Time Remaining:
0REVIEW
0 of 50 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
Name  Marked  Completed  Skipped 
Information
* These questions are only for students studying for the Optometry Admission Test (OAT). If you are taking the DAT, this section is not applicable to you. There is no physics on the DAT.
Be sure to check out our Physics Cheat Sheet!
1. You will not be timed.
2. Each question will be immediately scored and display the solution after you click “Check”, allowing you to improve your skills as you go along.
Good luck!
You have reached the maximum number of attempts for this practice test.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
Practice Test Summary
0 of 50 questions answered correctly
Time elapsed:
Time has elapsed
Raw score: you answered 0 of 0 (0) questions correct
Breakdown by Topic (Questions Correct / Total Questions)
Categories
 Energy and Momentum 0%
Categories marked in red should receive additional attention during your review.
Solutions

Question 1 of 50Physics  Energy & Momentum
1. Question
1. Superman (100 kg) flying at 100 √2 m/s at a 45^{o} angle Northeast, collides with Lex Luthor, who is attempting to escape in his hovercraft (240 kg total) traveling 100 m/s directly East. Superman sticks to the hovercraft containing Lex.
What is the final velocity of the Superman and the hovercraft stuck together?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Object 1 (m_{1}) (Superman) = 100 kg
 Velocity Object 1 (v_{1}) = 100 √2 m/s NE @ 45^{o}
 Mass Object 2 (m_{2}) (Hovercraft + Lex) = 240 kg
 Velocity Object 2 (v_{2}) = 100 m/s East
Solving for: Final velocity of stuck together object (“Object 3”)
Step 2: Next, let’s work backwards. What do we need in order to solve for the final velocity of object 3?
Let recall our inelastic collision equation: (i.e. the objects become stuck together):
Note: in this problem, v_{1} (Superman) has nonzero x and y components. Thus, you should separate the inelastic equation into its x and y components:
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
m_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}When we plug in our variables, we can solve for v_{3x} and v_{3y}. We can then use Pythagorean Theorem to solve for the net velocity v_{3}.
Step 3: Next, let’s solve for v_{3x}
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
m_{1} = 100 kg
v_{1x} = (cos 45)(100 √2 m/s) = (^{1}/_{√2})(100 √2 m/s) = 100 m/sm_{2} = 240 kg
v_{2x} = 100 m/sm_{1}v_{1x} + m_{2}v_{2x} = 10,000 kg ᐧ m/s + 24,000 kg ᐧ m/s = 34,000 kg ᐧ m/s
m_{3}v_{3x} = 34,000 kg ᐧ m/s
m_{3} = 100 kg + 240 kg = 340 kg
v_{3x} = (34,000 kg ᐧ m/s) / 340 kg = 100 m/s
Step 4: Next, let’s solve for v_{3y}
m_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}
m_{1} = 100 kg
v_{1x} = (sin 45)(100 √2 m/s) = (^{1}/_{√2})(100 √2 m/s) = 100 m/sm_{2} = 240 kg
v_{2y} = 0 m/sm_{1}v_{1x} + m_{2}v_{2x} = 10,000 kg ᐧ m/s
m_{3}v_{3y} = 10,000 kg ᐧ m/s
m_{3} = m_{1} + m_{2} = 100 kg + 240 kg = 340 kg
v_{3y} = (10,000 kg ᐧ m/s) / 340 kg = approximately 30 m/s (Use 333 kg to approximate!)
Step 5: Last, let’s solve for net velocity (v_{3}) using Pythagorean Theorem
v_{3} ≈ √(100^{2} + 30^{2}) = √(10,000 + 900) = √10,900 m/s = answer choice B
Note: If you find it helpful, you can estimate that √10,900 = slightly higher than 100 (~104). Only one of the options is close to this answer.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Object 1 (m_{1}) (Superman) = 100 kg
 Velocity Object 1 (v_{1}) = 100 √2 m/s NE @ 45^{o}
 Mass Object 2 (m_{2}) (Hovercraft + Lex) = 240 kg
 Velocity Object 2 (v_{2}) = 100 m/s East
Solving for: Final velocity of stuck together object (“Object 3”)
Step 2: Next, let’s work backwards. What do we need in order to solve for the final velocity of object 3?
Let recall our inelastic collision equation: (i.e. the objects become stuck together):
Note: in this problem, v_{1} (Superman) has nonzero x and y components. Thus, you should separate the inelastic equation into its x and y components:
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
m_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}When we plug in our variables, we can solve for v_{3x} and v_{3y}. We can then use Pythagorean Theorem to solve for the net velocity v_{3}.
Step 3: Next, let’s solve for v_{3x}
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
m_{1} = 100 kg
v_{1x} = (cos 45)(100 √2 m/s) = (^{1}/_{√2})(100 √2 m/s) = 100 m/sm_{2} = 240 kg
v_{2x} = 100 m/sm_{1}v_{1x} + m_{2}v_{2x} = 10,000 kg ᐧ m/s + 24,000 kg ᐧ m/s = 34,000 kg ᐧ m/s
m_{3}v_{3x} = 34,000 kg ᐧ m/s
m_{3} = 100 kg + 240 kg = 340 kg
v_{3x} = (34,000 kg ᐧ m/s) / 340 kg = 100 m/s
Step 4: Next, let’s solve for v_{3y}
m_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}
m_{1} = 100 kg
v_{1x} = (sin 45)(100 √2 m/s) = (^{1}/_{√2})(100 √2 m/s) = 100 m/sm_{2} = 240 kg
v_{2y} = 0 m/sm_{1}v_{1x} + m_{2}v_{2x} = 10,000 kg ᐧ m/s
m_{3}v_{3y} = 10,000 kg ᐧ m/s
m_{3} = m_{1} + m_{2} = 100 kg + 240 kg = 340 kg
v_{3y} = (10,000 kg ᐧ m/s) / 340 kg = approximately 30 m/s (Use 333 kg to approximate!)
Step 5: Last, let’s solve for net velocity (v_{3}) using Pythagorean Theorem
v_{3} ≈ √(100^{2} + 30^{2}) = √(10,000 + 900) = √10,900 m/s = answer choice B
Note: If you find it helpful, you can estimate that √10,900 = slightly higher than 100 (~104). Only one of the options is close to this answer.

Question 2 of 50Physics  Energy & Momentum
2. Question
2. Which of the following are units of Work?Correct / You marked this questionYou can think of work as the amount of energy it takes to displace an object from its original position. Let’s look at each answer option individually:
[A] Joule
This is the correct answer. Joule is a unit of energy. Work can be measured in joules.
[B] Watt
Watt is a unit of power. Power measures the amount of work done per unit time. Watt = joule/second
[C] Newton
Newton is a unit of force. Forces can cause work to be done, but the work done is measured in joules.
[D] Kg * m/s^{2}
Kg * m/s^{2} is equivalent to Newtons, a unit of force. Forces can cause work to be done, but the work done is measured in joules.
[E] Amperes
Amperes are the unit for electric current. One ampere is equivalent to 1 coulomb/second.
Incorrect / You marked this questionYou can think of work as the amount of energy it takes to displace an object from its original position. Let’s look at each answer option individually:
[A] Joule
This is the correct answer. Joule is a unit of energy. Work can be measured in joules.
[B] Watt
Watt is a unit of power. Power measures the amount of work done per unit time. Watt = joule/second
[C] Newton
Newton is a unit of force. Forces can cause work to be done, but the work done is measured in joules.
[D] Kg * m/s^{2}
Kg * m/s^{2} is equivalent to Newtons, a unit of force. Forces can cause work to be done, but the work done is measured in joules.
[E] Amperes
Amperes are the unit for electric current. One ampere is equivalent to 1 coulomb/second.

Question 3 of 50Physics  Energy & Momentum
3. Question
3. In a manufacturing facility, a robot is used to raise boxes into a shipping crate. It takes 4 seconds for the robot to use 400 Newtons of force (applied vertically) to raise a box vertically 4 meters. How much work does it take to do the same task in half the time?Correct / You marked this questionStep 1: Let’s write down our work formula, to determine the ratio of work done in Scenario 1 (regular time) versus work done in Scenario 2 (half the time)
We are told that we hold everything else constant (F, d, and θ in this case). Thus, if the robot performed the same task in half the time, the amount of work done would be the same!
To reiterate, Work is NOT dependent on time!
Step 2: Solve for the amount of work done in either scenario
W = F*d*cos(θ)
W = (400 N)(4m)
W = 1600 JoulesNote: in this problem, you are told that the box is lifted vertically, and the force is applied vertically. As F and d are in the same direction, θ = 0, and cos(θ) = 1.
Incorrect / You marked this questionStep 1: Let’s write down our work formula, to determine the ratio of work done in Scenario 1 (regular time) versus work done in Scenario 2 (half the time)
We are told that we hold everything else constant (F, d, and θ in this case). Thus, if the robot performed the same task in half the time, the amount of work done would be the same!
To reiterate, Work is NOT dependent on time!
Step 2: Solve for the amount of work done in either scenario
W = F*d*cos(θ)
W = (400 N)(4m)
W = 1600 JoulesNote: in this problem, you are told that the box is lifted vertically, and the force is applied vertically. As F and d are in the same direction, θ = 0, and cos(θ) = 1.

Question 4 of 50Physics  Energy & Momentum
4. Question
4. A 15 kg child is pushed up a frictionless slide. The slide is 5 meters in length and sloped at an angle 30^{o} relative to the horizontal. The pushing force causes the child to accelerate up the slide at 1 m/s^{2}. How much work is required to push the child up the slide?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: First, let’s write down our variables, a quick sketch, and what we are solving for:
Mass Child = 15 kg
θ = 30^{o}
g = 10 m/s^{2}
Length of Slide = 5 m
a = 1 m/s^{2}Solving for: the work is required to push the child up the slide
Step 2: Next, let’s work backwards. What do we need to solve for the the work is required to push the child up the slide?
We are given d, and can assume that d and F are in the same direction (i.e. both parallel to the slope of the slide).
Thus, if we solve for F, we simply will need to plug our values into the work equation above.
Step 3: Solve for F
Be careful here! Remember, the net force on the child in the direction parallel to the slope will be the pushing force MINUS the force of gravity pulling the child down the slide.
First, let’s solve for the force of gravity pulling the child down the slide:
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to W_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (15 kg)(10 m/s^{2})(cos 60)
W_{x} = F_{x} = 75 NWe know that F_{net} causes and acceleration of 1 m/s^{2}. Thus:
F_{net} = 15 kg * 1 m/s^{2} = 15 N
F_{net} = pushing force MINUS the force of gravity pulling the child down the slide
15 N = pushing force – 75 N
Pushing Force = 90 NStep 4: Plug in our F, d, and θ values to solve for work
W = F*d*cos(θ)
W = 90 N * 5 m * 1
W = 450 JKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1: First, let’s write down our variables, a quick sketch, and what we are solving for:
Mass Child = 15 kg
θ = 30^{o}
g = 10 m/s^{2}
Length of Slide = 5 m
a = 1 m/s^{2}Solving for: the work is required to push the child up the slide
Step 2: Next, let’s work backwards. What do we need to solve for the the work is required to push the child up the slide?
We are given d, and can assume that d and F are in the same direction (i.e. both parallel to the slope of the slide).
Thus, if we solve for F, we simply will need to plug our values into the work equation above.
Step 3: Solve for F
Be careful here! Remember, the net force on the child in the direction parallel to the slope will be the pushing force MINUS the force of gravity pulling the child down the slide.
First, let’s solve for the force of gravity pulling the child down the slide:
Remember that weight is a force, and F_{weight} = mass * g (where g = gravitational acceleration = 10 m/s^{2}). F_{x} is equivalent to W_{x} in our sketch.
To find the x component of W, we will need to do some geometry:
We know that θ_{1} + β_{1} + 90^{o} = 180^{o}. Thus β_{1} = 60^{o}.
β_{1} = β_{2} = 60^{o}
θ_{1} = θ_{2} = 30^{o}If you look at our sketch, you can see that the parallel component of weight (i.e W_{x}) will equal:
W_{x} = (mass)(g)(cosβ_{2})
W_{x} = (15 kg)(10 m/s^{2})(cos 60)
W_{x} = F_{x} = 75 NWe know that F_{net} causes and acceleration of 1 m/s^{2}. Thus:
F_{net} = 15 kg * 1 m/s^{2} = 15 N
F_{net} = pushing force MINUS the force of gravity pulling the child down the slide
15 N = pushing force – 75 N
Pushing Force = 90 NStep 4: Plug in our F, d, and θ values to solve for work
W = F*d*cos(θ)
W = 90 N * 5 m * 1
W = 450 JKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 5 of 50Physics  Energy & Momentum
5. Question
5. John pulls his 4 kg wagon with 10 N of force at 30^{o} above the horizontal. How much work is done to move the wagon 10 meters?Correct / You marked this questionStep 1: First, let’s write down our variables, a quick sketch, and what we are solving for:
 Force (F) = 10 N
 θ = 30^{o}
 Mass = 4 kg
Solving for: the amount of work that is done to move the wagon 10 meters
Step 2: Next, let’s work backwards. What do we need in order to solve for the the amount of work that is done to move the wagon 10 meters?
We are given F, d, and θ. We just need to plug them into our work equation to solve.
Step 3: Plug in our F, d, and θ values to solve for work
W = F*d*cos(θ)
W = 10 N * 10 m * cos(30^{o})
W = 50 √3 JKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1: First, let’s write down our variables, a quick sketch, and what we are solving for:
 Force (F) = 10 N
 θ = 30^{o}
 Mass = 4 kg
Solving for: the amount of work that is done to move the wagon 10 meters
Step 2: Next, let’s work backwards. What do we need in order to solve for the the amount of work that is done to move the wagon 10 meters?
We are given F, d, and θ. We just need to plug them into our work equation to solve.
Step 3: Plug in our F, d, and θ values to solve for work
W = F*d*cos(θ)
W = 10 N * 10 m * cos(30^{o})
W = 50 √3 JKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 6 of 50Physics  Energy & Momentum
6. Question
6. What is the power required to lift 100 kg couch (starting from rest on the floor) 20 m directly upwards in 4 s?Correct / You marked this questionStep 1: Recall our power equation:
We are given t. Thus, to solve for power, we must solve for work.
Step 2: Recall our work equation:
We are given d, and we can deduce that since the couch is being lifted directly upwards, cos(θ) = 1. Thus, if we can solve for F, we can solve for work (and power).
Step 3: Solve for F
Here is where we get stuck. How can we solve for the force required to lift the couch?
If we are not told the force directly, we must be given enough information to deduce the acceleration of the couch. From there, we could use the F = m * a equation to solve for F. We know there MUST be a nonzero value for acceleration, because the couch’s velocity starts at zero.
We are not given enough information to solve for acceleration, meaning we can’t solve for F.
We will have to use a different formula to calculate work or energy.
Recall the change in mechanical energy is work. Mechanical energy being separated into kinetic energy the energy of motion, and potential energy the energy of condition or position.
The formula for potential energy will give us the energy needed to overcome or work needed to be applied to lift an object a distance away from the Earth.
Plugging in our values:
PE = (100 kg)(10 m/s^{2})(20 m)
PE = 20,000 J = WPlugging in our values for the power:
P = (20,000 J)/(4 s)
P = 5,000 WIncorrect / You marked this questionStep 1: Recall our power equation:
We are given t. Thus, to solve for power, we must solve for work.
Step 2: Recall our work equation:
We are given d, and we can deduce that since the couch is being lifted directly upwards, cos(θ) = 1. Thus, if we can solve for F, we can solve for work (and power).
Step 3: Solve for F
Here is where we get stuck. How can we solve for the force required to lift the couch?
If we are not told the force directly, we must be given enough information to deduce the acceleration of the couch. From there, we could use the F = m * a equation to solve for F. We know there MUST be a nonzero value for acceleration, because the couch’s velocity starts at zero.
We are not given enough information to solve for acceleration, meaning we can’t solve for F.
We will have to use a different formula to calculate work or energy.
Recall the change in mechanical energy is work. Mechanical energy being separated into kinetic energy the energy of motion, and potential energy the energy of condition or position.
The formula for potential energy will give us the energy needed to overcome or work needed to be applied to lift an object a distance away from the Earth.
Plugging in our values:
PE = (100 kg)(10 m/s^{2})(20 m)
PE = 20,000 J = WPlugging in our values for the power:
P = (20,000 J)/(4 s)
P = 5,000 W 
Question 7 of 50Physics  Energy & Momentum
7. Question
7. A 500 Newton force causes 1000 kg train to move downhill at a constant velocity of 10 m/s. What is the train’s momentum?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, and what were are solving for
 Pushing Force = 500 N
 Train Mass = 1000 kg
 Velocity = 10 m/s
Solving for: train’s momentum
Step 2: Let’s work backwards. What do we need to solve for the train’s momentum?
Recall our momentum equation:
We are given mass and velocity. The 500 N force is irrelevant for solving for momentum. Simply plug mass and velocity into our momentum equation.
Step 3: Plug mass and velocity in momentum equation
p = mv
p = (1000 kg)(10 m/s)
p = 10,000 kg ⋅ m/sNote: even if you totally forgot the momentum equation, look at your answer options. All have units of kg ⋅ m/s. You can make an educated guess that momentum = mass (kg) * velocity (m/s).
Incorrect / You marked this questionStep 1: Let’s write down our variables, and what were are solving for
 Pushing Force = 500 N
 Train Mass = 1000 kg
 Velocity = 10 m/s
Solving for: train’s momentum
Step 2: Let’s work backwards. What do we need to solve for the train’s momentum?
Recall our momentum equation:
We are given mass and velocity. The 500 N force is irrelevant for solving for momentum. Simply plug mass and velocity into our momentum equation.
Step 3: Plug mass and velocity in momentum equation
p = mv
p = (1000 kg)(10 m/s)
p = 10,000 kg ⋅ m/sNote: even if you totally forgot the momentum equation, look at your answer options. All have units of kg ⋅ m/s. You can make an educated guess that momentum = mass (kg) * velocity (m/s).

