AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #14 : Conservation Of Energy

2 objects(named object A and object B) of equal masses and initial kinetic energy collide onto one another. During the collision, object A loses \(\displaystyle \frac{1}{2}\) of its kinetic energy, which object B gains. Assume mass of both objects remain unchanged. 

By what factor will the velocity of the object A change after the collision? 

Possible Answers:

Velocity will change by a factor of \(\displaystyle \frac{1}{4}\)

Velocity will change by a factor of \(\displaystyle \sqrt\frac{1}{2}\)

Velocity will change by a factor of \(\displaystyle \sqrt 2\)

Velocity will change by a factor of \(\displaystyle \frac{1}{2}\)

Correct answer:

Velocity will change by a factor of \(\displaystyle \sqrt\frac{1}{2}\)

Explanation:

Kinetic energy \(\displaystyle E_k\) is related to velocity \(\displaystyle v\) and mass \(\displaystyle m\) by

\(\displaystyle E_k=\frac{1}{2}mv^2\)

Object A's kinetic energy has decreased by \(\displaystyle \frac{1}{2}\). We know that its mass hasn't changed. 

Relate velocity to kinetic energy, while mass is not changing

\(\displaystyle \sqrt E_k\propto v\)

Since \(\displaystyle E_k\) has changed by a factor of \(\displaystyle \frac{1}{2}\)\(\displaystyle v\) will change by a factor of

\(\displaystyle \sqrt\frac{1}{2}\propto v\)

Example Question #161 : Newtonian Mechanics

Coaster

A roller coaster cart is starting at rest at a height of \(\displaystyle 40m\). It goes down the track and around a loop. Assuming there is no friction between the track and the cart, what is the velocity of the cart when it has come out of the loop at the bottom of the track in miles per hour?

Possible Answers:

None of these

\(\displaystyle v=1753 \frac{miles}{hour}\)

\(\displaystyle v=12.3\frac{miles}{hour}\)

\(\displaystyle v=62.6\frac{miles}{hour}\)

Correct answer:

\(\displaystyle v=62.6\frac{miles}{hour}\)

Explanation:

Conservation of energy is a very powerful tool for solving some mechanics problems. If we used Newtonian mechanics this would be a lot more difficult. The conservation of energy for the cart denotes:

\(\displaystyle K_{i}+U_{i}=K_{f}+U_{f}\)

Where \(\displaystyle K\) is the kinetic energy of the system and \(\displaystyle U\) is the potential energy.  \(\displaystyle K\) and \(\displaystyle U\) are defined by:

\(\displaystyle K=\frac{1}{2}m v^{2}\)

\(\displaystyle U=mgh\)

When the cart is at the top of the track it is at rest. This means that there is no initial kinetic energy. When the cart is at the bottom of the track it's height is zero, so it has no final potential energy. Therefore,

\(\displaystyle U_{i}=K_{f} \rightarrow mgh=\frac{1}{2}mv^{2}\)

Notice that the mass cancels. Solve for \(\displaystyle v\):

\(\displaystyle v^{2}=2gh \rightarrow v=\sqrt{2gh}=28\frac{m}{s}\)

Convert to miles per hour:

\(\displaystyle 28 \frac{m}{s}\left(\frac{6.21\times 10^{-4}miles}{m} \right )\left (\frac{60s}{1min}\right ) \left(\frac{60min}{1hour} \right )=62.6\frac{miles}{hour}\)

Example Question #202 : Ap Physics 1

You are in Paris, France, holding on to a tennis ball of mass \(\displaystyle 58.1 grams\).

 

You throw it straight up. It leaves your hand two meters above the ground at a speed of \(\displaystyle 15 m/s\).

 

What is the maximum height above the ground the ball obtains?

 

You may use\(\displaystyle \frac{9.8 m}{s^2}\) as your acceleration.

Possible Answers:

\(\displaystyle 13.5 meters\)

None of these

\(\displaystyle 20.1 meters\)

\(\displaystyle 10.6 meters\)

\(\displaystyle 12.2 meters\)

Correct answer:

\(\displaystyle 13.5 meters\)

Explanation:

We will use conservation of energy to help us.

