AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #31 : Motion In One Dimension

A rocket ship moves at \(\displaystyle 200\frac{m}{s}\)when a part breaks off and falls straight down. The rocket continues to move at the same velocity after the part breaks off.

How far below the ship is the part after 5 seconds?

\(\displaystyle g=10\frac{m}{s^{2}}\)

Possible Answers:

\(\displaystyle 1000m\)

\(\displaystyle 125m\)

\(\displaystyle 875m\)

\(\displaystyle 75m\)

\(\displaystyle 275m\)

Correct answer:

\(\displaystyle 125m\)

Explanation:

We use the equation:

\(\displaystyle x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\) 

With \(\displaystyle x_{0}=0m\), \(\displaystyle v_{0}=200\frac{m}{s}\), \(\displaystyle a=-10\frac{m}{s^{2}}\) , and \(\displaystyle t=5s\) we have:

\(\displaystyle x=0+200*5+\frac{1}{2}*-10*5^{2}\)

\(\displaystyle x=875m\)

The part moves 875 meters upward in the time it breaks from the ship. Then, we use:

\(\displaystyle x=vt\) with \(\displaystyle v=200\frac{m}{s}\) and \(\displaystyle t=5s\) to find that the ship has moved 1000 meters during the same 5 second period.

So the ship moved upward 1000 meters in the 5 second time period while the part moved 875 meters upward so to find the distance between the ship and the part as the problem statement requested we subtract:

\(\displaystyle 1000m-875m=125m\)

Example Question #271 : Newtonian Mechanics

A car moving to the right at \(\displaystyle 20\frac{m}{s}\) accelerates to the right at \(\displaystyle 3\frac{m}{s^{2}}\) for 14 meters.

What is the cars velocity after the acceleration period?

Possible Answers:

\(\displaystyle 20\frac{m}{s}\)

\(\displaystyle 22\frac{m}{s}\)

\(\displaystyle 26\frac{m}{s}\)

\(\displaystyle 29\frac{m}{s}\)

\(\displaystyle 18\frac{m}{s}\)

Correct answer:

\(\displaystyle 22\frac{m}{s}\)

Explanation:

We use the equation:

\(\displaystyle v_{f}^{2}=v_{0}^{2}+2ax\) 

Where the initial velocity is \(\displaystyle v_{o}=20\frac{m}{s}\), the acceleration is \(\displaystyle a=3\frac{m}{s^{2}}\) , and the displacement during the acceleration \(\displaystyle x=14m\).

Plug in those values into our equation and solve.

\(\displaystyle v_f^2=(20\frac{m}{s})^2+2*(3\frac{m}{s^2})*14m\)

\(\displaystyle v_f=\sqrt{400+84}=\sqrt{484}=22\frac{m}{s}\)

Example Question #34 : Motion In One Dimension

Calculate the acceleration, \(\displaystyle a\) of a race car that can go a distance of \(\displaystyle 1320ft\) in \(\displaystyle 3.5s\), assuming the acceleration is constant and the car starts from rest.

\(\displaystyle 1ft\approx 0.3048m\)

Possible Answers:

\(\displaystyle 215.5 \frac{m}{s^2}\)

\(\displaystyle 5.4\frac{m}{s^2}\)

\(\displaystyle 229.7\frac{m}{s^2}\)

\(\displaystyle 65.7 \frac{m}{s^2}\)

Correct answer:

\(\displaystyle 65.7 \frac{m}{s^2}\)

Explanation:

However unrealistic, if the acceleration of the race car were constant, we could calculate a value. We need everything in SI units, so we have to convert the distance to \(\displaystyle m\)

\(\displaystyle 1320ft\left(\frac{0.3048m}{1ft}\right )=402.3m\)

Using a kinematic equation, we can directly calculate the acceleration \(\displaystyle a\),

\(\displaystyle \Delta x = \frac{1}{2} a t^2 \Rightarrow a = \frac{2 \Delta x}{t^2}=\frac{2(402.3m)}{(3.5s)^2}=65.7\frac{m}{s^2}\)

Example Question #311 : Ap Physics 1

A block is placed at a height of \(\displaystyle 18m\) on a ramp at an incline of \(\displaystyle 24^o\). It slides down from rest with an acceleration of \(\displaystyle 2.2\frac{m}{s^2}\) due to a frictional force. What is the velocity of the block when it reaches the bottom of incline?

24_deg._incline_block

Possible Answers:

\(\displaystyle v_f=12\frac{m}{s}\)

\(\displaystyle v_f=16\frac{m}{s}\)

\(\displaystyle v_f=5.1\frac{m}{s}\)

\(\displaystyle v_f=8.9\frac{m}{s}\)

\(\displaystyle v_f=14\frac{m}{s}\)

Correct answer:

\(\displaystyle v_f=14\frac{m}{s}\)

Explanation:

To find the final velocity we use the following kinematic equation.

