it back to the ground. However, the force acting on the object never changes.
24.(C)
In order for the mass to remain stationary, the vertical component of T2 (T2 sin θ) must be equal to w. Since T2 is the hypotenuse of a right triangle, it is greater than T2 sin θ. Therefore, T2 must be greater than w.
25.(B)
Determine the acceleration of the entire system.
The string is pulling the 1 kg mass. Sum the forces for this mass only.
26.(C)
Since the object returns to its original position, displacement is zero. Direction is continuously changing, and this means that velocity is changing. Therefore, answers I and III are true. Speed is a measure of distance divided by time. Even though displacement is zero, the distance traveled is not zero. This eliminates answer II. Acceleration changes direction during the motion but is never zero, eliminating answer IV.
27.(E)
Four forces are present: the force of gravity acting downward, the normal force acting perpendicular to the incline, tension pulling the object up the incline, and friction acting to oppose the motion.
28.(B)
Velocity is tangent to the path in the direction of motion. Centripetal acceleration is always directed toward the center of rotation and is perpendicular to the velocity.
29.(C)
The friction force creates the centripetal force needed to keep the object moving in a circle.
30.(E)
Only two forces are acting on the car as it is in the loop: the force of gravity, Fg, and the normal force, N. The force of gravity always acts downward. The normal force always acts perpendicularly to the plane of the surface upon which the mass rests. In this case, that would be downward. The velocity is directed in the −x-direction, but that is not a force. Instead, it is the result of the inertia of the car.
31.(A)
Apparent weight is a measure of the normal force acting on a mass. The centripetal force, Fc, is the sum of the two forces acting on the passenger in the car. Setting Fc equal to the sum of Fg and N yields:
In this case, the passengers would feel weightless at the instant the roller coaster reaches the top of the loop.
32.(D)
33.(D)
The work–kinetic energy theorem states that work is equal to a change in kinetic energy. Use the work found in the previous answer.
The object is initially at rest, vi = 0.
34.(E)
The work done by gravity is equal to the change in gravitational potential energy, ΔU, as the object moves through the vertical distance of 0.2 m above point Q.
35.(A)
This can be solved using conservation of energy. The 6 joules of gravitational potential energy at point P become kinetic energy, K, at point Q.
36.(C)
Both times, the object is lifted to the same height h at constant velocity. To overcome gravity and move the object at constant speed, the upward force must be equal to the force of gravity. Therefore, each time the needed force is the same regardless of the time during which the force acts. Moving the object the same vertical distance h requires the same amount of work against the force of gravity. However, in order to do this in half the amount of time, twice the power is needed.
37.(C)
Momentum is conserved before and after the collision.
38.(D)
Impulse, J, is equal to the force multiplied by the change in time. It is also equal to the mass multiplied by the change in velocity, which is known as the change in momentum. It is not equal to the momentum itself but, rather, the change in momentum.
39.(C)
Mass is not affected by changes in the acceleration of gravity. Even though the problem asks for the mass on the Moon, the weight on Earth can be used to determine the mass on Earth. This value is equal to the mass on the Moon.
40.(B)
Try each possibility in Newton’s law of gravitation.
Choice B results in the largest possible gravity.
41.(E)
According to Newton’s third law, the two charges pull on each other with the same force.
42.(E)
Coulomb’s law is very similar to Newton’s law of universal gravitation. When the –1 C charge is at point P, the radial distance will be of what it was originally. The result is consistent with the inverse square law.
43.(B)
Identify the charges as charge 1, q1, and charge 2, q2. The electric field points toward the negative charge, so the electric field of charge 1, E1, points toward charge 1. The electric field points away from the positive charge, so the electric field of charge 2, E2, points away from charge 2. These two vectors can be added using vector addition to find the total electric field due to both charges.
The resulting electric field points in the −x direction.
44.(D)
Since the spheres are positively charged, they will repel one another. As they begin to move farther away from one another, the force acting on them will decrease according to Coulomb’s law. A reduction in force leads to a reduction in acceleration. Velocity, however, will continue to increase because even though acceleration is decreasing, it continues to act in the direction of motion and continues to increase the speed of the charged spheres.
45.(D)
Protons and electrons have the same magnitude of charge, e, but opposite signs. As a result, the charged electric plates apply an equal electric force on the similarly charged proton and electron, but in opposite directions. The acceleration of each particle is dependent upon its mass. Electrons have a much smaller mass than protons. As a result, the same force applied to an electron will cause the electron to have a greater acceleration than the more massive proton.
46.(C)
This problem can be solved using conservation of energy. The electric potential energy is converted into kinetic energy.
47.(C)
Capacitance is proportional to the area of the plates divided by the distance between them. Doubling both area and distance will result in the capacitance remaining the same.
48.(B)
The resistors can be added in parallel to determine their total resistance.
Apply Ohm’s law, V = IR, to find the current.
