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AP Physics C: Mechanics : Motion

Study concepts, example questions & explanations for AP Physics C: Mechanics

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All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : Using Spring Equations

All angles in this problem are expressed in radians.

An object oscillates horizontally, and its displacement from equilibrium can be found using the equation:

$x=(2.00m)cos(\pi t+\frac{\pi }{2})$

where is in seconds. What is the velocity of the object at t=3.00s ?

Possible Answers:

$2.00\frac{m}{s}$

$-2\pi \frac{m}{s}$

$2\pi \frac{m}{s}$

$0.00\frac{m}{s}$

$\pi\frac{m}{s}$

Correct answer:

$2\pi \frac{m}{s}$

Explanation:

To find the equation for velocity, we take the first derivative of the position function. Don't forget the chain rule for the inside of the cosine!

$v=\frac{d(x)}{dt}$

$v=(-2.00\pi \frac{m}{s})sin(\pi t+\frac{\pi }{2})$

$v=(-2.00\pi \frac{m}{s})sin(\pi (2.00s))+\frac{\pi }{2})$

$v=2\pi \frac{m}{s}$

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Example Question #41 : Mechanics Exam

A mass on the end of a spring oscillates back and forth. The period of an oscillating spring is measured to be 2.27 seconds. The mass is measured to be 0.300 kg. What is the spring's spring constant?

Possible Answers:

$0.422\frac{kg}{s^2}$

$1.07\frac{kg}{s^2}$

$0.873\frac{kg}{s^2}$

$0.575\frac{kg}{s^2}$

$0.943\frac{kg}{s^2}$

Correct answer:

$0.575\frac{kg}{s^2}$

Explanation:

To solve this problem, we can use the following equation:

$\small T=2\pi\sqrt{\frac{m}{k}}$

Here, is the period of an oscillation, is the mass of the oscillating thing in kg, and is the spring constant in $\frac{kg}{s^2}$ .

Rearranging the equation to solve for the spring constant, , we get:

$k\cdot =\frac{m}{(\frac{T}{2\pi})^2}$

Plugging in the two known values, we get:

$k\cdot =\frac{0.300kg}{(\frac{2.27s}{2\pi})^2}$

Solving for , we get $0.575\frac{kg}{s^2}$ .

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Example Question #41 : Mechanics Exam

A $1.5\: kg$ block on a frictionless table is connected to a horizontal spring with constant $450\: N/m$ . If the block is released from rest when the spring is stretched a distance of $7\: cm$ ,

what is its speed when the spring is compressed a distance of $3\: cm$ ?

Possible Answers:

$1.32\:m/s$

$1.2\:m/s$

$1.1\:m/s$

$0.849\:m/s$

$0.995\:m/s$

Correct answer:

$1.1\:m/s$

Explanation:

Assuming this is a frictionless table, we don't have to take the work done by friction into account.

This is a conservation of energy problem. In this problem we have to realize that the potential energy of the spring at $7\:cm$ is equal to the kinetic energy of the spring at $3\:cm$ + the potential energy of the spring at $3\:cm$ .

This can be shown in the following equation:

$\frac{1}{2}kx_7^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx_3^2$

If we solve for , we get the following equation:

$v = \sqrt{2\cdot\frac{\frac{1}{2}kx_7^2-\frac{1}{2}kx_3^2}{m}}$

Remember to convert the distances given in centimeters to meters.

If we plug in all the variables, we get $v = 1.1\:m/s$

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Example Question #1 : Using Pendulum Equations

A simple pendulum of length swings at a radius from a fixed point on the ceiling. As it moves, the pendulum creates an angle $\theta$ with the line that extends vertically downwards from the fixed point. At what angle does the mass attached to the pendulum have the highest kinetic energy?

Possible Answers:

${30^\circ}$

$0^\circ$

$90^\circ$

$45^\circ$

Correct answer:

$0^\circ$

Explanation:

The correct answer is $0^\circ$ . Since the pendulum is at the bottom of its motion at this point, it has the lowest amount of energy given to gravitational potential and thus, the highest kinetic energy.

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Example Question #1 : Using Pendulum Equations

If a simple pendulum is constructed using a rope of negligible weight and a large steel ball weighing 100N . What is the period of the pendulum?

Possible Answers:

$2\pi\cdot\sqrt{10}s$

Not enough information

2 s

200s

$2\pi s$

Correct answer:

2 s

Explanation:

Use the equation for the period of a simple pendulum:

$T=2\pi\sqrt\frac{L}{g}$

Here, is the period in seconds, is the length of the pendulum in meters, and is the acceleration due to gravity in $\frac{m}{s^2}$ .

