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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #1 : Other Harmonic Systems

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If a mass on the end of a string of length 8 cm is pulled 30 degrees away from vertical, what will its speed be when the string is vertically alligned if the mass is released from rest?

$g=9.81\frac{m}{s^2}$

Possible Answers:

$1.7\frac{m}{s}$

$1.3\frac{m}{s}$

$2.1\frac{m}{s}$

$0.46\frac{m}{s}$

$4.5\frac{m}{s}$

Correct answer:

$0.46\frac{m}{s}$

Explanation:

We can determine the vertical components by using some geometry.

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Since this problem boils down to being a problem of conservation of energy, we can state that

$E_{i}=E_{f}\rightarrow PE_{i}+KE_{i}=PE_{f}+KE_{f}$

Since the mass is released from rest and we can state at the bottom of its arc it will be at height 0 m, this equation can be simplified:

$PE_{i}=KE_{f} \rightarrow mgh=\frac{1}{2}mv^{2} \rightarrow gh=\frac{1}{2}v^{2}$

is equal to the height of the mass above 0 m; therefore, by referring to the diagram, we can determine that $h=R-Rcos\Theta$ . By plugging this in and rearranging the equation, we can solve for the speed of the mass when the string is vertically aligned:

$v=\sqrt{2gR(1-cos\Theta)}=\sqrt{2(9.81\frac{m}{s^2})(8\cdot10^{-2}m)(1-cos30^{o})}=0.46\frac{m}{s}$

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Example Question #1 : Other Harmonic Systems

A ball with mass 5 kg is attached to a spring and is released 10 meters from equilibrium. After some time, the ball passes the equilibrium point moving at $7\frac{m}{s}$ . What is the spring constant () of the spring?

Possible Answers:

$1.225\frac{kg}{s^2}$

$0.82\frac{kg}{s^2}$

$2.45 \frac{kg}{s^2}$

$3.5\frac{kg}{s^2}$

$0.41\frac{kg}{s^2}$

Correct answer:

$2.45 \frac{kg}{s^2}$

Explanation:

When the object is released, it has no kinetic energy (it isn't moving) and a potential energy of

$P.E.=\frac{1}{2}kx^2=\frac{1}{2}k(10m)^2=50m^2\cdot k$

When the object passes through the equilibrium point, it has no potential energy ( x=0 ) and a kinetic energy of

$K.E.=\frac{1}{2}mv^2=\frac{1}{2}(5kg)(7\frac{m}{s})^2=122.5\frac{kg\cdot m^2}{s^2}$

Due to the conservation of energy, these two quantities must be equal to each other:

$50m^2\cdot k=122.5\frac{kg*m^2}{s^2}$

$k = \frac{122.5\frac{kg\cdot m^2}{s^2}}{50m^2}=2.45\frac{kg}{s^2}$

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Example Question #2 : Other Harmonic Systems

A 30 kg weight attached to a spring is at equilibrium lying horizontally on a table. The spring is lifted up and is stretched by 80 cm before the weight is lifted off the table. What is this spring's spring constant ()?

Possible Answers:

$460\frac{kg}{s^2}$

$3.68\frac{kg}{s^2}$

$37.5\frac{kg}{s^2}$

$368\frac{kg}{s^2}$

$0.375\frac{kg}{s^2}$

Correct answer:

$368\frac{kg}{s^2}$

Explanation:

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

F_g=mg

$294N=(30kg)(9.81\frac{m}{s^2})$

The force applied by the spring in the opposite direction must be equal to this:

-kx=-294N

k(0.8m) = 294N

$k=368\frac{N}{m}=368\frac{kg}{s^2}$

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Example Question #991 : Ap Physics 1

Instead of using a second hand to count seconds, a watchmaker decides to construct a simple harmonic system involving a mass and a spring suspended from a table.

Ignoring the effects of gravity, if the spring the watchmaker selects has a spring constant of $0.01\frac{kg}{s^2}$ , how large of a mass in kilograms should he attach to the spring such that the harmonic system does not oscillate too slow or too fast?

Possible Answers:

$\frac{9}{\pi^2}kg$

$\frac{1}{3}kg$

$\frac{1}{6}kg$

$\frac{3}{\pi^2}kg$

$\frac{9}{10}kg$

Correct answer:

$\frac{9}{\pi^2}kg$

Explanation:

The angular velocity of the harmonic system is equal to the square root of the spring constant over the mass:

$\omega=\sqrt{\frac{k}{m}}$

Since we need the angular velocity to be $\frac{2\pi}{60}$ and we are given the spring constant as $0.01\frac{kg}{s^2}$ , then we can set up an equation:

$\frac{2\pi}{60}=\sqrt{\frac{0.01}{m}}$

$\frac{4\pi^2}{60^2}=\frac{0.01}{m}$

$m=\frac{60^2*0.01}{4\pi^2}=\frac{9}{\pi^2}kg$

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Example Question #5 : Other Harmonic Systems

The position of a 1kg mass in an oscillating spring-mass system is given by the following equation:

$x(t)=2sin(4\pi t)+4$ , where is measured in seconds , and is measured in meters .

What is the velocity equation of the system as a function of time?

Possible Answers:

$v(t)=8\pi cos( t)$

$v(t)=2 cos(4\pi t)$

$v(t)=2 sin(4\pi t)$

$v(t)=8\pi sin(4\pi t)$

$v(t)=8\pi cos(4\pi t)$

Correct answer:

$v(t)=8\pi cos(4\pi t)$

Explanation:

The velocity equation can be found by differentiating the position equation.