Question 8 of 50Physics  Energy & Momentum
8. Question
8. It requires 5000 J of work to move a 1000 kg train to downhill at a constant velocity of 10 m/s. What is the train’s kinetic energy?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, and what were are solving for
 Work = 5000 J
 Train Mass = 1000 kg
 Velocity = 10 m/s
Solving for: train’s kinetic energy
Step 2: Let’s work backwards. What do we need to solve for the train’s kinetic energy?
Recall our kinetic energy equation:
We are given mass and velocity. The 5000 J of work irrelevant for solving for kinetic energy. Simply plug mass and velocity into our kinetic energy equation.
Step 3: Plug mass and velocity in kinetic energy equation
KE = ^{1}/_{2}mv^{2}
KE = ^{1}/_{2}(1000 kg)(10 m/s)^{2}
KE = 50,000 JIncorrect / You marked this questionStep 1: Let’s write down our variables, and what were are solving for
 Work = 5000 J
 Train Mass = 1000 kg
 Velocity = 10 m/s
Solving for: train’s kinetic energy
Step 2: Let’s work backwards. What do we need to solve for the train’s kinetic energy?
Recall our kinetic energy equation:
We are given mass and velocity. The 5000 J of work irrelevant for solving for kinetic energy. Simply plug mass and velocity into our kinetic energy equation.
Step 3: Plug mass and velocity in kinetic energy equation
KE = ^{1}/_{2}mv^{2}
KE = ^{1}/_{2}(1000 kg)(10 m/s)^{2}
KE = 50,000 J 
Question 9 of 50Physics  Energy & Momentum
9. Question
9. A boy drops a rock off the edge of a cliff into the water below. The cliff is 100 m high. The rock has a mass of 10 kg.
After the rock has dropped 20 meters, how much potential energy has been converted to kinetic energy?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what were are solving for
Mass = 10 kg
Height Cliff = 100 meters
Gravitational Acceleration = 10 m/s^{2}Solving for: the amount of potential energy that has been converted into kinetic energy after 20 m drop
Step 2: Next, let’s work backwards. What do we need in order to solve for the amount of potential energy that has been converted into kinetic energy after 20 m drop?
There are two ways you could go about solving this problem. Let’s define the height at the top of the cliff H_{1} and the height after dropping 20 meters H_{2}.
We know that: Total Mechanical Energy (E_{mechanical}) = KE + PE
As such, you could either:
A) Solve for the potential energy at H_{1} and H_{2}, and determine the differential
B) Solve for the kinetic energy at H_{1} and H_{2}, and determine the differentialOption A requires less calculation and is more efficient.
Step 3: Solve the potential energy at H_{1} and H_{2}, and determine the differential
Recall our potential energy equation:
At H_{1}:
PE = 10 kg * 10 m/s^{2} * 100 m
PE = 10,000 JAt H_{2}:
PE = 10 kg * 10 m/s^{2} * 80 m
PE = 8,000 JPE differential = 10,000 J – 8,000 J = 2,000 J
Note: Unless told otherwise, you can safely assume that 100% of the potential energy differential has been converted to kinetic energy.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what were are solving for
Mass = 10 kg
Height Cliff = 100 meters
Gravitational Acceleration = 10 m/s^{2}Solving for: the amount of potential energy that has been converted into kinetic energy after 20 m drop
Step 2: Next, let’s work backwards. What do we need in order to solve for the amount of potential energy that has been converted into kinetic energy after 20 m drop?
There are two ways you could go about solving this problem. Let’s define the height at the top of the cliff H_{1} and the height after dropping 20 meters H_{2}.
We know that: Total Mechanical Energy (E_{mechanical}) = KE + PE
As such, you could either:
A) Solve for the potential energy at H_{1} and H_{2}, and determine the differential
B) Solve for the kinetic energy at H_{1} and H_{2}, and determine the differentialOption A requires less calculation and is more efficient.
Step 3: Solve the potential energy at H_{1} and H_{2}, and determine the differential
Recall our potential energy equation:
At H_{1}:
PE = 10 kg * 10 m/s^{2} * 100 m
PE = 10,000 JAt H_{2}:
PE = 10 kg * 10 m/s^{2} * 80 m
PE = 8,000 JPE differential = 10,000 J – 8,000 J = 2,000 J
Note: Unless told otherwise, you can safely assume that 100% of the potential energy differential has been converted to kinetic energy.

Question 10 of 50Physics  Energy & Momentum
10. Question
10. A 5 meter tall man applies a 100 Newton force at 30^{o} relative to the horizontal to lift a 10 kg box from the ground to 2 meters above his head. How much work is required to maintain the box at this height above his head?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables and what we are solving for.
 Height Man = 5 m
 Total Distance Lifted (d) = 7 meters
 Lifting Force = 100 N
 Gravitational Acceleration = 10 m/s^{2}
Solving for: the work required to maintain the box above his head
Step 2: Next, let’s work backwards. What do we need in order to solve for the work required to maintain the box above his head?
Recall our work equation:
If we read the question carefully, it asks for the amount of work required to maintain the box at this height above his head.
To keep any object in the same location (i.e NO displacement) requires NO work (i.e work = 0 J), as d = 0.
Strategy Tip: It is fairly common for the OAT to ask questions like this, that really make you pay attention to what the problem is truly asking for. Although you are limited on time, a careful read of each question is a good time investment.
Incorrect / You marked this questionStep 1: Let’s write down our variables and what we are solving for.
 Height Man = 5 m
 Total Distance Lifted (d) = 7 meters
 Lifting Force = 100 N
 Gravitational Acceleration = 10 m/s^{2}
Solving for: the work required to maintain the box above his head
Step 2: Next, let’s work backwards. What do we need in order to solve for the work required to maintain the box above his head?
Recall our work equation:
If we read the question carefully, it asks for the amount of work required to maintain the box at this height above his head.
To keep any object in the same location (i.e NO displacement) requires NO work (i.e work = 0 J), as d = 0.
Strategy Tip: It is fairly common for the OAT to ask questions like this, that really make you pay attention to what the problem is truly asking for. Although you are limited on time, a careful read of each question is a good time investment.

Question 11 of 50Physics  Energy & Momentum
11. Question
11. How much work is required to accelerate a 400 kg car starting from rest at 10 m/s^{2} for 2 seconds? The coefficient of kinetic friction and static friction for the ground are both 0.5
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Car = 400 kg
 v_{i} = 0 m/s
 Acceleration (a )= 10 m/s^{2}
 Time (t) = 5 seconds
Solving for: the amount of work required to accelerate the car
Step 2: Next, let’s work backwards. What do we need in order to solve for the work required to accelerate the car in this scenario?
Recall our work equation:
We can assume that F and d are in the same direction, implying that cos(θ) = 1.
Thus, we must solve for both F and d in order to be able to solve for work.
Step 3: Solve for F
F_{TOTAL} = force required to overcome friction force (F_{f} ) (BOTH static and kinetic) + force required to accelerate the car (F_{ACCELERATION}) which will be = m*a
The force required to overcome the friction force will be:
F_{f} = µ_{s}F_{N} + µ_{k}F_{N}
F_{f} = 0.5(m*g) + 0.5(m*g)
F_{f} = 0.5(400 kg * 10 m/s^{2}) + 0.5(400 kg * 10 m/s^{2})
F_{f} = 4000 NYou can think of it this way: if 4000 N of force is applied, you can solve the rest of the problem as if it was a frictionless surface.
F_{ACCELERATION} = m * a
F_{ACCELERATION} = 400 kg * 10 m/s^{2}
F_{ACCELERATION} = 4000 NF_{TOTAL} = F_{f} + F_{ACCELERATION} = 8000 N
Step 4: Solve for d
Recall our displacement equation:
d = v_{i}t + ^{1}/_{2}at^{2}
d = 0 + ^{1}/_{2}(10 m/s^{2})(4 seconds)
d = 20 mStep 5: Solve for work
W = F*d*cos(θ)
W = (8,000 N)(20 m)
W = 160,000 JStrategy Tip: A fairly safe rule of thumb is that if the problem does not mention anything about friction, you can disregard friction. If it specifically calls it out (e.g. in this problem, providing you with the coefficients of static and kinetic friction) you should take friction into account.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Car = 400 kg
 v_{i} = 0 m/s
 Acceleration (a )= 10 m/s^{2}
 Time (t) = 5 seconds
Solving for: the amount of work required to accelerate the car
Step 2: Next, let’s work backwards. What do we need in order to solve for the work required to accelerate the car in this scenario?
Recall our work equation:
We can assume that F and d are in the same direction, implying that cos(θ) = 1.
Thus, we must solve for both F and d in order to be able to solve for work.
Step 3: Solve for F
F_{TOTAL} = force required to overcome friction force (F_{f} ) (BOTH static and kinetic) + force required to accelerate the car (F_{ACCELERATION}) which will be = m*a
The force required to overcome the friction force will be:
F_{f} = µ_{s}F_{N} + µ_{k}F_{N}
F_{f} = 0.5(m*g) + 0.5(m*g)
F_{f} = 0.5(400 kg * 10 m/s^{2}) + 0.5(400 kg * 10 m/s^{2})
F_{f} = 4000 NYou can think of it this way: if 4000 N of force is applied, you can solve the rest of the problem as if it was a frictionless surface.
F_{ACCELERATION} = m * a
F_{ACCELERATION} = 400 kg * 10 m/s^{2}
F_{ACCELERATION} = 4000 NF_{TOTAL} = F_{f} + F_{ACCELERATION} = 8000 N
Step 4: Solve for d
Recall our displacement equation:
d = v_{i}t + ^{1}/_{2}at^{2}
d = 0 + ^{1}/_{2}(10 m/s^{2})(4 seconds)
d = 20 mStep 5: Solve for work
W = F*d*cos(θ)
W = (8,000 N)(20 m)
W = 160,000 JStrategy Tip: A fairly safe rule of thumb is that if the problem does not mention anything about friction, you can disregard friction. If it specifically calls it out (e.g. in this problem, providing you with the coefficients of static and kinetic friction) you should take friction into account.