We can plug in the given quantities for length and acceleration to solve for the period.

$T=2\pi\sqrt\frac{1}{9.8}\approx2s$

Note that the mass of the rope as negligible. We also need not incorporate the extraneous information regarding the weight of the steel ball attached to the pendulum. This does not influence the period of this simple pendulum.

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Example Question #1 : Interpreting Harmonic Motion Diagrams

Two springs big

A block of mass $\small m$ is attached to two springs, each of whose spring constant is $\small k$ . The ends of the springs are fixed, and the block is free to move back and forth. It is released from rest at an initial amplitude, and its period is measured to be $\small T$ . What would the period be if the spring on the right side were to be moved to the other side, attached along side of the other spring?

Possible Answers:

$\small 2T$

$\small T$

$T\sqrt{2}$

$\small \frac{T}{\sqrt{2}}$

$\small \frac{T}{2}$

Correct answer:

$\small T$

Explanation:

Because the springs are effectively in a parallel arrangement already, moving one does not change the effective spring constant, and therefore does not affect the period.

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Example Question #1 : Circular And Rotational Motion

Rotatingball

A ball of mass is tied to a rope and moves along a horizontal circular path of radius as shown in the diagram (view from above). The maximum tension the rope can stand before breaking is given by $T_{max}$ . Which of the following represents the ball's linear velocity given that the rope does not break?

Possible Answers:

$v\geq\sqrt{rT_{max}}$

$v\geq\sqrt{\frac{rT_{max}}{m}}$

$v\leq\sqrt{\frac{rT_{max}}{m}}$

$v\leq\sqrt{rT_{max}}$

None of these

Correct answer:

$v\leq\sqrt{\frac{rT_{max}}{m}}$

Explanation:

This is a centripetal force problem. In this case the tension on the rope is the centripetal force that keeps the ball moving on a circle.

$T=F_C=\frac{mv^2}{r}$

If we want for the rope not to break, then the tension should never exceed $T_{max}$ .

$T_{max}\geq\frac{mv^2}{r}$

Now we just solve for velocity:

$v\leq\sqrt{\frac{rT_{max}}{m}}}$

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Example Question #1 : Understanding Linear Rotational Equivalents

What is the rotational equivalent of mass?

Possible Answers:

Angular momentum

Radius

Moment of inertia

Torque

Correct answer:

Moment of inertia

Explanation:

The correct answer is moment of inertia. For linear equations, mass is what resists force and causes lower linear accelerations. Similarly, in rotational equations, moment of inertia resists torque and causes lower angular accelerations.

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Example Question #1 : Circular And Rotational Motion

In rotational kinematics equations, what quantity is analogous to force in linear kinematics equations?

Possible Answers:

Torque

Angular acceleration

Impulse

Moment of inertia

Correct answer:

Torque

Explanation:

Just as force causes linear acceleration, torque causes angular acceleration. This can be seen most in the linear-rotational comparison of Newton's second law:

F=ma

$\tau=I\alpha$

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Example Question #1 : Understanding Linear Rotational Equivalents

A boot is put in a 1 m stick which is attached to a rotor. The rotor turns with an angular velocity of $30\pi \frac{rad}{s}$ . What is the linear velocity of the boot?

Possible Answers:

$15\frac{m}{s}$

$15\pi\frac{m}{s}$

$30\frac{m}{s}$

$30\pi\frac{m}{s}$

$3\pi\frac{m}{s}$

Correct answer:

$30\pi\frac{m}{s}$

Explanation:

Linear (tangential) velocity, $v_{T}$ is given by the following equation:

$v_{T}=\omega r$

Here, $\omega$ is the angular velocity in radians per second and is the radius in meters.

r=1m

Solve.

$\omega = 30\pi \frac{rad}{s}$

$\omega = 30\pi\frac{m}{s}$

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Motion

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #6 : Using Spring Equations

Example Question #41 : Mechanics Exam

Example Question #41 : Mechanics Exam

Example Question #1 : Using Pendulum Equations

Example Question #1 : Using Pendulum Equations

Example Question #1 : Interpreting Harmonic Motion Diagrams

Example Question #1 : Circular And Rotational Motion

Example Question #1 : Understanding Linear Rotational Equivalents

Example Question #1 : Circular And Rotational Motion

Example Question #1 : Understanding Linear Rotational Equivalents

All AP Physics C: Mechanics Resources

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