$v(t)=x'(t)=\frac{d}{dt}(2sin(4\pi t)+4)=2*4\pi cos(4\pi t)$

$=8\pi cos(4\pi t)$

Here, we made use of the chain rule in taking the derivative.

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Example Question #1 : Circular And Rotational Motion

A horizontally mounted wheel of radius is initially at rest, and then begins to accelerate constantly until it has reached an angular velocity $\omega$ after 5 complete revolutions. What was the angular acceleration of the wheel?

Possible Answers:

$\frac{\omega ^{2}}{10\pi}$

$\frac{\omega ^{2}}{10\pi r}$

$\frac{\omega ^{2}}{20\pi }$

$\frac{\omega}{10\pi }$

$\frac{\omega ^{2}}{5\pi r}$

Correct answer:

$\frac{\omega ^{2}}{20\pi }$

Explanation:

You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:

$v_{f}^{2}=v_{i}^{2}+2ad$

Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:

$\omega _{f}^{2}=\omega _{i}^{2}+2\alpha \theta$

The wheel starts at rest, so the initial angular velocity, $\omega _{i}$ , is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of $2\pi$ radians. So, we can convert the total revolutions to an angular distance to get:

$\theta=5\:rev\cdot \frac{2\pi }{1\:rev}=10\pi$

The final angular velocity was given as $\omega$ in the text of the question. So, we should use the above equation to solve for the angular acceleration, $\alpha$ .

$\omega ^{2}=0+2\alpha\cdot 10\pi$

$\alpha=\frac{\omega ^{2}}{20\pi }$

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Example Question #2 : Angular Velocity And Acceleration

An object moves at a constant speed of $9.0 \:\frac{m}{s}$ in a circular path of radius of 1.5 m. What is the angular acceleration of the object?

Possible Answers:

$14\:\frac{rad}{s^2}$

$36\:\frac{rad}{s^2}$

$54\:\frac{rad}{s^2}$

$81\:\frac{rad}{s^2}$

$6.0\:\frac{rad}{s^2}$

Correct answer:

$36\:\frac{rad}{s^2}$

Explanation:

For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is

$a=\alpha r$

Linear acceleration is given by , angular acceleration is $\alpha$ , and the radius of the circular path is .

For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by

$a_{centripetal}=\frac{v^{2}}{r}$

We know the linear velocity is $9.0 \:\frac{m}{s}$ , and the radius is 1.5 m, so we can find the linear acceleration...

$a_{c}=\frac{(9.0\:\frac{m}{s})^{2}}{1.5\:m}=54\:\frac{m}{s^2}$

Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...

$\alpha=\frac{a}{r}=\frac{54\:\frac{m}{s^2}}{1.5\Lm}=36\:\frac{rad}{s^2}$

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Example Question #1 : Circular And Rotational Motion

If it takes a bike wheel 3 seconds to complete one revolution, what is the wheel's angular velocity?

Possible Answers:

$0.84s^{-1}$

$3.37s^{-1}$

$2.09s^{-1}$

$1.28s^{-1}$

Correct answer:

$2.09s^{-1}$

Explanation:

The definition of angular velocity is $\omega=\frac{\Delta \Theta }{\Delta t}$ .

By identifying the given information to be $\Delta \Theta =2\pi$ and $\Delta t =3s$ , we can plug this into the equation to calculate the angular velocity:

$\omega=\frac{2\pi}{3 s}=2.09s^-1$

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Example Question #2 : Circular And Rotational Motion

What is the angular velocity of the second hand of a clock?

Possible Answers:

$\frac{60 rad}{\pi s}$

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

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

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

$\frac{\pi rad}{60s}$

Correct answer:

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

Explanation:

The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering $2\pi rad$ in one minute, or 60s. The formula for angular velocity is:

$\omega = \frac{\theta}{t}$

So the angular velocity, $\omega$ is $\frac{2\pi rad}{60s}$ , which simplifies to our answer, $\frac{\pi rad}{30s}$

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Example Question #1 : Angular Velocity And Acceleration

What is the difference in the angular velocity of the second hand of radius 1cm on a wristwatch, compared to the second hand of radius 5m on a large clock tower?

Possible Answers:

The clocktower second hand has an angular velocity that is 20 times faster than that of the wristwatch

The clocktower second hand has an angular velocity that is 500 times slower than that of the wristwatch

The clocktower second hand has an angular velocity that is 5 times faster than that of the wristwatch

No difference

The clocktower second hand has an angular velocity that is 500 times faster than that of the wristwatch

Correct answer:

No difference

Explanation:

The angular velocity should not change based on the radius of the second hand. No matter what size the second hand, it will still cover one revolution every minute or 60s. The linear velocity will be greater and the angular momentum will also be greater for the clocktower, but its angular velocity will be the same. This can be seen by looking at the equation for angular velocity:

$\omega=\frac{\theta}{t}$

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Other Harmonic Systems

Example Question #1 : Other Harmonic Systems

Example Question #2 : Other Harmonic Systems

Example Question #991 : Ap Physics 1

Example Question #5 : Other Harmonic Systems

Example Question #1 : Circular And Rotational Motion

Example Question #2 : Angular Velocity And Acceleration

Example Question #1 : Circular And Rotational Motion

Example Question #2 : Circular And Rotational Motion

Example Question #1 : Angular Velocity And Acceleration

All AP Physics 1 Resources

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