Question 12 of 50Physics  Energy & Momentum
12. Question
12. Two trains are traveling on a track. Train 1 (300 kg moving at 100 m/s) collides with Train 2 (1200 kg moving at 10 m/s). The trains stick together. What is the velocity of the combined trains after the collision?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Train 1 = 300 kg
 v_{1} = 100 m/s
 Mass Train 2 = 1200 kg
 v_{2} = 10 m/s
Solving for: the velocity of the combined trains after the collision
Step 2: Next, let’s work backwards. What do we need to determine the velocity of the combined trains after the collision?
This is an excellent situation to recall our inelastic collisions equation:
Remember that inelastic collisions are ones where the objects “stick together.” We are given m_{1}, v_{1}, m_{2} and v_{2}. Thus, we need to solve for m_{3}, plug the rest of our values into our inelastic collision equation, then solve for v_{3}.
Step 3: Solve for m_{3}, plug the rest of our values into our inelastic collision equation, then solve for v_{3}.
m_{3} = m_{1} + m_{2}
m_{3} = 300 kg + 1200 kg
m_{3} = 1500 kgm_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(300 kg)(100 m/s) + (1200 kg)(10 m/s) = (1500 kg)v_{3}
v_{3} = 28 m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Train 1 = 300 kg
 v_{1} = 100 m/s
 Mass Train 2 = 1200 kg
 v_{2} = 10 m/s
Solving for: the velocity of the combined trains after the collision
Step 2: Next, let’s work backwards. What do we need to determine the velocity of the combined trains after the collision?
This is an excellent situation to recall our inelastic collisions equation:
Remember that inelastic collisions are ones where the objects “stick together.” We are given m_{1}, v_{1}, m_{2} and v_{2}. Thus, we need to solve for m_{3}, plug the rest of our values into our inelastic collision equation, then solve for v_{3}.
Step 3: Solve for m_{3}, plug the rest of our values into our inelastic collision equation, then solve for v_{3}.
m_{3} = m_{1} + m_{2}
m_{3} = 300 kg + 1200 kg
m_{3} = 1500 kgm_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(300 kg)(100 m/s) + (1200 kg)(10 m/s) = (1500 kg)v_{3}
v_{3} = 28 m/s 
Question 13 of 50Physics  Energy & Momentum
13. Question
13. How much work is required to accelerate a 200 kg car starting from 10 m/s to 20 m/s? Neglect friction and assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables what we are solving for.
 Mass Car = 200 kg
 v_{i} = 10 m/s
 v_{f} = 20 m/s
Solving for: the amount of work required to accelerate the car
Step 2: Next, let’s work backwards. What do we need in order to solve for the work required to accelerate the car from
10 m/s to 20 m/s?We have two potential work equations we can use here. As you are not given information about the force applied or displacement, we can deduce that we should use the equation that relates work to kinetic energy:
Step 3: Solve for the change in KE to solve for the amount of work required
W = ΔKE
W = KE_{FINAL} – KE_{INITIAL}
W = ^{1}/_{2}m(v_{FINAL})^{2} – ^{1}/_{2}m(v_{INTIAL})^{2}
W = ^{1}/_{2}(200 kg)(20 m/s)^{2} – ^{1}/_{2}(200 kg)(10 m/s)^{2}
W = 30,000 JIncorrect / You marked this questionStep 1: Let’s write down our variables what we are solving for.
 Mass Car = 200 kg
 v_{i} = 10 m/s
 v_{f} = 20 m/s
Solving for: the amount of work required to accelerate the car
Step 2: Next, let’s work backwards. What do we need in order to solve for the work required to accelerate the car from
10 m/s to 20 m/s?We have two potential work equations we can use here. As you are not given information about the force applied or displacement, we can deduce that we should use the equation that relates work to kinetic energy:
Step 3: Solve for the change in KE to solve for the amount of work required
W = ΔKE
W = KE_{FINAL} – KE_{INITIAL}
W = ^{1}/_{2}m(v_{FINAL})^{2} – ^{1}/_{2}m(v_{INTIAL})^{2}
W = ^{1}/_{2}(200 kg)(20 m/s)^{2} – ^{1}/_{2}(200 kg)(10 m/s)^{2}
W = 30,000 J 
Question 14 of 50Physics  Energy & Momentum
14. Question
14. Which of the following divers has the most total mechanical energy? Neglect air resistance.
Diver A: a 100 kg man standing on a diving board 25 m tall
Diver B: a 120 kg man who jumped off of a diving board 15 m tall, just before he hits the water
Diver C: a 110 kg man who jumped off of a diving board that is 23 m tall, and is in free fall 10 meters from hitting the water.
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down a quick sketch:
Step 2: Let’s work backwards. How do we solve for which person has the greatest total mechanical energy?
This is a good example of when taking a moment to pause before diving into the calculations (no pun intended) will help you. Let’s recall our total mechanical energy equation:
Diver A has only potential energy. Diver B has converted all of his potential energy to kinetic energy. Diver C has converted some of his potential energy into kinetic energy. But rather than solve for kinetic energy (time consuming!) we can simply solve for the potential energy each has when standing on the diving board. This will equal each diver’s total mechanical energy, which doesn’t change, even as PE is converted to KE.
Step 3: Calculate each diver’s PE (and thus Total Mechanical Energy) when standing on diving board:
Diver A
PE = mgh
PE = (100 kg)(10 m/s^{2})(25 m)
PE = 25,000 JDiver B
PE = mgh
PE = (120 kg)(10 m/s^{2})(15 m)
PE = 18,000 JDiver C
PE = mgh
PE = (110 kg)(10 m/s^{2})(23 m)
PE = 25,300 JThus, Diver C has the most total mechanical energy.
Incorrect / You marked this questionStep 1: Let’s write down a quick sketch:
Step 2: Let’s work backwards. How do we solve for which person has the greatest total mechanical energy?
This is a good example of when taking a moment to pause before diving into the calculations (no pun intended) will help you. Let’s recall our total mechanical energy equation:
Diver A has only potential energy. Diver B has converted all of his potential energy to kinetic energy. Diver C has converted some of his potential energy into kinetic energy. But rather than solve for kinetic energy (time consuming!) we can simply solve for the potential energy each has when standing on the diving board. This will equal each diver’s total mechanical energy, which doesn’t change, even as PE is converted to KE.
Step 3: Calculate each diver’s PE (and thus Total Mechanical Energy) when standing on diving board:
Diver A
PE = mgh
PE = (100 kg)(10 m/s^{2})(25 m)
PE = 25,000 JDiver B
PE = mgh
PE = (120 kg)(10 m/s^{2})(15 m)
PE = 18,000 JDiver C
PE = mgh
PE = (110 kg)(10 m/s^{2})(23 m)
PE = 25,300 JThus, Diver C has the most total mechanical energy.

Question 15 of 50Physics  Energy & Momentum
15. Question
15. Two balls collide on a frictionless pool table and bounce off of a each other. The white ball (1 kg) traveling at 10 m/s hits the black ball (1 kg) at rest which shoots forward at 8 m/s.
What is the magnitude of the velocity with which the white ball rebounds off of the black ball?Correct / You marked this questionStep 1: Let’s recall our elastic collision equation.
Remember, elastic collisions are ones in which the objects “bounce off” of each other.
Step 2: Let’s plug our values into the elastic collision equation:
m_{1}v_{1 INITIAL} + m_{2}v_{2 INITIAL} = m_{1}v_{1 FINAL} + m_{2}v_{2 FINAL}
(1 kg)(10 m/s) + (1 kg)(0 m/s) = (1 kg)(v_{1 FINAL}) + (1 kg)(8 m/s)
10 kg*m/s = 8 kg*m/s + (1 kg)(v_{1 FINAL})
v_{1 FINAL} = 2 m/sIncorrect / You marked this questionStep 1: Let’s recall our elastic collision equation.
Remember, elastic collisions are ones in which the objects “bounce off” of each other.
Step 2: Let’s plug our values into the elastic collision equation:
m_{1}v_{1 INITIAL} + m_{2}v_{2 INITIAL} = m_{1}v_{1 FINAL} + m_{2}v_{2 FINAL}
(1 kg)(10 m/s) + (1 kg)(0 m/s) = (1 kg)(v_{1 FINAL}) + (1 kg)(8 m/s)
10 kg*m/s = 8 kg*m/s + (1 kg)(v_{1 FINAL})
v_{1 FINAL} = 2 m/s 
Question 16 of 50Physics  Energy & Momentum
16. Question
16. A car has a momentum of 120 kg ⋅ m/s. What would the car’s momentum be if it had ^{1}/_{2} the mass and 3x the velocity?Correct / You marked this questionStep 1: Let’s recall our momentum equation:
Step 2: Plug in the altered mass and velocity values
Original:
mv = 120 kg ⋅ m/sAltered:
(^{1}/_{2}m)(3v) = ^{3}/_{2} * 120 kg ⋅ m/s = 180 kg ⋅ m/sIncorrect / You marked this questionStep 1: Let’s recall our momentum equation:
Step 2: Plug in the altered mass and velocity values
Original:
mv = 120 kg ⋅ m/sAltered:
(^{1}/_{2}m)(3v) = ^{3}/_{2} * 120 kg ⋅ m/s = 180 kg ⋅ m/s 
Question 17 of 50Physics  Energy & Momentum
17. Question
17. A car has a momentum of 70 kg ⋅ m/s. What would the car’s momentum be if it had 2x the mass and ^{1}/_{2} the velocity?Correct / You marked this questionStep 1: Let’s recall our momentum equation:
Step 2: Plug in the altered mass and velocity values
Original:
mv = 70 kg ⋅ m/sAltered:
(2m)(^{1}/_{2}v) = ^{2}/_{2} * 70 kg ⋅ m/s = 70 kg ⋅ m/sIncorrect / You marked this questionStep 1: Let’s recall our momentum equation:
Step 2: Plug in the altered mass and velocity values
Original:
mv = 70 kg ⋅ m/sAltered:
(2m)(^{1}/_{2}v) = ^{2}/_{2} * 70 kg ⋅ m/s = 70 kg ⋅ m/s 
Question 18 of 50Physics  Energy & Momentum
18. Question
18. Mike, a 64 kg player for the Flyers hockey team is skating down the ice towards the goal at 4.4 m/s. A hockey player from the opposing team checks (impacts) Mike. Mike’s 8 kg helmet is knocked off. Helmetless, Mike is knocked backwards away from the goal at 1.8 m/s.
What is the magnitude of Mike’s momentum change due to the hit?Correct / You marked this questionStep 1: Let’s recall our momentum equation
The momentum change will equal Momentum Final (after the check) MINUS Momentum Initial (before the check)
Step 2: Solve for Final and Initial Momentum
Final Momentum
p = mv
p = (64 – 8 kg)(1.8 m/s) = – 100.8 kg ⋅ m/s AWAY FROM GOAL ()Initial Momentum
p = mv
p = (64 kg)(4.4 m/s) = 281.6 kg ⋅ m/s TOWARDS GOAL (+)Step 3: Solve for the magnitude of change in momentum
 – 100.8 kg ⋅ m/s MINUS 281.6 ⋅ m/s  = 382.4 kg ⋅ m/s
Incorrect / You marked this questionStep 1: Let’s recall our momentum equation
The momentum change will equal Momentum Final (after the check) MINUS Momentum Initial (before the check)
Step 2: Solve for Final and Initial Momentum
Final Momentum
p = mv
p = (64 – 8 kg)(1.8 m/s) = – 100.8 kg ⋅ m/s AWAY FROM GOAL ()Initial Momentum
p = mv
p = (64 kg)(4.4 m/s) = 281.6 kg ⋅ m/s TOWARDS GOAL (+)Step 3: Solve for the magnitude of change in momentum
 – 100.8 kg ⋅ m/s MINUS 281.6 ⋅ m/s  = 382.4 kg ⋅ m/s