49.(A)
Resistor 1 is wired in parallel, and the voltage drop across it will be equal to the voltage of the battery. Use Ohm’s law, V = IR, to solve for the current flowing through R1.
50.(E)
Power can be determined two ways. The current flowing through R3 can be found in the same manner as in the previous problem.
Then the power can be determined as follows:
Instead, it could have been determined directly using:
51.(B)
A bulb with less resistance will allow more current to flow. In households, the voltage is constant. So increasing the current will increase the power according to P = IV. Power is the rate of energy dissipation. Increasing the power increases a bulb’s brightness.
52.(E)
The right-hand rule states that the thumb of the right hand points in the direction of motion of a charged particle (the −x-direction in this case), the extended fingers point in the direction of the magnetic field (down the page in this case), and the palm of the hand points in the direction of the force (out of the page in this case). However, the particle is an electron. Electrons experience the complete opposite force. So the direction of force is into the page, –z-direction. As an alternative, you can use the left hand to determine the direction of magnetic force on negative charges.
53.(A)
In this case, the charged proton is moving parallel to the magnetic field. No magnetic force acts on the proton if it is moving completely parallel to the field.
54.(C)
The right-hand rule states that the thumb of the right hand points in the direction of either the motion of a charged particle or the current in a wire (the −z-direction in this case), the extended fingers point in the direction of the magnetic field (down the page in this case), and the palm of the hand points in the direction of the force (to the left of the page in this case, the –x-direction).
55.(D)
Reversing the magnet would reverse the direction of the current.
56.(B)
At position II, the total energy is in the form of kinetic energy, K, and potential energy is zero.
57.(E)
At position I, the constant force of gravity is greater than the variable force of the spring. At position III, the variable force of the spring is greater than the constant force of gravity. At position II, the constant force of gravity is equal to but in the opposite direction of the variable force of the spring. Therefore, there is no net force at position II.
58.(D)
The period of a spring depends on the mass and the spring constant.
The period of an oscillating spring-mass system is proportional to the square root of the suspended mass. Doubling mass m would result in increasing the period to T.
59.(B)
At the nodes, waves add destructively and there is zero amplitude.
60.(C)
Light intensity is directly proportional to the amplitude of a wave. Frequency and wavelength have no effect on the intensity of light.
61.(C)
Beat frequency is the difference between two interfering waveform frequencies.
fbeat = f1 − f2 = 12 Hz − 8.0 Hz = 4.0 Hz
62.(B)
As the index of refraction increases, the angle measured between the light ray and a normal line drawn perpendicular to the surface of the medium decreases. Medium 2 has the smallest angle and therefore the greatest index of refraction. Medium 1 has a slightly larger angle and a slightly smaller index of refraction. Medium 3 has the largest angle and the smallest index of refraction.
63.(A)
Interference is constructive when the path length difference is a whole number of wavelengths. The second bright maximum occurs at a path length difference of 2λ.
64.(E)
Each of these statements is true.
65.(E)
During this process, the ice must first melt at a constant temperature according to the formula:
Then the melted ice must rise in temperature by 50 K.
Keep in mind that even though the temperature is given in degrees Celsius, the difference in temperatures is required and the difference is the same for both the Celsius and Kelvin scales. The sum of melting the ice and then heating the liquid water is 100,000 J.
66.(C)
For this process, the 10 grams of water (0.01 kg) must be boiled at its boiling temperature of 100°C. The heat required is given by the following formula at that constant temperature:
67.(E)
Entropy always increases for an isolated system that is reaching equilibrium. This is the second law of thermodynamics.
68.(B)
The amount of heat, Q, transferred through an object is inversely proportional to the length, L, that that heat must transverse while moving through the object as shown in the equation below.
Doubling the distance that heat must travel cuts the amount of heat transferred in half.
69.(A)
Adding heat to a system of gas is positive and removing heat is negative. Doing work on a system is positive while work done by the system is negative. Use the first law of thermodynamics.
70.(A)
The current is directly proportional to the intensity of the light.
71.(E)
When the frequency of the light is increased, the energy of each photon is increased, E = hf. Photons with higher energies will emit electrons with higher kinetic energies, K = hf − ϕ. The ejected electrons arrive at the opposite plate of the photocell, creating a potential energy and a proportional potential difference between the plates of the photocell.
qV = K
Therefore, increasing the frequency increases both the kinetic energy of the ejected electrons and the resulting voltage of the photocell.
72.(E)
Fusion results when there is an increase in the atomic number. Hydrogen has an atomic number of 1, and helium has an atomic number of 2. The fusing together of hydrogen atoms produces helium.
73.(C)
Isotopes have the same elemental symbol but a different number of neutrons. Both of these forms of uranium have the same atomic number, 92, but their masses vary depending upon the number of neutrons. Uranium-238 has 146 neutrons, while uranium-235 has 143 neutrons.
74.(D)
Adding neutrons adds to only the strong force. Adding neutrons cannot add to the electrostatic force because neutrons have a neutral charge.
75.(E)
This is the definition of a transistor.