 

We will treat our initial situation as the moment the ball left the hand, and the final situation as the ball at the maximum height.

 

\(\displaystyle KE_{initial}+PE_{initial}=KE_{final}+PE_{final}\)

 

\(\displaystyle \frac{1}{2}mv_i}}^2+mgh_i=\frac{1}{2}mv_f}}^2+mgh_f\)

 

Mass cancels out

 

\(\displaystyle \frac{1}{2}v_i}}^2+gh_i=\frac{1}{2}v_f}}^2+gh_f\)

 

\(\displaystyle V_f\) will be zero at our maximum height

 

\(\displaystyle \frac{1}{2}v_i}}^2+gh_i=gh_f\)

 

Rearranging our equatin to solve for \(\displaystyle h_f\)

 

\(\displaystyle \frac{.5v_i^2+gh_i}{g}=h_f\)

 

We then plug in our values

\(\displaystyle \frac{.5\cdot15^2+9.8\cdot2}{9.8}=h_f\)


\(\displaystyle 13.5 meters=h_f\)

Example Question #203 : Ap Physics 1

An object of mass \(\displaystyle 35 kg\) falls from a \(\displaystyle 50m\) tall building.

Determine the velocity just before hitting the ground.

Possible Answers:

\(\displaystyle v=28.4\frac{m}{s}\)

\(\displaystyle v=95.5\frac{m}{s}\)

\(\displaystyle v=31.3\frac{m}{s}\)

\(\displaystyle v=100.3\frac{m}{s}\)

None of these

Correct answer:

\(\displaystyle v=31.3\frac{m}{s}\)

Explanation:

Use conservation of energy:

\(\displaystyle E_{initial}=E_{{final}}\)

\(\displaystyle KE_1+PE_1=KE_2+PE_2\)

At the moment of dropping, there will be no velocity and thus no kinetic energy. Right before hitting the ground, there will be no height, and this no potential energy.

\(\displaystyle PE_1=KE_2\)

\(\displaystyle mgh=.5mv^2\)

Solve for velocity:

\(\displaystyle v=\sqrt{2gh}\)

\(\displaystyle v=31.3\frac{m}{s}\)

Example Question #204 : Ap Physics 1

An object of mass \(\displaystyle 35 kg\) falls from a \(\displaystyle 50m\) tall building.

Determine the momentum just before hitting the ground.

Possible Answers:

\(\displaystyle P=1003\frac{kg*m}{s}\)

None of these

\(\displaystyle P=678\frac{kg*m}{s}\)

\(\displaystyle P=1096\frac{kg*m}{s}\)

\(\displaystyle P=2103\frac{kg*m}{s}\)

Correct answer:

\(\displaystyle P=1096\frac{kg*m}{s}\)

Explanation:

Use conservation of energy

\(\displaystyle E_{initial}=E_{{final}}\)

\(\displaystyle KE_1+PE_1=KE_2+PE_2\)

At the moment of dropping, there will be no velocity and thus no kinetic energy. Right before hitting the ground, there will be no height, and this no potential energy.

\(\displaystyle PE_1=KE_2\)

\(\displaystyle mgh=.5mv^2\)

Solve for velocity:

\(\displaystyle v=\sqrt{2gh}\)

\(\displaystyle v=31.3\frac{m}{s}\)

Use definition of momentum:

\(\displaystyle P=mv\)

\(\displaystyle P=35kg*31.3\frac{m}{s}\)

\(\displaystyle P=1096\frac{kg*m}{s}\)

Example Question #205 : Ap Physics 1

An object of mass \(\displaystyle 35 kg\) falls from a \(\displaystyle 50m\) tall building.

Determine the potential energy after the object has fallen \(\displaystyle 25m\).

Possible Answers:

\(\displaystyle PE=5589J\)

\(\displaystyle PE=4449J\)

\(\displaystyle PE=8575J\)

None of these

\(\displaystyle PE=3595J\)

Correct answer:

\(\displaystyle PE=8575J\)

Explanation:

\(\displaystyle H_i+\Delta H=H_f\)

\(\displaystyle 50m-25m=25m\)

Use the formula for potential energy due to gravity:

\(\displaystyle PE=mgh\)

\(\displaystyle PE=35kg*9.8\frac{m}{s^2}*25m\)

\(\displaystyle PE=8575J\)

Example Question #206 : Ap Physics 1

An object of mass \(\displaystyle 35 kg\) falls from a \(\displaystyle 50m\) tall building.