Question 19 of 50Physics  Energy & Momentum
19. Question
19. A car crashes into a brick wall with 1,000,000 Newtons of force. The car rapidly decelerates at 500 m/s^{2} to a complete stop in 0.05 seconds.
What was the car’s initial momentum before it hit the wall?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Force Crash = 1,000,000 N
 Acceleration = – 500 m/s^{2}
 Time (t) = 0.05 seconds
Solving for: the car’s initial momentum before it hit the wall
Step 2: Let’s work backwards. What do we need in order to solve for the car’s initial momentum before it hit the wall?
Let’s recall our momentum equation:
We are given neither the mass nor the velocity of the car before the crash. Thus, we must solve for both.
Step 3: Solve for the car’s mass.
We are given the force with which the car hits the wall, along with the acceleration (i.e deceleration) associated with this force. Thus, if we use our handydandy F = m*a equation, we can solve for the car’s mass:
F = m*a
1,000,000 N = m * 500 m/s^{2}
m = 2,000 kgNote: the negative acceleration (deceleration) of course won’t yield a negative mass value. It only dictates the direction of the acceleration.
Step 4: Solve for the car’s initial velocity
We are given the car’s deceleration from initial velocity to zero, and the time in which this takes place. Let’s use our average acceleration equation equation to solve for initial velocity:
a_{avg} = (v_{f} – v_{i}) / t
– 500 m/s^{2} = (0 – v_{i}) / 0.05 s
v_{i} = 500 m/s^{2} * 0.05 s
v_{i} = 25 m/sStep 5: Solve for the car’s initial momentum
p = mv
p = 2000 kg * 25 m/s
p = 50,000 kg ⋅ m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Force Crash = 1,000,000 N
 Acceleration = – 500 m/s^{2}
 Time (t) = 0.05 seconds
Solving for: the car’s initial momentum before it hit the wall
Step 2: Let’s work backwards. What do we need in order to solve for the car’s initial momentum before it hit the wall?
Let’s recall our momentum equation:
We are given neither the mass nor the velocity of the car before the crash. Thus, we must solve for both.
Step 3: Solve for the car’s mass.
We are given the force with which the car hits the wall, along with the acceleration (i.e deceleration) associated with this force. Thus, if we use our handydandy F = m*a equation, we can solve for the car’s mass:
F = m*a
1,000,000 N = m * 500 m/s^{2}
m = 2,000 kgNote: the negative acceleration (deceleration) of course won’t yield a negative mass value. It only dictates the direction of the acceleration.
Step 4: Solve for the car’s initial velocity
We are given the car’s deceleration from initial velocity to zero, and the time in which this takes place. Let’s use our average acceleration equation equation to solve for initial velocity:
a_{avg} = (v_{f} – v_{i}) / t
– 500 m/s^{2} = (0 – v_{i}) / 0.05 s
v_{i} = 500 m/s^{2} * 0.05 s
v_{i} = 25 m/sStep 5: Solve for the car’s initial momentum
p = mv
p = 2000 kg * 25 m/s
p = 50,000 kg ⋅ m/s 
Question 20 of 50Physics  Energy & Momentum
20. Question
20. A cyclist crashes directly into a pole with 3000 Newtons of force. The bike rapidly decelerates at 75 m/s^{2} to a complete stop in 0.2 seconds.
What was the cyclist’s initial momentum before hitting the pole?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Force Crash = 3000 N
 Acceleration = – 75 m/s^{2}
 Time (t) = 0.2 seconds
Solving for: the cyclist’s initial momentum before hitting the pole
Step 2: Let’s work backwards. What do we need in order to solve for the cyclist’s initial momentum before hitting the pole?
Let’s recall our momentum equation:
We are given neither the mass nor the velocity of the cyclist before the crash. Thus, we must solve for both.
Step 3: Solve for the cyclist’s mass.
We are given the force with which the cyclist hits the pole, along with the acceleration (i.e deceleration) associated with this force. Thus, if we use our handydandy F = m*a equation, we can solve for the cyclist’s mass:
F = m*a
3,000 N = m(75 m/s^{2})
m = 40 kgNote: the negative acceleration (deceleration) of course won’t yield a negative mass value. It only dictates the direction of the acceleration.
Step 4: Solve for the cyclist’s initial velocity
We are given the cyclist’s deceleration from initial velocity to zero, and the time in which this takes place. Let’s use our average acceleration equation equation to solve for initial velocity:
a_{avg} = (v_{f} – v_{i}) / t
75 m/s^{2} = (0 – v_{i}) / 0.2 s
v_{i} = 75 m/s^{2} * 0.2 s
v_{i} = 15 m/sStep 5: Solve for the cyclist’s initial momentum
p = mv
p = 40 kg * 15 m/s
p = 600 kg ⋅ m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Force Crash = 3000 N
 Acceleration = – 75 m/s^{2}
 Time (t) = 0.2 seconds
Solving for: the cyclist’s initial momentum before hitting the pole
Step 2: Let’s work backwards. What do we need in order to solve for the cyclist’s initial momentum before hitting the pole?
Let’s recall our momentum equation:
We are given neither the mass nor the velocity of the cyclist before the crash. Thus, we must solve for both.
Step 3: Solve for the cyclist’s mass.
We are given the force with which the cyclist hits the pole, along with the acceleration (i.e deceleration) associated with this force. Thus, if we use our handydandy F = m*a equation, we can solve for the cyclist’s mass:
F = m*a
3,000 N = m(75 m/s^{2})
m = 40 kgNote: the negative acceleration (deceleration) of course won’t yield a negative mass value. It only dictates the direction of the acceleration.
Step 4: Solve for the cyclist’s initial velocity
We are given the cyclist’s deceleration from initial velocity to zero, and the time in which this takes place. Let’s use our average acceleration equation equation to solve for initial velocity:
a_{avg} = (v_{f} – v_{i}) / t
75 m/s^{2} = (0 – v_{i}) / 0.2 s
v_{i} = 75 m/s^{2} * 0.2 s
v_{i} = 15 m/sStep 5: Solve for the cyclist’s initial momentum
p = mv
p = 40 kg * 15 m/s
p = 600 kg ⋅ m/s 
Question 21 of 50Physics  Energy & Momentum
21. Question
21. An engineering team is designing a new airbag. The airbag will need to bring the driver’s motion from full velocity to a complete stop. The airbag must be able to function for a driver of 200 kg, exert a maximum force of 20,000 N, and cause the stop in at most 0.5 seconds.
What is the maximum driver velocity for which the airbag would be able to meet the requirements above?Correct / You marked this questionStep 1: Let’s write down our variables, and what we are solving for:
 Maximum Force = 20,000 N
 Driver Mass = 200 kg
 Time (t) = 0.5 seconds
Solving for: the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem
Step 2: Let’s work backwards. What do we need in order to solve for the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem?
We are given the maximum force that should be exerted from the airbag onto the driver. We know by Newton’s 3^{rd} Law of Motion that the magnitude of this force would be equal to the force of the driver onto the airbag.
We are also given the specific driver mass. We can solve for the maximum acceleration (i.e. deceleration) that can be experienced by the driver relative to a the maximum force. We simply use our F = m*a equation.
From there, we can deduce maximum velocity from maximum acceleration.
Step 3: Solve for maximum acceleration
We know that the driver is actually decelerating, so whatever value we get for acceleration should be negative.
F = m*a
20,000 N = (200 kg) (a)
a = – 100 m/s^{2}Sanity Check: Remember, we are solving for MAXIMUM acceleration.
Refer to the above F = m*a equation: The driver’s mass is fixed. So if the Force were any smaller, the acceleration would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of acceleration.
Step 4: Solve for maximum velocity
a_{avg} = (v_{f} – v_{i}) / t
– 100 m/s^{2} = (0 m/s – v_{INITIAL}) / (0.5 seconds)
v_{INITIAL} = 50 m/sSanity Check: Remember, we are solving for MAXIMUM initial velocity. We proved above that 100 m/s^{2} is the maximum magnitude of acceleration.
Refer to the above equation for a_{avg}: If the acceleration were smaller in magnitude, the initial velocity would also be smaller in magnitude. If the time were less than 0.5 seconds, the initial velocity would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of initial velocity.
Incorrect / You marked this questionStep 1: Let’s write down our variables, and what we are solving for:
 Maximum Force = 20,000 N
 Driver Mass = 200 kg
 Time (t) = 0.5 seconds
Solving for: the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem
Step 2: Let’s work backwards. What do we need in order to solve for the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem?
We are given the maximum force that should be exerted from the airbag onto the driver. We know by Newton’s 3^{rd} Law of Motion that the magnitude of this force would be equal to the force of the driver onto the airbag.
We are also given the specific driver mass. We can solve for the maximum acceleration (i.e. deceleration) that can be experienced by the driver relative to a the maximum force. We simply use our F = m*a equation.
From there, we can deduce maximum velocity from maximum acceleration.
Step 3: Solve for maximum acceleration
We know that the driver is actually decelerating, so whatever value we get for acceleration should be negative.
F = m*a
20,000 N = (200 kg) (a)
a = – 100 m/s^{2}Sanity Check: Remember, we are solving for MAXIMUM acceleration.
Refer to the above F = m*a equation: The driver’s mass is fixed. So if the Force were any smaller, the acceleration would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of acceleration.
Step 4: Solve for maximum velocity
a_{avg} = (v_{f} – v_{i}) / t
– 100 m/s^{2} = (0 m/s – v_{INITIAL}) / (0.5 seconds)
v_{INITIAL} = 50 m/sSanity Check: Remember, we are solving for MAXIMUM initial velocity. We proved above that 100 m/s^{2} is the maximum magnitude of acceleration.
Refer to the above equation for a_{avg}: If the acceleration were smaller in magnitude, the initial velocity would also be smaller in magnitude. If the time were less than 0.5 seconds, the initial velocity would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of initial velocity.

Question 22 of 50Physics  Energy & Momentum
22. Question
22. An car manufacturing team is designing a new airbag. The airbag will need to bring the driver’s motion from full velocity to a complete stop. The airbag must be able to function for a driver of 150 kg, exert a maximum force of 30,000 N, and cause the stop in at most 0.2 seconds.
What is the maximum driver velocity in which the airbag would be able to meet the requirements aboveCorrect / You marked this questionStep 1: Let’s write down our variables, and what we are solving for:
 Maximum Force = 30,000 N
 Driver Mass = 150 kg
 Time (t) = 0.2 seconds
Solving for: the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem
Step 2: Let’s work backwards. What do we need in order to solve for the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem?
We are given the maximum force that should be exerted from the airbag onto the driver. We know by Newton’s 3^{rd} Law of Motion that the magnitude of this force would be equal to the force of the driver onto the airbag.
We are also given the specific driver mass. We can solve for the maximum acceleration (i.e. deceleration) that can be experienced by the driver relative to a the maximum force. We simply use our F = m*a equation.
From there, we can deduce maximum velocity from maximum acceleration.
Step 3: Solve for maximum acceleration
We know that the driver is actually decelerating, so whatever value we get for acceleration should be negative.
F = m*a
30,000 N = (150 kg)(a)
a = – 200 m/s^{2}Sanity Check: Remember, we are solving for MAXIMUM acceleration.
Refer to the above F = m*a equation: The driver’s mass is fixed. So if the Force were any smaller, the acceleration would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of acceleration.
Step 4: Solve for maximum velocity
a_{avg} = (v_{f} – v_{i}) / t
– 200 m/s^{2} = (0 m/s – v_{INITIAL}) / (0.2 seconds)
v_{INITIAL} = 40 m/sSanity Check: Remember, we are solving for MAXIMUM initial velocity. We proved above that 200 m/s^{2} is the maximum magnitude of acceleration.
Refer to the above equation for a_{avg}: If the acceleration were smaller in magnitude, the initial velocity would also be smaller in magnitude. If the time were less than 0.2 seconds, the initial velocity would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of initial velocity.
Incorrect / You marked this questionStep 1: Let’s write down our variables, and what we are solving for:
 Maximum Force = 30,000 N
 Driver Mass = 150 kg
 Time (t) = 0.2 seconds
Solving for: the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem
Step 2: Let’s work backwards. What do we need in order to solve for the maximum driver velocity in which the airbag would be able to meet the requirements listed in the problem?
We are given the maximum force that should be exerted from the airbag onto the driver. We know by Newton’s 3^{rd} Law of Motion that the magnitude of this force would be equal to the force of the driver onto the airbag.
We are also given the specific driver mass. We can solve for the maximum acceleration (i.e. deceleration) that can be experienced by the driver relative to a the maximum force. We simply use our F = m*a equation.
From there, we can deduce maximum velocity from maximum acceleration.
Step 3: Solve for maximum acceleration
We know that the driver is actually decelerating, so whatever value we get for acceleration should be negative.
F = m*a
30,000 N = (150 kg)(a)
a = – 200 m/s^{2}Sanity Check: Remember, we are solving for MAXIMUM acceleration.
Refer to the above F = m*a equation: The driver’s mass is fixed. So if the Force were any smaller, the acceleration would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of acceleration.
Step 4: Solve for maximum velocity
a_{avg} = (v_{f} – v_{i}) / t
– 200 m/s^{2} = (0 m/s – v_{INITIAL}) / (0.2 seconds)
v_{INITIAL} = 40 m/sSanity Check: Remember, we are solving for MAXIMUM initial velocity. We proved above that 200 m/s^{2} is the maximum magnitude of acceleration.
Refer to the above equation for a_{avg}: If the acceleration were smaller in magnitude, the initial velocity would also be smaller in magnitude. If the time were less than 0.2 seconds, the initial velocity would also be smaller in magnitude. Thus, we have solved for the maximum magnitude of initial velocity.

Question 23 of 50Physics  Energy & Momentum
23. Question
23. Two balls collide on a frictionless pool table and bounce off of a each other. The green ball (1 kg) is moving at 8 m/s towards the far end of the table. The white ball (1 kg) is moving at 18 m/s in the same direction as the green ball. The white ball hits the green ball and bounces backwards at 2 m/s.
What is the magnitude of the velocity with which the green ball moves forward after being hit by the white ball?Correct / You marked this questionStep 1: Let’s recall our elastic collision equation.
Remember, elastic collisions are ones in which the objects “bounce off” of each other.
Step 2: Let’s plug our values into the elastic collision equation:
m_{1}v_{1 INITIAL} + m_{2}v_{2 INITIAL} = m_{1}v_{1 FINAL} + m_{2}v_{2 FINAL}
(1 kg)(8 m/s) + (1 kg)(18 m/s) = (1 kg)(v_{1 FINAL}) + (1 kg)(2 m/s)
26 kg*m/s = 2 kg*m/s + (1 kg)(v_{1 FINAL})
v_{1 FINAL} = 28 m/sIncorrect / You marked this questionStep 1: Let’s recall our elastic collision equation.
Remember, elastic collisions are ones in which the objects “bounce off” of each other.
Step 2: Let’s plug our values into the elastic collision equation:
m_{1}v_{1 INITIAL} + m_{2}v_{2 INITIAL} = m_{1}v_{1 FINAL} + m_{2}v_{2 FINAL}
(1 kg)(8 m/s) + (1 kg)(18 m/s) = (1 kg)(v_{1 FINAL}) + (1 kg)(2 m/s)
26 kg*m/s = 2 kg*m/s + (1 kg)(v_{1 FINAL})
v_{1 FINAL} = 28 m/s 
Question 24 of 50Physics  Energy & Momentum
24. Question
24. A students spits a 0.01 kg spitball at 25 m/s at a scrunched up 0.04 kg wad of paper on at rest on his desk. The spitball and wad of paper stick together and slide across his desk. The coefficient of kinetic friction between the desk and the wad of paper is 0.5. Ignore static friction.
How far does the wad of paper (with spitball attached) slide across the student’s desk?
Neglect air resistance and assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Spitball = 0.01 kg
 Velocity Spitball = 25 m/s
 Mass Wad of Paper = 0.04 kg
 Velocity Wad of Paper = 0 m/s
 Coefficient of Kinetic Friction (μ_{k}) = 0.5
Solving for: how far the wad of paper (with spitball attached) slides across the student’s desk
Step 2: Let’s work backwards. How to we solve for how far the wad of paper (with spitball attached) slides across the student’s desk?
We can deduce that this situation is an inelastic collision (where the objects stick together). Thus, the combined wad of paper + spitball will have an initial velocity that we can solve for using our inelastic collision equation.
We also know that the force of friction will cause the wad of paper + spitball to decelerate as it moves across the table. We can use our F = m*a to solve for this deceleration.
Finally, having solved for initial velocity and acceleration, and knowing that the wad of paper + spitball eventually comes to a rest, we can use our Final Velocity equation to determine the distance traveled.
Step 3: Use our inelastic collision equation, solve for the velocity of the combined object (wad of paper + spitball)
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.01 kg)(25 m/s) + (0.04 kg)(0 m/s) = (0.01 + 0.04 kg)v_{3}
0.25 kg m/s = 0.05 kg (v_{3})
v_{3} = 5 m/sStep 4: Solve for the frictional force in the opposite direction of velocity
Let’s recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m *g)
F_{f} = (0.5)(0.05 kg *10 m/s^{2})
F_{f} = 0.25 NStep 5: Use F = m*a to solve for the deceleration of the combined object due to friction
F_{f} = m * a
0.25 N = 0.05 kg * a
a = 5 m/s^{2}Referencing our sketch: This implies that the wad of paper + spitball will decelerate at 5 m/s^{2} (i.e a = 5 m/s^{2}).
Step 6: Use our equation for Final Velocity to solve for distance traveled
v_{f}^{2} = v_{i}^{2} + 2ad
0 = (5 m/s)^{2} + 2 (5 m/s^{2}) d
0 = (25 m^{2}/s^{2}) + 2 ( 5 m/s^{2}) d
– 25 m^{2}/s^{2} = 10 m/s^{2} * (d)
d = 2.5 mNote: the initial velocity (v_{i}) in this equation is referring to the initial velocity of the combined mass (v_{3}), NOT of the spitball!
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Spitball = 0.01 kg
 Velocity Spitball = 25 m/s
 Mass Wad of Paper = 0.04 kg
 Velocity Wad of Paper = 0 m/s
 Coefficient of Kinetic Friction (μ_{k}) = 0.5
Solving for: how far the wad of paper (with spitball attached) slides across the student’s desk
Step 2: Let’s work backwards. How to we solve for how far the wad of paper (with spitball attached) slides across the student’s desk?
We can deduce that this situation is an inelastic collision (where the objects stick together). Thus, the combined wad of paper + spitball will have an initial velocity that we can solve for using our inelastic collision equation.
We also know that the force of friction will cause the wad of paper + spitball to decelerate as it moves across the table. We can use our F = m*a to solve for this deceleration.
Finally, having solved for initial velocity and acceleration, and knowing that the wad of paper + spitball eventually comes to a rest, we can use our Final Velocity equation to determine the distance traveled.
Step 3: Use our inelastic collision equation, solve for the velocity of the combined object (wad of paper + spitball)
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.01 kg)(25 m/s) + (0.04 kg)(0 m/s) = (0.01 + 0.04 kg)v_{3}
0.25 kg m/s = 0.05 kg (v_{3})
v_{3} = 5 m/sStep 4: Solve for the frictional force in the opposite direction of velocity
Let’s recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m *g)
F_{f} = (0.5)(0.05 kg *10 m/s^{2})
F_{f} = 0.25 NStep 5: Use F = m*a to solve for the deceleration of the combined object due to friction
F_{f} = m * a
0.25 N = 0.05 kg * a
a = 5 m/s^{2}Referencing our sketch: This implies that the wad of paper + spitball will decelerate at 5 m/s^{2} (i.e a = 5 m/s^{2}).
Step 6: Use our equation for Final Velocity to solve for distance traveled
v_{f}^{2} = v_{i}^{2} + 2ad
0 = (5 m/s)^{2} + 2 (5 m/s^{2}) d
0 = (25 m^{2}/s^{2}) + 2 ( 5 m/s^{2}) d
– 25 m^{2}/s^{2} = 10 m/s^{2} * (d)
d = 2.5 mNote: the initial velocity (v_{i}) in this equation is referring to the initial velocity of the combined mass (v_{3}), NOT of the spitball!