\(\displaystyle v_f^2=v_o^2+2a\Delta x\)

The block starts from rest, so \(\displaystyle v_o=0\). The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

\(\displaystyle \Delta x=\frac{18m}{sin24^\circ}\)

\(\displaystyle \Delta x = 44.25m\)

We know that the acceleration is \(\displaystyle a=2.2\frac{m}{s^2}\).

Plug in the values to find the velocity at the bottom of the incline.

\(\displaystyle v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)\)

\(\displaystyle v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}\)

Example Question #37 : Motion In One Dimension

Object A is traveling at \(\displaystyle < 10,0>\frac{m}{s}\). It is hit from behind by object B traveling at \(\displaystyle < 15,0>\frac{m}{s}\). If they both have a mass of \(\displaystyle 25kg\), determine their final velocity if they stick together.

Possible Answers:

\(\displaystyle < 0,12.5>\frac{m}{s}\)

\(\displaystyle < 15.0,0>\frac{m}{s}\)

\(\displaystyle < 15,15>\frac{m}{s}\)

\(\displaystyle < 0,0>\frac{m}{s}\)

\(\displaystyle < 12.5,0>\frac{m}{s}\)

Correct answer:

\(\displaystyle < 12.5,0>\frac{m}{s}\)

Explanation:

\(\displaystyle \overrightarrow{P}_{Total}=\overrightarrow{P}_1+\overrightarrow{P}_2...\)

Definition of momentum:

\(\displaystyle \overrightarrow{P}=m\overrightarrow{v}\)

Combine equations:

\(\displaystyle \overrightarrow{P}_{Total}=m_1\overrightarrow{v}_1+m_2\overrightarrow{v}_2\)

Plug in values:

\(\displaystyle \overrightarrow{P}_{Total}=25*< 10,0>+25*< 15,0>\)

\(\displaystyle \overrightarrow{P}_{Total}=< 250,0>+< 375,0>\)

\(\displaystyle \frac{\overrightarrow{P}_{Total}}{m}=\frac{< 625,0>kg\frac{m}{s}}{50kg}\)

\(\displaystyle \overrightarrow{v}=< 12.5,0>\frac{m}{s}\)

Example Question #311 : Ap Physics 1

A ball is dropped from a building of height 45m. How long does it take for the ball to hit the ground? Ignore air resistance.

Possible Answers:

\(\displaystyle 10s\)

\(\displaystyle 2s\)

\(\displaystyle 9s\)

\(\displaystyle 3s\)

\(\displaystyle 5s\)

Correct answer:

\(\displaystyle 3s\)

Explanation:

Let's first list the values we know.

\(\displaystyle a=10\frac{m}{s^2}\)

\(\displaystyle v_o=0\frac{m}{s}\)

\(\displaystyle d=45m\)

We are looking for the time it takes for this ball to reach the ground, thus we are looking for \(\displaystyle t\). The kinematics equation we need to use is:

\(\displaystyle d=v_ot +\frac{1}{2}at^2\)

Since the initial velocity is zero, the \(\displaystyle v_ot\) term disappears. Thus we are left with:

\(\displaystyle 45m=\frac{1}{2}*10\frac{m}{s^2}t^2\)

Solve for time.

\(\displaystyle t=3s\)

We only take the positive value in our calculation since time cannot be negative. 

Example Question #41 : Linear Motion And Momentum

Suppose that two people are ice skating. Person A has a mass of \(\displaystyle 100\:kg\) and person B has a mass of \(\displaystyle 50\:kg\). At one point, both of them are standing on the ice together while not moving. If both of them push away from each other and slide across the frictionless ice, what will be the speed of person B compared to person A?

Possible Answers:

The speed of Person B will be equal to the speed of Person A

The speed of Person B will be twice as much as the speed of Person A

The speed of Person B will be half as much as the speed of Person A

The speed of Person B will be one-fourth the speed of Person A

Correct answer:

The speed of Person B will be twice as much as the speed of Person A

Explanation:

In this question, we're presented with a scenario in which two people are skating on ice. Initially, the two of them are standing still, and then they push away from each other. We then have to figure out the relative velocities of the two people.

To answer this question, we'll need to consider the momentum of the two skaters. Due to the conservation of momentum, we know that the initial momentum of the two skaters together will be equal to the momentum of the two skaters after they have pushed away from each other. We can express this in equation form as follows.

\(\displaystyle (m_{A}+m_{B})v_{i}=m_{A}v_{A}+m_{B}\)

Initially, the two skaters are standing still. Thus, the initial velocity \(\displaystyle v_{i}\) is equal to \(\displaystyle 0\). This means that the initial momentum of these skaters is \(\displaystyle 0\). Since the initial momentum is equal to \(\displaystyle 0\), then the final momentum must also be equal to zero. Therefore, we can rewrite the above equation as follows.