Determine the kinetic energy of the object 1.5S after being dropped.

Possible Answers:

\(\displaystyle 3.8*10^3J\)

\(\displaystyle 8.5*10^3J\)

None of these

\(\displaystyle 2.2*10^3J\)

\(\displaystyle 3.5*10^3J\)

Correct answer:

\(\displaystyle 3.8*10^3J\)

Explanation:

First, use the velocity update kinematic equation:

\(\displaystyle v_i+a\Delta t=v_f\)

\(\displaystyle 0\frac{m}{s}+9.8\frac{m}{s^2}1.5s=v_f\)

\(\displaystyle 14.7\frac{m}{s}=v_f\)

Then use the definition of kinetic energy:

\(\displaystyle KE=.5mv^2\)

\(\displaystyle KE=.5*35*14.7^2\)

\(\displaystyle KE=3.8*10^3J\)

Example Question #22 : Conservation Of Energy

A solid disk of mass \(\displaystyle M\) and radius \(\displaystyle R\) rolls down a frictionless incline of height \(\displaystyle h\) and angle \(\displaystyle \theta\). The disk starts from rest at the top of the ramp, and the moment of inertia of the disk is \(\displaystyle I=\frac{1}{2}MR^2\)

If the mass rolls without slipping, what is the linear velocity of the mass at the bottom of the ramp in terms of \(\displaystyle M\)\(\displaystyle R\)\(\displaystyle h\)\(\displaystyle g\), and \(\displaystyle \theta\)?

Possible Answers:

\(\displaystyle \sqrt{gh}\)

\(\displaystyle \sqrt{2gh}\)

\(\displaystyle \sqrt{\frac{4gh}{3}}\)

\(\displaystyle \sqrt{\frac{4Rgh}{3}}\)

\(\displaystyle \sqrt{Rgh}\)

Correct answer:

\(\displaystyle \sqrt{\frac{4gh}{3}}\)

Explanation:

Since the ramp is frictionless and the disk rolls without slipping, we can infer two things: one conservation of energy applies here, and two we have the condition that \(\displaystyle v=R\omega\), where \(\displaystyle v\) is the linear velocity, and \(\displaystyle \omega\) is the angular velocity. For convenience, we will set the potential energy to 0 at the bottom of the ramp, since then for the final energy of the disk will not contain any potential energy terms. Now the initial energy of the disk \(\displaystyle E_i\) will be all potential energy, since the disk starts rolling from rest. Thus \(\displaystyle E_i=Mgh\), its gravitational potential energy. At the bottom of the ramp, it will not have any potential energy, but it will have two types of kinetic energy: translational kinetic energy due to the fact that the disk itself is moving down the ramp, and rotational kinetic energy due to the fact that the disk is rotating about its center. Thus its final energy \(\displaystyle E_f\) is given by 

\(\displaystyle E_f=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2\) 

Where the first term is the translational kinetic energy contribution, and the second term is the rotational kinetic energy contribution. Now, applying conservation of energy gives us:

\(\displaystyle E_i=E_f\)

\(\displaystyle Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2\)

Since we are solving for \(\displaystyle v\) the linear velocity, we use the fact that \(\displaystyle \omega=\frac{v}{R}\) and also we replace the moment of inertia \(\displaystyle I\) with \(\displaystyle \frac{1}{2}MR^2\) to obtain the following:

\(\displaystyle Mgh=\frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2\)

\(\displaystyle Mgh=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2\)

\(\displaystyle Mgh=\frac{3}{4}Mv^2\)

Now we solve for \(\displaystyle v\) by multiplying each side by \(\displaystyle \frac{4}{3M}\) and then taking the square root. We obtain: 

\(\displaystyle \frac{4gh}{3}=v^2\)

\(\displaystyle v=\sqrt{\frac{4gh}{3}}\)

Thus the linear velocity is given by \(\displaystyle \sqrt{\frac{4gh}{3}}\)

Example Question #207 : Ap Physics 1

A spring-loaded pop gun fires a dart out of the muzzle at \(\displaystyle 20\frac{m}{s}\). The spring inside is compressed a distance of \(\displaystyle 20cm\) and the dart has a mass of \(\displaystyle 100g\).