Question 25 of 50Physics  Energy & Momentum
25. Question
25. A teenager throws a 0.2 kg dart at 30 m/s at a 0.4 kg tissue box on at rest on a table. The dart sticks into the tissue box, and the combined object slides across his table. The coefficient of kinetic friction between the table and the tissue box is 0.5. Static friction is negligible.
How far does the tissue box (with dart inside) slide across the table?
Neglect air resistance and assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Dart = 0.2 kg
 Velocity Dart = 30 m/s
 Mass Tissue Box = 0.4kg
 Velocity Tissue Box = 0 m/s
 Coefficient of Kinetic Friction (μ_{k}) = 0.5
Solving for: how far the tissue box (with dart inside) slides across the table
Step 2: Let’s work backwards. How do we solve for how far the tissue box (with dart inside) slides across the table?
We can deduce that this situation is an inelastic collision (where the objects stick together). Thus, the combined tissue box + dart will have an initial velocity that we can solve for using our inelastic collision equation.
We also know that the force of friction will cause the tissue box + dart to decelerate as it moves across the table. We can use our
F = m*a equation to solve for this deceleration.Finally, having solved for initial velocity and acceleration, and knowing that the tissue box + dart eventually comes to a rest, we can use our Final Velocity equation to determine the distance traveled.
Step 3: Use our inelastic collision equation, solve for the velocity of the combined object (tissue box + dart)
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.2 kg)(30 m/s) + (0.4 kg)(0 m/s) = (0.2 + 0.4 kg) (v_{3})
6 kg*m/s = 0.6 kg (v_{3})
v_{3} = 10 m/sStep 4: Solve for the frictional force in the opposite direction of velocity
Let’s recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m*g)
F_{f} = (0.5)(0.6 kg * 10 m/s^{2})
F_{f} = 3 NStep 5: Use F = m*a to solve for the deceleration of the combined object due to friction
F_{f} = m * a
3 N = 0.6 kg * a
a = 5 m/s^{2}Referencing our sketch: This implies that the tissue box + dart will decelerate at 5 m/s^{2} (i.e a = 5 m/s^{2}).
Step 6: Use our equation for Final Velocity to solve for distance traveled
v_{f}^{2} = v_{i}^{2} + 2ad
0 = (10 m/s)^{2} + 2 (5 m/s^{2}) d
0 = (100 m^{2}/s^{2}) + 2(5 m/s^{2}) d
– 100 m^{2}/s^{2} = – 10 m/s^{2} (d)
d = 10 mNote: the initial velocity (v_{i}) in this equation is referring to the initial velocity of the combined mass (v_{3}), NOT of the dart!
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Dart = 0.2 kg
 Velocity Dart = 30 m/s
 Mass Tissue Box = 0.4kg
 Velocity Tissue Box = 0 m/s
 Coefficient of Kinetic Friction (μ_{k}) = 0.5
Solving for: how far the tissue box (with dart inside) slides across the table
Step 2: Let’s work backwards. How do we solve for how far the tissue box (with dart inside) slides across the table?
We can deduce that this situation is an inelastic collision (where the objects stick together). Thus, the combined tissue box + dart will have an initial velocity that we can solve for using our inelastic collision equation.
We also know that the force of friction will cause the tissue box + dart to decelerate as it moves across the table. We can use our
F = m*a equation to solve for this deceleration.Finally, having solved for initial velocity and acceleration, and knowing that the tissue box + dart eventually comes to a rest, we can use our Final Velocity equation to determine the distance traveled.
Step 3: Use our inelastic collision equation, solve for the velocity of the combined object (tissue box + dart)
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.2 kg)(30 m/s) + (0.4 kg)(0 m/s) = (0.2 + 0.4 kg) (v_{3})
6 kg*m/s = 0.6 kg (v_{3})
v_{3} = 10 m/sStep 4: Solve for the frictional force in the opposite direction of velocity
Let’s recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m*g)
F_{f} = (0.5)(0.6 kg * 10 m/s^{2})
F_{f} = 3 NStep 5: Use F = m*a to solve for the deceleration of the combined object due to friction
F_{f} = m * a
3 N = 0.6 kg * a
a = 5 m/s^{2}Referencing our sketch: This implies that the tissue box + dart will decelerate at 5 m/s^{2} (i.e a = 5 m/s^{2}).
Step 6: Use our equation for Final Velocity to solve for distance traveled
v_{f}^{2} = v_{i}^{2} + 2ad
0 = (10 m/s)^{2} + 2 (5 m/s^{2}) d
0 = (100 m^{2}/s^{2}) + 2(5 m/s^{2}) d
– 100 m^{2}/s^{2} = – 10 m/s^{2} (d)
d = 10 mNote: the initial velocity (v_{i}) in this equation is referring to the initial velocity of the combined mass (v_{3}), NOT of the dart!

Question 26 of 50Physics  Energy & Momentum
26. Question
26. A teenager throws a 0.4 kg dart at a 0.8 kg tissue box on at rest on a table. The dart sticks into the tissue box, and the combined object slides 10 m across his table. The coefficient of kinetic friction between the table and the tissue box is 0.5. Static friction is negligible.
What is the initial velocity of the dart?
Neglect air resistance and assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
Mass Dart = 0.4 kg
Mass Tissue Box = 0.8 kg
Velocity Tissue Box = 0 m/s
Distance Traveled = 10 m
Coefficient of Kinetic Friction (μ_{k}) = 0.4Solving for: the initial velocity of the dart
Step 2: Let’s work backwards. How to we solve for the initial velocity of the dart?
We can deduce that this situation is an inelastic collision (where the objects stick together). Thus, the combined tissue box + dart will have an initial velocity that we can solve for using our inelastic collision equation.
However, you will see that we have TWO unknown variables when we plug in our given variables.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.4 kg)(v_{1}) + (0.8 kg)(0 m/s) = (0.4 + 0.8 kg)(v_{3})This is a tricky situation. However, we can use our “final velocity” equation to deduce the initial velocity of the combined dart / tissue box. This corresponds to (v_{3}) in our inelastic collision equation above. We will then circle back to this step.
Step 3: Use our “final velocity” equation to deduce the initial velocity of the combined dart / tissue box
Okay, pause for a second. This can get a bit confusing, so remember exactly what you’re solving for.
In our final velocity equation above, v_{f} corresponds to the final velocity of the dart / tissue box when it has come to a stop. This is, of course, 0 m/s.
In our final velocity equation above, v_{i} corresponds to the initial velocity of the dart / tissue box. This corresponds to (v_{3}) in our inelastic collision equation above.
v_{f}^{2} = v_{i}^{2} + 2ad
(0 m/s)^{2} = v_{i}^{2} + 2a(10 m)Oh boy, we now have gotten to ANOTHER spot with two unknown variables. However, we know that the force of friction is the only force that causes the dart / tissue box to come to a stop. We can use our F = m*a equation to solve for a. We will then circle back to this step.
Step 4: Solve for the frictional force in the opposite direction of velocity
Let’s recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m*g)
F_{f} = (0.5)(1.2 kg * 10 m/s^{2})
F_{f} = 6 NStep 5: Use F = m*a to solve for the deceleration of the combined object due to friction
F_{f} = m * a
6 N = 1.2 kg * a
a = 5 m/s^{2}Referencing our sketch: This implies that the tissue box + dart will decelerate at 5 m/s^{2} (i.e a = – 5 m/s^{2}).
Step 6: Circle back to our final velocity equation to solve for v_{i} (i.e. v_{3})
v_{f}^{2} = v_{i}^{2} + 2ad
(0 m/s)^{2} = v_{i}^{2} + 2a(10m)
(0 m/s)^{2} = v_{i}^{2} + 2*(5 m/s^{2})(10m)
v_{i} = v_{3} = 10 m/sStep 7: Circle back to our inelastic collision equation to solve for initial velocity of the dart
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.4 kg)(v_{1}) + (0.8kg)(0 m/s) = (0.4 + 0.8 kg)(10 m/s)
(0.4 kg)(v_{1}) = (12 kg m/s)
v_{1} = 30 m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
Mass Dart = 0.4 kg
Mass Tissue Box = 0.8 kg
Velocity Tissue Box = 0 m/s
Distance Traveled = 10 m
Coefficient of Kinetic Friction (μ_{k}) = 0.4Solving for: the initial velocity of the dart
Step 2: Let’s work backwards. How to we solve for the initial velocity of the dart?
We can deduce that this situation is an inelastic collision (where the objects stick together). Thus, the combined tissue box + dart will have an initial velocity that we can solve for using our inelastic collision equation.
However, you will see that we have TWO unknown variables when we plug in our given variables.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.4 kg)(v_{1}) + (0.8 kg)(0 m/s) = (0.4 + 0.8 kg)(v_{3})This is a tricky situation. However, we can use our “final velocity” equation to deduce the initial velocity of the combined dart / tissue box. This corresponds to (v_{3}) in our inelastic collision equation above. We will then circle back to this step.
Step 3: Use our “final velocity” equation to deduce the initial velocity of the combined dart / tissue box
Okay, pause for a second. This can get a bit confusing, so remember exactly what you’re solving for.
In our final velocity equation above, v_{f} corresponds to the final velocity of the dart / tissue box when it has come to a stop. This is, of course, 0 m/s.
In our final velocity equation above, v_{i} corresponds to the initial velocity of the dart / tissue box. This corresponds to (v_{3}) in our inelastic collision equation above.
v_{f}^{2} = v_{i}^{2} + 2ad
(0 m/s)^{2} = v_{i}^{2} + 2a(10 m)Oh boy, we now have gotten to ANOTHER spot with two unknown variables. However, we know that the force of friction is the only force that causes the dart / tissue box to come to a stop. We can use our F = m*a equation to solve for a. We will then circle back to this step.
Step 4: Solve for the frictional force in the opposite direction of velocity
Let’s recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m*g)
F_{f} = (0.5)(1.2 kg * 10 m/s^{2})
F_{f} = 6 NStep 5: Use F = m*a to solve for the deceleration of the combined object due to friction
F_{f} = m * a
6 N = 1.2 kg * a
a = 5 m/s^{2}Referencing our sketch: This implies that the tissue box + dart will decelerate at 5 m/s^{2} (i.e a = – 5 m/s^{2}).
Step 6: Circle back to our final velocity equation to solve for v_{i} (i.e. v_{3})
v_{f}^{2} = v_{i}^{2} + 2ad
(0 m/s)^{2} = v_{i}^{2} + 2a(10m)
(0 m/s)^{2} = v_{i}^{2} + 2*(5 m/s^{2})(10m)
v_{i} = v_{3} = 10 m/sStep 7: Circle back to our inelastic collision equation to solve for initial velocity of the dart
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(0.4 kg)(v_{1}) + (0.8kg)(0 m/s) = (0.4 + 0.8 kg)(10 m/s)
(0.4 kg)(v_{1}) = (12 kg m/s)
v_{1} = 30 m/s 
Question 27 of 50Physics  Energy & Momentum
27. Question
27. Two cars crash into each other head on and stick together. Car 1 is 175 kg and moving at 16 m/s. Car 2 is 25 kg and moving at 32 m/s towards Car 1.
What is the final velocity of the combined cars?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 175 kg
 Velocity Car 1 = 16 m/s
 Mass Car 2 = 25 kg
 Velocity Car 2 = 32 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We are given the necessary values to solve for v_{3}.
Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for v_{3}
Important: remember, the cars crash head on, which implies that they are moving in OPPOSITE DIRECTIONS! As such, we need to set one direction as positive (+) and one direction as negative ().
Let’s set to the right in our diagram as positive, and to the left as negative.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(175 kg)(16 m/s) + (25 kg)(32 m/s) = (200 kg)(v_{3})
2800 kg*m/s – 800 kg*m/s = (200 kg)(v_{3})
v_{3} = 10 m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 175 kg
 Velocity Car 1 = 16 m/s
 Mass Car 2 = 25 kg
 Velocity Car 2 = 32 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We are given the necessary values to solve for v_{3}.
Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for v_{3}
Important: remember, the cars crash head on, which implies that they are moving in OPPOSITE DIRECTIONS! As such, we need to set one direction as positive (+) and one direction as negative ().
Let’s set to the right in our diagram as positive, and to the left as negative.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(175 kg)(16 m/s) + (25 kg)(32 m/s) = (200 kg)(v_{3})
2800 kg*m/s – 800 kg*m/s = (200 kg)(v_{3})
v_{3} = 10 m/s 
Question 28 of 50Physics  Energy & Momentum
28. Question
28. Two cars crash into each other head on and stick together. Car 1 is 80 kg and moving at 25 m/s. Car 2 is 20 kg and moving at 10 m/s towards Car 1.
What is the final velocity of the combined cars?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 80 kg
 Velocity Car 1 = 25 m/s
 Mass Car 2 = 20 kg
 Velocity Car 2 = 10 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We are given the necessary values to solve for v_{3}.
Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for v_{3}
Important: remember, the cars crash head on, which implies that they are moving in OPPOSITE DIRECTIONS! As such, we need to set one direction as positive (+) and one direction as negative ().
Let’s set to the right in our diagram as positive, and to the left as negative.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(80kg)(25 m/s) + (20 kg)(10 m/s) = (100 kg)(v_{3})
2000 kg*m/s – 200 kg*m/s = (100 kg)(v_{3})
v_{3} = 18 m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 80 kg
 Velocity Car 1 = 25 m/s
 Mass Car 2 = 20 kg
 Velocity Car 2 = 10 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We are given the necessary values to solve for v_{3}.
Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for v_{3}
Important: remember, the cars crash head on, which implies that they are moving in OPPOSITE DIRECTIONS! As such, we need to set one direction as positive (+) and one direction as negative ().
Let’s set to the right in our diagram as positive, and to the left as negative.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(80kg)(25 m/s) + (20 kg)(10 m/s) = (100 kg)(v_{3})
2000 kg*m/s – 200 kg*m/s = (100 kg)(v_{3})
v_{3} = 18 m/s 
Question 29 of 50Physics  Energy & Momentum
29. Question
29. Two cars crash into each other head on and stick together. Car 1 is 40 kg and moving at 50 m/s. Car 2 is 10 kg and moving at 15 m/s.
What is the final velocity of the combined cars?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 40 kg
 Velocity Car 1 = 50 m/s
 Mass Car 2 = 10 kg
 Velocity Car 2 = 15 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We are given the necessary values to solve for v_{3}.
Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for v_{3}
Important: remember, the cars crash head on, which implies that they are moving in OPPOSITE DIRECTIONS! As such, we need to set one direction as positive (+) and one direction as negative ().
Let’s set to the right in our diagram as positive, and to the left as negative.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(40kg)(50 m/s) + (10 kg)(15 m/s) = (50 kg)(v_{3})
2000 kg*m/s – 150 kg*m/s = (50 kg)(v_{3})
v_{3} = 37 m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 40 kg
 Velocity Car 1 = 50 m/s
 Mass Car 2 = 10 kg
 Velocity Car 2 = 15 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We are given the necessary values to solve for v_{3}.
Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for v_{3}
Important: remember, the cars crash head on, which implies that they are moving in OPPOSITE DIRECTIONS! As such, we need to set one direction as positive (+) and one direction as negative ().
Let’s set to the right in our diagram as positive, and to the left as negative.
m_{1}v_{1} + m_{2}v_{2} = m_{3}v_{3}
(40kg)(50 m/s) + (10 kg)(15 m/s) = (50 kg)(v_{3})
2000 kg*m/s – 150 kg*m/s = (50 kg)(v_{3})
v_{3} = 37 m/s 
Question 30 of 50Physics  Energy & Momentum
30. Question
30. Two cars crash into each other stick together. Car 1 is 10 kg and moving at 10 m/s, 30^{o} North of East. Car 2 is 20 kg and moving at 2.5√3 m/s due West.
What is are the x and y components of the final velocity (v_{3}) of the combined cars?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 10 kg
 Velocity Car 1 = 10 m/s
 Mass Car 2 = 20 kg
 Velocity Car 2 = 2.5 √3 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We know that this equation holds true if we parse it out into x and y components. That is:
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
m_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for the x and y components of v_{3}
Let’s define to the right in our diagram as positive, and to the left as negative. Let’s start with solving for the x component of v_{3}:
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
(10 kg)(10 m/s)(cos30) + (20 kg)(2.5√3 m/s) = (30 kg)(v_{3x})
50√3 kg*m/s – 50√3 kg*m/s = (30 kg)(v_{3x})
v_{3x} = 0 m/sm_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}
(10kg)(10 m/s)(sin30) + 0 kg = (30 kg)(v_{3y})
50 kg*m/s = (30 kg)(v_{3y})
v_{3y} = ^{5}/_{3} m/sKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car 1 = 10 kg
 Velocity Car 1 = 10 m/s
 Mass Car 2 = 20 kg
 Velocity Car 2 = 2.5 √3 m/s
Solving for: the final velocity of the cars after they have smashed together (v_{3})
Step 2: Let’s work backwards. How do we solve for the final velocity of the cars after they have smashed together?
As the cars “stick together” after the crash, we know this is an inelastic collision. Let’s whip out our inelastic collision equation:
We know that this equation holds true if we parse it out into x and y components. That is:
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
m_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}Step 3: Plug in the mass and velocity values into our inelastic collision equation to solve for the x and y components of v_{3}
Let’s define to the right in our diagram as positive, and to the left as negative. Let’s start with solving for the x component of v_{3}:
m_{1}v_{1x} + m_{2}v_{2x} = m_{3}v_{3x}
(10 kg)(10 m/s)(cos30) + (20 kg)(2.5√3 m/s) = (30 kg)(v_{3x})
50√3 kg*m/s – 50√3 kg*m/s = (30 kg)(v_{3x})
v_{3x} = 0 m/sm_{1}v_{1y} + m_{2}v_{2y} = m_{3}v_{3y}
(10kg)(10 m/s)(sin30) + 0 kg = (30 kg)(v_{3y})
50 kg*m/s = (30 kg)(v_{3y})
v_{3y} = ^{5}/_{3} m/sKnowing the sin and cos values of 0, 30, 45, 60, and 90 is very important throughout the OAT. Commit them to memory.