\(\displaystyle 0=m_{A}v_{A}+m_{B}v_{B}\)

\(\displaystyle -m_{A}v_{A}=m_{B}v_{B}\)

\(\displaystyle -(\frac{m_{A}}{m_{B}})v_{A}=v_{B}\)

Now, if we plug in the values for the masses of each person, we can obtain our answer.

\(\displaystyle -(\frac{100\: kg}{50\: kg})v_{A}=v_{B}\)

\(\displaystyle -2v_{A}=v_{B}\)

A few things are worth mentioning. The negative sign in the above expression means that the velocities of both people are in opposite directions, which we would expect considering that the two are pushing away from one another. Furthermore, we can see that the speed of person B is twice as much as the speed of person A because person B's mass was half that of person A's mass.

Example Question #42 : Linear Motion And Momentum

When striking, a mosquito can accelerate from rest to a max speed of \(\displaystyle 12.1\frac{m}{s}\) in \(\displaystyle 0.119 seconds\). What is the mosquito's acceleration during this attack? For this question ignore the effects of both air resistance and the free fall acceleration due to gravity.

Possible Answers:

\(\displaystyle 107\frac{m}{s}\)

\(\displaystyle 111\frac{m}{s}\)

\(\displaystyle 102\frac{m}{s}\)

\(\displaystyle 105\frac{m}{s}\)

\(\displaystyle 109\frac{m}{s}\)

Correct answer:

\(\displaystyle 102\frac{m}{s}\)

Explanation:

Use the equation for acceleration. 

\(\displaystyle a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}\)

Plug in the known values and solve for the acceleration.

\(\displaystyle a=\frac{12.1\frac{m}{s}-o\frac{m}{s}}{0.119s-0s}\)

\(\displaystyle a=101.6\frac{m}{s^{2}}\)

The mosquito has an incredible acceleration of \(\displaystyle 102m/s^{2}\).

Example Question #41 : Linear Motion And Momentum

A rock is dropped from the top of Rampart High School radio tower at a height of \(\displaystyle 50 m\). Neglecting air resistance, how long does it take the rock to contact the ground below? 

Possible Answers:

\(\displaystyle 2.2 s\)

\(\displaystyle 2.6 s\)

\(\displaystyle 3.2 s\)

\(\displaystyle 4.2 s\)

\(\displaystyle 2.4 s\)

Correct answer:

\(\displaystyle 3.2 s\)

Explanation:

Use a kinematic equation to solve for the final time.

\(\displaystyle X_f=X_i+V_o(\Delta t)+\frac{1}{2}(a)(\Delta t)^{^{2}}\)

Plug in the known values and solve for the final time.

\(\displaystyle 0m=50m+0\frac{m}{s}(\Delta t)+\frac{1}{2}(-9.8\frac{m}{s^{2}})(\Delta t)^{2}\)

\(\displaystyle -50m=\frac{1}{2}(-9.8\frac{m}{s^{2}})(t_{f}-0s)\)

\(\displaystyle t_{f}=3.2 s\)

The rock will take \(\displaystyle 3.2 s\) to fall from the top of the building to the ground below.

Example Question #311 : Ap Physics 1

A squirrel in a tree drops an acorn from a height of \(\displaystyle 10 m\) above your head. Neglecting air resistance, how fast will the acorn be traveling when it makes contact with your head?

Possible Answers:

\(\displaystyle 20\frac{m}{s}\)

\(\displaystyle 5\frac{m}{s}\)

\(\displaystyle 25\frac{m}{s}\)

\(\displaystyle 15\frac{m}{s}\)

\(\displaystyle 10\frac{m}{s}\)

Correct answer:

\(\displaystyle 15\frac{m}{s}\)

Explanation:

Use a kinematic equation to solve for the time the acorn takes to fall to your head.

\(\displaystyle X_f=X_i+V_o(\Delta t)+\frac{1}{2}(a)(\Delta t)^{2}\)

Plug in the known values and solve for the final time.

\(\displaystyle 0m=10m+0\frac{m}{s}(\Delta t)+\frac{1}{2}(-9.8\frac{m}{s^{2}})(\Delta t)^{2}\)

\(\displaystyle -10m=(-4.9\frac{m}{s^{2}})(t_{f}-0s)^{2}\)

\(\displaystyle t_{f}=1.42 s\)

Use a second kinematic equation and solve for the final velocity.

\(\displaystyle V_f=V_o+a(\Delta t)\)

Plug in the known values and solve for the final velocity.

\(\displaystyle V_{f}=0\frac{m}{s}+(-9.8\frac{m}{s^{2}})1.42s\)

\(\displaystyle V_{f}=13.9\frac{m}{s}\)

When rounding and using significant figures the final answer would be \(\displaystyle 15 s\).

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