What is the spring constant for the spring in the popgun?

Possible Answers:

\(\displaystyle 1000\frac{N}{m}\)

\(\displaystyle 100\frac{N}{m}\)

\(\displaystyle 1000000\frac{N}{m}\)

\(\displaystyle 0.1\frac{N}{m}\)

\(\displaystyle 10000\frac{N}{m}\)

Correct answer:

\(\displaystyle 1000\frac{N}{m}\)

Explanation:

The popgun compresses a spring and stores potential energy in the spring. The spring then releases and fires the dart, converting all of the potential energy in the spring into kinetic energy which launches the dart out of the gun at \(\displaystyle 20\frac{m}{s}\). Therefore, we can use conservation of energy to find the value of the spring constant. The initial energy is given by

\(\displaystyle E_i=\frac{1}{2}kx^2\)

Where \(\displaystyle k\) is the spring constant and \(\displaystyle x\) is the compression distance. Because the dart is sitting in the gun, there is no initial kinetic energy. The final energy is given by 

\(\displaystyle E_f=\frac{1}{2}mv^2\)

Where \(\displaystyle m\) is the mass of the dart, and \(\displaystyle v\) is the velocity of the dart. From the givens in the problem, we know that 

\(\displaystyle m=100g=0.1kg\)\(\displaystyle x=20cm=0.2m\), and \(\displaystyle v=20\frac{m}{s}\) 

Now we apply conservation of energy. We have:

\(\displaystyle E_i=E_f\)

\(\displaystyle \frac{1}{2}kx^2=\frac{1}{2}mv^2\)

To solve for \(\displaystyle k\), we multiply each side of the equation by \(\displaystyle \frac{2}{x^2}\) to obtain:

\(\displaystyle k=\frac{mv^2}{x^2}=\frac{(0.1kg)(20\frac{m}{s})^2}{(0.2m)^2}=1000\frac{N}{m}\)

Therefore the spring constant is \(\displaystyle k=1000\frac{N}{m}\).

Example Question #21 : Conservation Of Energy

An trampolinist's motion is tracked by a software to determine their motion. The trampolinist weighs \(\displaystyle 50kg\).

Suppose the trampolinist is traveling up at a speed of \(\displaystyle 2\frac{m}{s}\) at \(\displaystyle 7m\) up, reaches a peak of \(\displaystyle 10m\), and comes back down to the same height of \(\displaystyle 7m\) coming down at \(\displaystyle 1.5\frac{m}{s}\). Determine the energy lost to the environment. 

Possible Answers:

\(\displaystyle 87.5J\)

\(\displaystyle 43.75J\)

\(\displaystyle 12.5J\)

\(\displaystyle 0J\)

Correct answer:

\(\displaystyle 43.75J\)

Explanation:

Energy must be conserved. To determine energy lost to the environment, we simply subtract the energies when the trampolinist was going up to the energy they have coming down.

Since the trampolinist is at the same height, the gravitational potential energy is the same. However, their kinetic energy has changed, and this change is equivalent to the energy gained by the environment:

\(\displaystyle E_{final}-E_{initial}=\frac{1}{2}mv_{final}^2-\frac{1}{2}mv_{initial}^2=E_{lost}\), where \(\displaystyle m\) is mass and \(\displaystyle v\) is their velocity. 

Since we know both velocities and masses, we plug into the above equation:

\(\displaystyle E_{lost}=\frac{1}{2}50kg(\frac{1.5m}{s})^2^ - \frac{1}{2}50kg(\frac{2m}{s})^2=-43.75J\)

Since that much energy was lost, that must have been gained by the environment:

\(\displaystyle E_{gained}=-E_{lost}=43.75J\)

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