Question 31 of 50Physics  Energy & Momentum
31. Question
31. A car’s engine uses 500 Newtons of force to move at a net constant velocity of 25 m/s down the road. The car is 1200 kg. How much engine power is required to move at this velocity?Correct / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
 Force = 500 N
 Velocity = 25 m/s
 Mass Car = 1200 kg
Solving for: power required to achieve velocity of 25 m/s
Step 2: Let’s work backwards. How do we solve for the power required to achieve this velocity?
Hmm. Looks like we’ll need to solve for work. Let’s recall our work equation:
We can assume that the Force is in the same direction as displacement in this case, so cos(θ) = 1. We are given the Force (F) but not d. We are, however, given velocity. Let’s recall our velocity equation:
Interesting! What if we rewrote our Power equation as:
P = W/t = (Fd)/t = F * velocity
Step 3: Plug in our given variables to our “new” power equation:
P = Fv
P = (500 N)(25 m/s)
P = 12,500 WNote: you may have thought that no work was being done, because the car was traveling at constant velocity. However, we are explicitly told that 500 N of force is used by the engine to achieve this velocity. And we are specifically asked how much engine power is required to achieve this velocity.
Your thinking would have been correct if the question asked how much net power is required to for this velocity. In this case, acceleration would have been zero, and thus the net force on the car would also have been zero. This would imply that work and thus power would also be zero.
Incorrect / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
 Force = 500 N
 Velocity = 25 m/s
 Mass Car = 1200 kg
Solving for: power required to achieve velocity of 25 m/s
Step 2: Let’s work backwards. How do we solve for the power required to achieve this velocity?
Hmm. Looks like we’ll need to solve for work. Let’s recall our work equation:
We can assume that the Force is in the same direction as displacement in this case, so cos(θ) = 1. We are given the Force (F) but not d. We are, however, given velocity. Let’s recall our velocity equation:
Interesting! What if we rewrote our Power equation as:
P = W/t = (Fd)/t = F * velocity
Step 3: Plug in our given variables to our “new” power equation:
P = Fv
P = (500 N)(25 m/s)
P = 12,500 WNote: you may have thought that no work was being done, because the car was traveling at constant velocity. However, we are explicitly told that 500 N of force is used by the engine to achieve this velocity. And we are specifically asked how much engine power is required to achieve this velocity.
Your thinking would have been correct if the question asked how much net power is required to for this velocity. In this case, acceleration would have been zero, and thus the net force on the car would also have been zero. This would imply that work and thus power would also be zero.

Question 32 of 50Physics  Energy & Momentum
32. Question
32. A car’s engine uses 1200 Newtons and 60,000 W of power to maintain a certain velocity. The car is 1000 kg. What is this velocity?Correct / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
 Force = 1200 N
 Mass Car = 1000 kg
 Power = 60,000 W
Solving for: velocity attained by the car
Step 2: Let’s work backwards. How do we solve for velocity attained by the car?
We will need to manipulate our power equation such that it includes velocity. Let’s recall our power equation:
Let’s recall our work equation:
We can assume that the force is in the same direction as displacement in this case, so cos(θ) = 1. Let’s recall our velocity equation:
Interesting! What if we rewrote our Power equation as:
P = W/t = (Fd)/t = F * velocity
Step 3: Plug in our given variables to our “new” power equation to solve for velocity
P = Fv
60,000 W = (1200 N) * v
v = 50 m/sNote: you may have thought that no work was being done, because the car was traveling at constant velocity. However, we are explicitly told that 1200 N of force is used by the engine to achieve this velocity. And we are specifically asked how much engine power is required to achieve this velocity.
Your thinking would have been correct if the question asked how much net power is required to for this velocity. In this case, acceleration would have been zero, and thus the net force on the car would also have been zero. This would imply that work and thus power would also be zero.
Incorrect / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
 Force = 1200 N
 Mass Car = 1000 kg
 Power = 60,000 W
Solving for: velocity attained by the car
Step 2: Let’s work backwards. How do we solve for velocity attained by the car?
We will need to manipulate our power equation such that it includes velocity. Let’s recall our power equation:
Let’s recall our work equation:
We can assume that the force is in the same direction as displacement in this case, so cos(θ) = 1. Let’s recall our velocity equation:
Interesting! What if we rewrote our Power equation as:
P = W/t = (Fd)/t = F * velocity
Step 3: Plug in our given variables to our “new” power equation to solve for velocity
P = Fv
60,000 W = (1200 N) * v
v = 50 m/sNote: you may have thought that no work was being done, because the car was traveling at constant velocity. However, we are explicitly told that 1200 N of force is used by the engine to achieve this velocity. And we are specifically asked how much engine power is required to achieve this velocity.
Your thinking would have been correct if the question asked how much net power is required to for this velocity. In this case, acceleration would have been zero, and thus the net force on the car would also have been zero. This would imply that work and thus power would also be zero.

Question 33 of 50Physics  Energy & Momentum
33. Question
33. A 1000 kg train is moving with 10,000 kg m/s of constant momentum at a constant velocity of 50 m/s on a track. How much power is required to maintain this velocity?Correct / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
 Momentum = 10,000 kg m/s
 Mass Train = 1000 kg
 Velocity = 50 m/s
Solving for: the amount of power required to maintain velocity of 50 m/s
Step 2: Let’s work backwards. How do we solve for the amount of power required?
Let’s recall our power equation.
Let’s recall our work equation:
Let’s recall our force equation:
We know that the train is moving at a constant velocity, meaning acceleration equals zero. How much force is required to cause zero acceleration? Zero, of course!
How much work is done if the force equals zero? You got it, zero.
And how much power is required to do zero work? Bingo, 0 W
Incorrect / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
 Momentum = 10,000 kg m/s
 Mass Train = 1000 kg
 Velocity = 50 m/s
Solving for: the amount of power required to maintain velocity of 50 m/s
Step 2: Let’s work backwards. How do we solve for the amount of power required?
Let’s recall our power equation.
Let’s recall our work equation:
Let’s recall our force equation:
We know that the train is moving at a constant velocity, meaning acceleration equals zero. How much force is required to cause zero acceleration? Zero, of course!
How much work is done if the force equals zero? You got it, zero.
And how much power is required to do zero work? Bingo, 0 W

Question 34 of 50Physics  Energy & Momentum
34. Question
34. There are two paths to push a box to the top of a hill. On Path A, you push the box up 10 m at 30^{o} where the coefficient of kinetic friction is 0.4. On Path B, you push the box up a 10 m hill at 30^{o} where the coefficient of kinetic friction is 0.9.
Which path requires more work, against friction, to be done in order to push the box up the hill?Correct / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
Path A
Distance Traveled (d) = 10 m
θ = 30^{o}
µ_{k} = 0.4Path B
Distance Traveled (d) = 10 m
θ = 30^{o}
µ_{k} = 0.9
Solving for: which path requires more work to be doneStep 2: Let’s work backwards. What do we need to solve for the amount of work to be done for each path?
Let’s recall our work equation:
From here, let’s compare work in Path A versus work in Path B with respect to our work equation.
Step 3: Compare Work in Path A vs. Path B
W = F*d*cos(θ)
 We know that d is the same for A & B
 We know that θ is the same for A & B
Is F the same for A & B?
Step 4: Determine F for Path A & B
Let’s recall our friction force equation:
We know that the normal force (F_{N}) will be the same for Path A & Path B, as we are pushing the same box at the same angle.
However, the friction coefficients are NOT the same for Path A & B. Thus, the friction force will be UNEQUAL. As such, work will also be unequal.
Path B has the greater coefficient of friction, and thus work will be greater for Path B
Note: You may be thinking: wait, isn’t work “path independent?” meaning that the amount of work required to displace an object is independent of the path taken. In some cases, yes, but in other cases, no. To simplify for the OAT, you can assume work to be “path independent” when no friction or air resistance is involved. If friction or air resistance is involved, it is likely the case that work depends on path.
Incorrect / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
Path A
Distance Traveled (d) = 10 m
θ = 30^{o}
µ_{k} = 0.4Path B
Distance Traveled (d) = 10 m
θ = 30^{o}
µ_{k} = 0.9
Solving for: which path requires more work to be doneStep 2: Let’s work backwards. What do we need to solve for the amount of work to be done for each path?
Let’s recall our work equation:
From here, let’s compare work in Path A versus work in Path B with respect to our work equation.
Step 3: Compare Work in Path A vs. Path B
W = F*d*cos(θ)
 We know that d is the same for A & B
 We know that θ is the same for A & B
Is F the same for A & B?
Step 4: Determine F for Path A & B
Let’s recall our friction force equation:
We know that the normal force (F_{N}) will be the same for Path A & Path B, as we are pushing the same box at the same angle.
However, the friction coefficients are NOT the same for Path A & B. Thus, the friction force will be UNEQUAL. As such, work will also be unequal.
Path B has the greater coefficient of friction, and thus work will be greater for Path B
Note: You may be thinking: wait, isn’t work “path independent?” meaning that the amount of work required to displace an object is independent of the path taken. In some cases, yes, but in other cases, no. To simplify for the OAT, you can assume work to be “path independent” when no friction or air resistance is involved. If friction or air resistance is involved, it is likely the case that work depends on path.

Question 35 of 50Physics  Energy & Momentum
35. Question
35. There are two paths to reach the 10^{th} floor. On Path A, you take the spiral stairs. On Path B, you take an elevator.
Which path requires more work to be done (on you) to reach the 10^{th} floor?
Neglect friction and air resistance. Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionThis is a conceptual question.
When friction and air resistance are not involved, we can generally assume that work is “path independent,” meaning that the amount of work required to displace an object is independent of the path taken. Thus, all else equal (i.e. moving the same object from the same start point to the same endpoint), Path A and Path B will require the same amount of work.
Note: work is not ALWAYS path independent. If friction or air resistance is involved, it is likely the case that work depends on path taken.
Incorrect / You marked this questionThis is a conceptual question.
When friction and air resistance are not involved, we can generally assume that work is “path independent,” meaning that the amount of work required to displace an object is independent of the path taken. Thus, all else equal (i.e. moving the same object from the same start point to the same endpoint), Path A and Path B will require the same amount of work.
Note: work is not ALWAYS path independent. If friction or air resistance is involved, it is likely the case that work depends on path taken.

Question 36 of 50Physics  Energy & Momentum
36. Question
36. What is the minimum amount of work required to lift a 900 kg elevator 10,000 m from the 3^{rd} floor to the top story of a building?
Neglect friction and air resistance. Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Elevator = 900 kg
 Height Building = 10,000 m
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the amount of work required to lift the elevator to the top of the building
Step 2: Let’s work backwards. What do we need to solve of the work required to lift the elevator to the top of the building?
Let’s recall our work equation:
A downward force (weight) is pulling on the elevator. This force = m*g.
The smallest force required to move the elevator will cause the least amount of acceleration to the elevator, which would be zero acceleration. If the elevator is ascending with no acceleration, it must be the case that the elevator is moving with constant velocity. In this case, the net force on the elevator is zero.
In order for the net force to be zero, the upward ascending force must be equal in magnitude to the downward force (weight).
Step 3: Solve for upward force, then plug into our work equation
Upward force will be equal in magnitude to the weight of the elevator.
F = m*g
F = (900 kg)(10 m/s^{2})
F = 9,000 NThus, work will equal:
W = F*d*cos(θ)
W = (9,000 N)(10,000 m)(1)
W = 90,000,000 JIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Elevator = 900 kg
 Height Building = 10,000 m
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the amount of work required to lift the elevator to the top of the building
Step 2: Let’s work backwards. What do we need to solve of the work required to lift the elevator to the top of the building?
Let’s recall our work equation:
A downward force (weight) is pulling on the elevator. This force = m*g.
The smallest force required to move the elevator will cause the least amount of acceleration to the elevator, which would be zero acceleration. If the elevator is ascending with no acceleration, it must be the case that the elevator is moving with constant velocity. In this case, the net force on the elevator is zero.
In order for the net force to be zero, the upward ascending force must be equal in magnitude to the downward force (weight).
Step 3: Solve for upward force, then plug into our work equation
Upward force will be equal in magnitude to the weight of the elevator.
F = m*g
F = (900 kg)(10 m/s^{2})
F = 9,000 NThus, work will equal:
W = F*d*cos(θ)
W = (9,000 N)(10,000 m)(1)
W = 90,000,000 J 
Question 37 of 50Physics  Energy & Momentum
37. Question
37. What is the minimum amount of work required to accelerate a 500 kg elevator 2 m/s^{2} up 100 m?
Neglect friction and air resistance. Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Elevator = 500 kg
 Height Ascension = 100 m
 Acceleration of Elevator = 2 m/s^{2}
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the amount of work required to accelerate the elevator 100 m
Step 2: Let’s work backwards. What do we need to solve of the amount of work required to accelerate the elevator the elevator 2 m/s^{2} up 100 m?
Let’s recall our work equation:
A downward force (weight) is pulling on the elevator. This force = m*g. An upward force must overcome the force of gravity to accelerate the elevator upwards.
Step 3: Solve for upward force, then plug into our work equation
The force required to accelerate the elevator upwards has to overcome the force of gravity. Thus:
F = ma
F = (500 kg)(10 m/s^{2} + 2 m/s^{2})
F = 6,000 NStep 3: Plug the upward force into our work equation:
W = F*d*cos(θ)
W = (6,000 N)(100 m)(1)
W = 600,000 JIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Elevator = 500 kg
 Height Ascension = 100 m
 Acceleration of Elevator = 2 m/s^{2}
 Gravitational Acceleration (g) = 10 m/s^{2}
Solving for: the amount of work required to accelerate the elevator 100 m
Step 2: Let’s work backwards. What do we need to solve of the amount of work required to accelerate the elevator the elevator 2 m/s^{2} up 100 m?
Let’s recall our work equation:
A downward force (weight) is pulling on the elevator. This force = m*g. An upward force must overcome the force of gravity to accelerate the elevator upwards.
Step 3: Solve for upward force, then plug into our work equation
The force required to accelerate the elevator upwards has to overcome the force of gravity. Thus:
F = ma
F = (500 kg)(10 m/s^{2} + 2 m/s^{2})
F = 6,000 NStep 3: Plug the upward force into our work equation:
W = F*d*cos(θ)
W = (6,000 N)(100 m)(1)
W = 600,000 J 
Question 38 of 50Physics  Energy & Momentum
38. Question
38. A rocket of mass 10 x 10^{11} kg launched into space and has attained the velocity needed to escape the gravitational attraction of Planet X. The mass of Planet X is 2 x 10^{24} kg and distance from Planet X is 6.67 x 10^{11} m What is the kinetic energy of the satellite?Correct / You marked this questionStep 1: First, let’s write down our variables and what we are solving for:
 Mass Planet X = 2 x 10^{24} kg
 Mass Satellite = 10 x 10^{11} kg
 Distance of Planet X = 6.67 x 10^{11} m
Solving for: the kinetic energy of the rocket at escape velocity
Step 2: Let’s work backwards. What do we need in order to solve for the the kinetic energy of the satellite at escape velocity?
Let’s recall our kinetic energy equation:
We are given the mass of the satellite. Thus, we will need to solve for velocity.
Step 3: Solve for escape velocity.
Recall our escape velocity formula:
ProTip: remember, you are solving for kinetic energy, which uses v^{2}. The escape velocity equation solves for v = √… so v^{2} will just equal what is under the radical!
Note: for escape velocity, make sure you are using the mass of the PLANET and not of the object attempting to escape!
v^{2} = 400 m^{2}/s^{2}
Step 4. Plug our v^{2} value back into the kinetic energy equation
KE = ^{1}/_{2}mv^{2}
KE = ^{1}/_{2}(10 x 10^{11} kg)(400 m^{2}/s^{2})
KE = 2 x 10^{14} JIncorrect / You marked this questionStep 1: First, let’s write down our variables and what we are solving for:
 Mass Planet X = 2 x 10^{24} kg
 Mass Satellite = 10 x 10^{11} kg
 Distance of Planet X = 6.67 x 10^{11} m
Solving for: the kinetic energy of the rocket at escape velocity
Step 2: Let’s work backwards. What do we need in order to solve for the the kinetic energy of the satellite at escape velocity?
Let’s recall our kinetic energy equation:
We are given the mass of the satellite. Thus, we will need to solve for velocity.
Step 3: Solve for escape velocity.
Recall our escape velocity formula:
ProTip: remember, you are solving for kinetic energy, which uses v^{2}. The escape velocity equation solves for v = √… so v^{2} will just equal what is under the radical!
Note: for escape velocity, make sure you are using the mass of the PLANET and not of the object attempting to escape!
v^{2} = 400 m^{2}/s^{2}
Step 4. Plug our v^{2} value back into the kinetic energy equation
KE = ^{1}/_{2}mv^{2}
KE = ^{1}/_{2}(10 x 10^{11} kg)(400 m^{2}/s^{2})
KE = 2 x 10^{14} J 
Question 39 of 50Physics  Energy & Momentum
39. Question
39. A satellite of mass 10 x 10^{11} kg launched into space and is orbiting around Planet X. The mass of Planet X is 1 x 10^{24} kg and the satellite has a distance of 6.67 x 10^{11} m from Planet X.
What is the kinetic energy of the satellite?Correct / You marked this questionStep 1: First, let’s write down our variables and what we are solving for:
Mass Planet X = 1 x 10^{24} kg
Mass Rocket = 10 x 10^{11} kg
Distance of Planet X = 6.67 x 10^{11} mSolving for: the kinetic energy of the satellite at orbital velocity
Step 2: Let’s work backwards. What do we need in order to solve for the the kinetic energy of the satellite at orbital velocity?
Let’s recall our kinetic energy equation:
We are given the mass of the satellite. Thus, we will need to solve for velocity.
Step 3: Solve for orbital velocity
Recall our orbital velocity formula:
ProTip: remember, you are solving for kinetic energy, which uses v^{2}. The escape orbital equation solves for v = √… so v^{2} will just equal what is under the radical!
Note: for orbital velocity, make sure you are using the mass of the PLANET and not of the object orbiting!
v^{2} = 100 m^{2}/s^{2}
Step 4. Plug our v^{2} value back into the kinetic energy equation
KE = ^{1}/_{2}mv^{2}
KE = ^{1}/_{2}(10 x 10^{11} kg )(100 m^{2}/s^{2})
KE = 5 x 10^{13} JIncorrect / You marked this questionStep 1: First, let’s write down our variables and what we are solving for:
Mass Planet X = 1 x 10^{24} kg
Mass Rocket = 10 x 10^{11} kg
Distance of Planet X = 6.67 x 10^{11} mSolving for: the kinetic energy of the satellite at orbital velocity
Step 2: Let’s work backwards. What do we need in order to solve for the the kinetic energy of the satellite at orbital velocity?
Let’s recall our kinetic energy equation:
We are given the mass of the satellite. Thus, we will need to solve for velocity.
Step 3: Solve for orbital velocity
Recall our orbital velocity formula:
ProTip: remember, you are solving for kinetic energy, which uses v^{2}. The escape orbital equation solves for v = √… so v^{2} will just equal what is under the radical!
Note: for orbital velocity, make sure you are using the mass of the PLANET and not of the object orbiting!
v^{2} = 100 m^{2}/s^{2}
Step 4. Plug our v^{2} value back into the kinetic energy equation
KE = ^{1}/_{2}mv^{2}
KE = ^{1}/_{2}(10 x 10^{11} kg )(100 m^{2}/s^{2})
KE = 5 x 10^{13} J 
Question 40 of 50Physics  Energy & Momentum
40. Question
40. Evan is in his gokart (180 kg total) cruising at 30 m/s. He hits the brakes to slow down to 20 m/s. What is the absolute value of the work done to cause this change in velocity?Correct / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
Mass GoKart = 180 kg
Initial Velocity GoKart = 30 m/s
Final Velocity GoKart = 20 m/sSolving for: amount of work done to cause velocity change
Step 2: Let’s work backwards. How do we solve for the work done to cause the velocity change?
Let’s recall our work equation that uses kinetic energy:
If we can solve for initial kinetic energy and final kinetic energy, we can solve for work
Step 3: Solve for ΔKE
Let’s recall our kinetic energy equation:
KE (initial) = ^{1}/_{2}(180 kg)(30 m/s)^{2}
KE (final) = ^{1}/_{2}(180 kg)(20 m/s)^{2}
ΔKE = ^{1}/_{2}(180 kg)(30 m/s)^{2} – ^{1}/_{2}(180 kg)(20 m/s)^{2}
ΔKE = W = 45,000 JIncorrect / You marked this questionStep 1: Let’s write down our variables and what we are solving for:
Mass GoKart = 180 kg
Initial Velocity GoKart = 30 m/s
Final Velocity GoKart = 20 m/sSolving for: amount of work done to cause velocity change
Step 2: Let’s work backwards. How do we solve for the work done to cause the velocity change?
Let’s recall our work equation that uses kinetic energy:
If we can solve for initial kinetic energy and final kinetic energy, we can solve for work
Step 3: Solve for ΔKE
Let’s recall our kinetic energy equation:
KE (initial) = ^{1}/_{2}(180 kg)(30 m/s)^{2}
KE (final) = ^{1}/_{2}(180 kg)(20 m/s)^{2}
ΔKE = ^{1}/_{2}(180 kg)(30 m/s)^{2} – ^{1}/_{2}(180 kg)(20 m/s)^{2}
ΔKE = W = 45,000 J 
Question 41 of 50Physics  Energy & Momentum
41. Question
41. Mike pushes an 18 kg box 9 m across the floor with a force of 80 Newtons. The coefficient of friction between the box and the floor is 0.5.
What is the absolute value of the work done by friction on the box?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Box = 18 kg
 Displacement = 9 m
 Pushing Force = 80 N
 µ_{k} = 0.5
Solving for: the amount of work done by friction on the box
Step 2: Next, let’s work backwards. What do we need in order to solve the work done by friction on the box?
Recall our work equation:
We can assume that F and d are in the same direction, implying that cos(θ) = 1.
We are given d. Thus, we must determine the force due to friction (F_{f})
Step 3: Solve for Friction Force (F_{f})
Recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m*g)
F_{f} = (0.5)(18 kg)(10 m/s^{2})
F_{f} = 90 NStep 4: Solve for the work done by friction on the box
W = F*d*cos(θ)
W = (90 N)(9 m)(1)
W = 810 JIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Box = 18 kg
 Displacement = 9 m
 Pushing Force = 80 N
 µ_{k} = 0.5
Solving for: the amount of work done by friction on the box
Step 2: Next, let’s work backwards. What do we need in order to solve the work done by friction on the box?
Recall our work equation:
We can assume that F and d are in the same direction, implying that cos(θ) = 1.
We are given d. Thus, we must determine the force due to friction (F_{f})
Step 3: Solve for Friction Force (F_{f})
Recall our frictional force equation:
F_{f} = µ_{k}F_{N}
F_{f} = (0.5)(m*g)
F_{f} = (0.5)(18 kg)(10 m/s^{2})
F_{f} = 90 NStep 4: Solve for the work done by friction on the box
W = F*d*cos(θ)
W = (90 N)(9 m)(1)
W = 810 J 
Question 42 of 50Physics  Energy & Momentum
42. Question
42. A 12 kg ball is dropped from 20 meters high. What is the ball’s velocity when it hits the ground?
Neglect air resistance and assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Ball = 12 kg
 Height = 20 m
 Initial Velocity = 0 m/s
Solving for: the ball’s final velocity when it hits the ground
Step 2: Let’s work backwards. What do we need to solve for the ball’s final velocity?
There are a couple of ways you could solve this problem. You will, of course, want to use the simplest & quickest one.
At first glance, you may want to whip out a kinematics equation. You could get to the correct answer this way, but there’s a quicker route.
Notice the conservation of energy. As there is no air resistance, the potential energy at the apex will equal the kinetic energy when the ball hits the ground. Let’s solve using this approach.
Step 3: Set PE (top) equal to KE (bottom) to solve for final velocity
PE (top) = KE (bottom)
mgh = ^{1}/_{2}m(v_{FINAL})^{2}
gh = ^{1}/_{2}(v_{FINAL})^{2}
(10 m/s^{2})(20 m) = ^{1}/_{2}(v_{FINAL})^{2}
400 m^{2}/s^{2} = (v_{FINAL})^{2}
(v_{FINAL}) = 20 m/sIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Ball = 12 kg
 Height = 20 m
 Initial Velocity = 0 m/s
Solving for: the ball’s final velocity when it hits the ground
Step 2: Let’s work backwards. What do we need to solve for the ball’s final velocity?
There are a couple of ways you could solve this problem. You will, of course, want to use the simplest & quickest one.
At first glance, you may want to whip out a kinematics equation. You could get to the correct answer this way, but there’s a quicker route.
Notice the conservation of energy. As there is no air resistance, the potential energy at the apex will equal the kinetic energy when the ball hits the ground. Let’s solve using this approach.
Step 3: Set PE (top) equal to KE (bottom) to solve for final velocity
PE (top) = KE (bottom)
mgh = ^{1}/_{2}m(v_{FINAL})^{2}
gh = ^{1}/_{2}(v_{FINAL})^{2}
(10 m/s^{2})(20 m) = ^{1}/_{2}(v_{FINAL})^{2}
400 m^{2}/s^{2} = (v_{FINAL})^{2}
(v_{FINAL}) = 20 m/s 
Question 43 of 50Physics  Energy & Momentum
43. Question
43. A 50 kg ball is launched from the ground directly vertically at 20 m/s. How high will the ball be at the top of its ascent?
Neglect air resistance and assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Ball = 50 kg
 Initial Velocity = 20 m/s
Solving for: the ball’s maximum height
Step 2: Let’s work backwards. What do we need to solve for the ball’s maximum height?
There are a couple of ways you could solve this problem. You will, of course, want to use the simplest & quickest one.
At first glance, you may want to whip out a kinematics equation. You could get to the correct answer this way, but there’s a quicker route.
Notice the conservation of energy. As there is no air resistance, the potential energy at the apex will equal the kinetic energy when the ball leaves the ground. Let’s solve using this approach.
Step 3: Set PE (top) equal to KE (bottom) to solve for final velocity
PE (top) = KE (bottom)
mgh = ^{1}/_{2}m(v_{INITIAL})^{2}
gh = ^{1}/_{2}(v_{INITIAL})^{2}
(10 m/s^{2})(h) = ^{1}/_{2}(20 m/s)^{2}
h = 20 mNote: notice that mass “cancels out” in going from step “ii” to step “iii” above. This implies that the max height in this problem is not dependent on mass – it would be the same for any mass ball, assuming it was shot upwards with the same velocity!
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Ball = 50 kg
 Initial Velocity = 20 m/s
Solving for: the ball’s maximum height
Step 2: Let’s work backwards. What do we need to solve for the ball’s maximum height?
There are a couple of ways you could solve this problem. You will, of course, want to use the simplest & quickest one.
At first glance, you may want to whip out a kinematics equation. You could get to the correct answer this way, but there’s a quicker route.
Notice the conservation of energy. As there is no air resistance, the potential energy at the apex will equal the kinetic energy when the ball leaves the ground. Let’s solve using this approach.
Step 3: Set PE (top) equal to KE (bottom) to solve for final velocity
PE (top) = KE (bottom)
mgh = ^{1}/_{2}m(v_{INITIAL})^{2}
gh = ^{1}/_{2}(v_{INITIAL})^{2}
(10 m/s^{2})(h) = ^{1}/_{2}(20 m/s)^{2}
h = 20 mNote: notice that mass “cancels out” in going from step “ii” to step “iii” above. This implies that the max height in this problem is not dependent on mass – it would be the same for any mass ball, assuming it was shot upwards with the same velocity!

Question 44 of 50Physics  Energy & Momentum
44. Question
44. A 10 kg tennis ball is launched from the ground directly vertically at 100 m/s. Due to air resistance, it loses 5,000 J of energy on its ascent.
How high will the ball be at the top of its ascent?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Ball = 10 kg
 Initial Velocity = 10 m/s
 Energy Lost = 5,000 J
Solving for: the ball’s maximum height
Step 2: Let’s work backwards. What do we need to solve for the ball’s maximum height?
At first glance, you may want to whip out a kinematics equation. However, the loss of 5,000 J of energy on the ascent should clue you in that this is a conservation of energy problem.
The kinetic energy when the ball leaves the ground will equal the potential energy at the ball’s apex PLUS the energy loss due to air resistance.
Step 3: Set PE (top) + Energy Lost equal to KE (bottom) to solve for max height
PE (top) + 5,000 J = KE (bottom)
mgh + 5,000 J = ^{1}/_{2}m(v_{INITIAL})^{2}
(h)(10 kg)(10 m/s^{2}) + 5000 J = ^{1}/_{2}(10kg)(100 m/s)^{2}
(h)(10 kg)(10 m/s^{2}) + 5000 J = 50,000 J
h = 45,000 J / (100 kg m/s^{2})
h = 450 mIncorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch and what we are solving for.
 Mass Ball = 10 kg
 Initial Velocity = 10 m/s
 Energy Lost = 5,000 J
Solving for: the ball’s maximum height
Step 2: Let’s work backwards. What do we need to solve for the ball’s maximum height?
At first glance, you may want to whip out a kinematics equation. However, the loss of 5,000 J of energy on the ascent should clue you in that this is a conservation of energy problem.
The kinetic energy when the ball leaves the ground will equal the potential energy at the ball’s apex PLUS the energy loss due to air resistance.
Step 3: Set PE (top) + Energy Lost equal to KE (bottom) to solve for max height
PE (top) + 5,000 J = KE (bottom)
mgh + 5,000 J = ^{1}/_{2}m(v_{INITIAL})^{2}
(h)(10 kg)(10 m/s^{2}) + 5000 J = ^{1}/_{2}(10kg)(100 m/s)^{2}
(h)(10 kg)(10 m/s^{2}) + 5000 J = 50,000 J
h = 45,000 J / (100 kg m/s^{2})
h = 450 m 
Question 45 of 50Physics  Energy & Momentum
45. Question
45. If you perform the same amount of work in half the time, the associated power will be:Correct / You marked this questionThis is a conceptual question
Recall the power equation:
If we we take ^{1}/_{2} t, power will be 2x as much.
Incorrect / You marked this questionThis is a conceptual question
Recall the power equation:
If we we take ^{1}/_{2} t, power will be 2x as much.

Question 46 of 50Physics  Energy & Momentum
46. Question
46. If you perform the same amount of work in three times as much time, the associated power will be:Correct / You marked this questionThis is a conceptual question
Recall the power equation:
If we we triple t, power will be ^{1}/_{3} as much.
Incorrect / You marked this questionThis is a conceptual question
Recall the power equation:
If we we triple t, power will be ^{1}/_{3} as much.

Question 47 of 50Physics  Energy & Momentum
47. Question
47. You are pushing a 100 kg box. If you do the same amount of work on the box, but push the box 2x as far in the same amount of time, the associated power will be:Correct / You marked this questionThis is a conceptual question
Recall the power equation:
The questions specifically states that we hold work and time the same. As such, it must be the case that power is the same.
You may be thinking: how could you do the same amount of work in the same amount of time but push the box 2x as far? A valid thought. One scenario that comes to mind is if the friction has been decreased. We are not given additional information, so we must work off of what we know to be true.
Incorrect / You marked this questionThis is a conceptual question
Recall the power equation:
The questions specifically states that we hold work and time the same. As such, it must be the case that power is the same.
You may be thinking: how could you do the same amount of work in the same amount of time but push the box 2x as far? A valid thought. One scenario that comes to mind is if the friction has been decreased. We are not given additional information, so we must work off of what we know to be true.

Question 48 of 50Physics  Energy & Momentum
48. Question
48. A 30 kg “offroading” truck is driving 15 m/s. It hits a rough patch of dirt, with a coefficient of friction of 0.5, and is brought to a stop.
What would be the relative stopping distance if all else was equal, but the coefficient of friction was 0.25?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car = 30 kg
 v_{INITIAL} = 15 m/s
 v_{FINAL} = 0 m/s
 Original µ_{k} = 0.5
Solving for: the relative stopping distance if all else was equal, but the coefficient of friction was ^{1}/_{2} of the original
Step 2: Next, let’s work backwards. What do we need in order to solve for the relative stopping distance if all else was equal, but the coefficient of friction was ^{1}/_{2} of the original?
Here we will need to use the fact that we have two different work equations to “work” with (pun intended)
Set the two equations equations equal to each other to determine what would happen if we cut the coefficient of friction in half.
Step 3: But first … rewrite our F*d*cos(θ) to include friction
We can assume that friction is the only force bringing the truck to a stop. Thus, the amount of work done to stop the car will be:
W = F*d*cos(θ)
W = µ_{k}F_{N}*dStep 3: Set ΔKE = F*d*cos(θ)
As the car comes to a full stop (i.e. v_{FINAL} = 0) we know that KE (final) = 0. Thus, ΔKE will be equal to ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}F_{N} * d = ΔKE
W = µ_{k}F_{N} * d = ^{1}/_{2}m(v_{INITIAL})^{2}We are told the we are holding all else equal. This means that v_{INITIAL} is equal in both scenarios. Thus, you can see that if the coefficient of kinetic friction is ^{1}/_{2}, the corresponding stopping distance (d) will be 2x.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car = 30 kg
 v_{INITIAL} = 15 m/s
 v_{FINAL} = 0 m/s
 Original µ_{k} = 0.5
Solving for: the relative stopping distance if all else was equal, but the coefficient of friction was ^{1}/_{2} of the original
Step 2: Next, let’s work backwards. What do we need in order to solve for the relative stopping distance if all else was equal, but the coefficient of friction was ^{1}/_{2} of the original?
Here we will need to use the fact that we have two different work equations to “work” with (pun intended)
Set the two equations equations equal to each other to determine what would happen if we cut the coefficient of friction in half.
Step 3: But first … rewrite our F*d*cos(θ) to include friction
We can assume that friction is the only force bringing the truck to a stop. Thus, the amount of work done to stop the car will be:
W = F*d*cos(θ)
W = µ_{k}F_{N}*dStep 3: Set ΔKE = F*d*cos(θ)
As the car comes to a full stop (i.e. v_{FINAL} = 0) we know that KE (final) = 0. Thus, ΔKE will be equal to ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}F_{N} * d = ΔKE
W = µ_{k}F_{N} * d = ^{1}/_{2}m(v_{INITIAL})^{2}We are told the we are holding all else equal. This means that v_{INITIAL} is equal in both scenarios. Thus, you can see that if the coefficient of kinetic friction is ^{1}/_{2}, the corresponding stopping distance (d) will be 2x.

Question 49 of 50Physics  Energy & Momentum
49. Question
49. A 22 kg “offroading” truck is driving 22 m/s. It hits a rough patch of dirt, with a coefficient of friction of 0.22, and is brought to a stop.
If all else were equal, what would the relative stopping distance be if the truck were moving at 44 m/s?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
Mass Car = 22 kg
Original v_{INITIAL} = 22 m/s
v_{FINAL} = 0 m/s
µ_{k} = 0.22Solving for: the relative stopping distance if all else was equal, but the initial velocity was doubled
Step 2: Next, let’s work backwards. What do we need in order to solve for the relative stopping distance if all else was equal, but the initial velocity was doubled?
Here we will need to use the fact that we have two different work equations to “work” with (pun intended)
Set the two equations equations equal to each other to determine what would happen if we doubled the initial velocity.
Step 3: But first … rewrite our F*d*cos(θ) to include friction
We can assume that friction is the only force bringing the truck to a stop. Thus, the amount of work done to stop the car will be:
W = F*d*cos(θ)
W = µ_{k}F_{N}*dStep 3: Set ΔKE = F*d*cos(θ)
As the car comes to a full stop (i.e. v_{FINAL} = 0) we know that KE (final) = 0. Thus, ΔKE will be equal to ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}F_{N} * d = ΔKE
W = µ_{k}F_{N} * d = ^{1}/_{2}m(v_{INITIAL})^{2}We are told the we are holding all else equal. Thus, you can see that if the initial velocity is doubled, the corresponding stopping distance (d) will be 4x the original stopping distance.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
Mass Car = 22 kg
Original v_{INITIAL} = 22 m/s
v_{FINAL} = 0 m/s
µ_{k} = 0.22Solving for: the relative stopping distance if all else was equal, but the initial velocity was doubled
Step 2: Next, let’s work backwards. What do we need in order to solve for the relative stopping distance if all else was equal, but the initial velocity was doubled?
Here we will need to use the fact that we have two different work equations to “work” with (pun intended)
Set the two equations equations equal to each other to determine what would happen if we doubled the initial velocity.
Step 3: But first … rewrite our F*d*cos(θ) to include friction
We can assume that friction is the only force bringing the truck to a stop. Thus, the amount of work done to stop the car will be:
W = F*d*cos(θ)
W = µ_{k}F_{N}*dStep 3: Set ΔKE = F*d*cos(θ)
As the car comes to a full stop (i.e. v_{FINAL} = 0) we know that KE (final) = 0. Thus, ΔKE will be equal to ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}F_{N} * d = ΔKE
W = µ_{k}F_{N} * d = ^{1}/_{2}m(v_{INITIAL})^{2}We are told the we are holding all else equal. Thus, you can see that if the initial velocity is doubled, the corresponding stopping distance (d) will be 4x the original stopping distance.

Question 50 of 50Physics  Energy & Momentum
50. Question
50. A 80 kg “offroading” truck is driving 12 m/s. It hits a rough patch of dirt, with a coefficient of friction of 0.64, and is brought to a stop. If all else were equal, what would the relative stopping distance be if the truck’s mass were 40 kg?Correct / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car = 80 kg
 Original v_{INITIAL} = 12 m/s
 v_{FINAL} = 0 m/s
 µ_{k} = 0.64
Solving for: the relative stopping distance if all else was equal, but the truck’s mass was halved?
Step 2: Next, let’s work backwards. What do we need in order to solve for the relative stopping distance if all else was equal, but the truck’s mass was halved?
Here we will need to use the fact that we have two different work equations to “work” with (pun intended)
Set the two equations equations equal to each other to determine what would happen if we doubled the mass of the truck.
Step 3: But first … rewrite our F*d*cos(θ) to include friction
We can assume that friction is the only force bringing the truck to a stop. Thus, the amount of work done to stop the car will be:
W = F*d*cos(θ)
W = µ_{k}F_{N}*d
W = µ_{k}mg*dStep 3: Set ΔKE = F*d*cos(θ)
As the car comes to a full stop (i.e. v_{FINAL} = 0) we know that KE (final) = 0. Thus, ΔKE will be equal to ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}F_{N}*d = ΔKE
W = µ_{k}F_{N}*d = ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}mg*d = ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}g*d = ^{1}/_{2}(v_{INITIAL})^{2}We are told the we are holding all else equal. Thus, you can see that if the mass is not directly impacting the distance, the corresponding stopping distance (d) will be the same.
Incorrect / You marked this questionStep 1: Let’s write down our variables, a quick sketch, and what we are solving for:
 Mass Car = 80 kg
 Original v_{INITIAL} = 12 m/s
 v_{FINAL} = 0 m/s
 µ_{k} = 0.64
Solving for: the relative stopping distance if all else was equal, but the truck’s mass was halved?
Step 2: Next, let’s work backwards. What do we need in order to solve for the relative stopping distance if all else was equal, but the truck’s mass was halved?
Here we will need to use the fact that we have two different work equations to “work” with (pun intended)
Set the two equations equations equal to each other to determine what would happen if we doubled the mass of the truck.
Step 3: But first … rewrite our F*d*cos(θ) to include friction
We can assume that friction is the only force bringing the truck to a stop. Thus, the amount of work done to stop the car will be:
W = F*d*cos(θ)
W = µ_{k}F_{N}*d
W = µ_{k}mg*dStep 3: Set ΔKE = F*d*cos(θ)
As the car comes to a full stop (i.e. v_{FINAL} = 0) we know that KE (final) = 0. Thus, ΔKE will be equal to ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}F_{N}*d = ΔKE
W = µ_{k}F_{N}*d = ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}mg*d = ^{1}/_{2}m(v_{INITIAL})^{2}
W = µ_{k}g*d = ^{1}/_{2}(v_{INITIAL})^{2}We are told the we are holding all else equal. Thus, you can see that if the mass is not directly impacting the distance, the corresponding stopping distance (d) will be the same.
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 ANSWERED
 MARKED