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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #51 : Motion In Two Dimensions

A ball rolls off of a 200m high cliff at a constant velocity of $20\frac{m}{s}$ . Neglecting air friction, what is the acceleration of the ball in the vertical and horizontal direction just before it hits the ground?

Possible Answers:

Vertical: $0\frac{m}{s^{2}}$

Horizontal: $0\frac{m}{s^{2}}$

Vertical: $9.8\frac{m}{s^{2}}$

Horizontal: $9.8\frac{m}{s^{2}}$

Vertical: $-9.8\frac{m}{s^{2}}$

Horizontal: $0\frac{m}{s^{2}}$

Vertical: $0\frac{m}{s^{2}}$

Horizontal: $-9.8\frac{m}{s^{2}}$

Correct answer:

Vertical: $-9.8\frac{m}{s^{2}}$

Horizontal: $0\frac{m}{s^{2}}$

Explanation:

The acceleration due to gravity only affects the object in the vertical direction, which is always $-9.8\frac{m}{s^{2}}$ . Because there is no external force acting on the ball in the horizontal direction there is no acceleration.

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Example Question #53 : Motion In Two Dimensions

During time period , a rocket ship deep in space of mass $2.55*10^5\textup{ kg}$ travels from $<0,10>\textup{ km}$ to $<5,5>\textup{ km}$ . During time period , the rocket fires. During time period , the rocket travels from $<-15,0>\textup{ km}$ to $<-30,-5>\textup{ km}$ .

Time periods , , and took $10\textup{ s}$ each.

Determine the momentum during time period .

Possible Answers:

None of these

$<3.985*10^8,-3.985*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<1.275*10^8,1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<-1.275*10^8,1.275*10^8>\frac{\textup{kg} \cdot \textup{m}}{\textup{s}}$

$<1.275*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

$<1.275*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Explanation:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

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Example Question #51 : Motion In Two Dimensions

During time period , a rocket ship deep in space of mass 2.55*10^5kg travels from <0,10>km to <5,5>km . During time period , the rocket fires. During time period , the rocket travels from <-15,0>km to <-30,-5>km .

Time periods , , and all took 10seconds.

Determine the momentum during time period .

Possible Answers:

$\overrightarrow{P}=<-3.825*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

None of these

$<3.825*10^8,1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<3.985*10^8,3.985*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$<8.265*10^8,7.600*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

$\overrightarrow{P}=<-3.825*10^8,-1.275*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Explanation:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Converting to and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

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Example Question #54 : Motion In Two Dimensions

Suppose that a person has walked $24\: m$ north and $12\: m$ east. What is this person's net displacement from where he initially started?

Possible Answers:

$36.0\: m$

$18.0\:m$

$3.46\:m$

$26.8\:m$

$12.0\:m$

Correct answer:

$26.8\:m$

Explanation:

To answer this question, it's helpful if we draw the path this person takes.

Vt physics 11 23 15 vector chart

Now, it's important to realize the difference between distance and displacement. The distance this person has traveled is the total amount of walking he has done. So in this case, the distance traveled would be:

$24\:m+12\:m=36\:m$

However, this is not the answer to our question. The question is asking for the displacement, which is the net change in position of this person. We can draw it as follows:

Vt physics 11 23 15 vector displacement

Notice that in the above drawing, we have a right triangle. This will actually simplify things a bit, because we can use trigonometry to find the angle of displacement (shown above in blue). Then, we can use that information to find the displacement (shown above in green).

To find the angle of displacement, we can use the following expression:

$tan(\Theta )=\frac{opposite}{adjacent}=\frac{12\: m}{24\: m}=0.5$

$\Theta=tan^{-1}(0.5)=26.565^{o}$

Now that we have the angle, we can calculate the displacement using further trigonometry.

$sin(\Theta )=\frac{opposite}{hypotenuse}=\frac{12\: m}{Displacement}$

$sin(26.565)=\frac{12\: m}{Displacement}$

$Displacement=\frac{12\: m}{sin(26.565^{o})}=\frac{12\: m}{0.447}=26.8\:m$

An alternative way to solve this problem, and arguably much easier, is to just use the Pythagorean theorem since we have a right triangle.

$Displacement=\sqrt{(12\: m)^{2}+(24\: m)^{2}}=26.8\:m$

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Example Question #57 : Motion In Two Dimensions

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

Determine the angle between the initial velocity of the football and the ground.

Possible Answers:

$\theta=33^\circ$

$\theta=77^\circ$

None of these

$\theta=18^\circ$

$\theta=66^\circ$

Correct answer:

$\theta=66^\circ$

Explanation:

Using:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$cos(\theta)=\frac{v_x}{v}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Combining equations

$cos(\theta)=\frac{d_x/t}{v}$

Plugging in values:

$cos(\theta)=\frac{45.7/5}{22.35}$

$\theta=67^\circ$

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Example Question #55 : Motion In Two Dimensions

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

Determine the initial vector velocity of the football.

Possible Answers:

None of these

$\overrightarrow{v}=<44.1,22.3>\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$

$\overrightarrow{v}=<20.4,9.14>\frac{m}{s}$

$\overrightarrow{v}=<17.5,16.1>\frac{m}{s}$

Correct answer:

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$

Explanation:

Determining horizontal component of velocity:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

Using

v_x^2+v_y^2=v^2

Solving for v_y

$v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$

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Example Question #53 : Motion In Two Dimensions

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

Determine the speed at the apex of the throw.

Possible Answers:

$10.2\frac{m}{s}$

$9.81\frac{m}{s}$

None of these

$9.14\frac{m}{s}$

$9.44\frac{m}{s}$

Correct answer:

$9.14\frac{m}{s}$

Explanation:

At the apex, the only velocity will be in the direction as the ball is at it's maximum height.

Using:

d_x=v_x*t

Solving for velocity

$v_x=\frac{d_x}{t}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

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Example Question #152 : Linear Motion And Momentum

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

Determine the speed of the football 4 seconds after the throw.

Possible Answers:

None of these

$<-9.14,-19.2>\frac{m}{s}$

$<9.14,-10.1>\frac{m}{s}$

$<9.14,-19.2>\frac{m}{s}$

$<9.14,19.2>\frac{m}{s}$

Correct answer:

$<9.14,-19.2>\frac{m}{s}$

Explanation:

Determining horizontal component of velocity:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

Using

v_x^2+v_y^2=v^2

Solving for v_y

$v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$

Using

v_i+a*t=v_f

<9.14,20.4>+<0,-9.8>*4=v_f

$<9.14,-19.2>\frac{m}{s}=v_f$

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Example Question #151 : Linear Motion And Momentum

Two space ships are racing deep in space. At t=0 , Spaceship A is at the origin while traveling at $<600,400> \frac{m}{s}$ . It is coasting, with it’s rockets off. Spaceship B is at <-600,-400>m . It’s traveling at $<150,100>\frac{m}{s}$ . It’s rockets are firing with a constant force of <1.5*10^6N, 1*10^6N> . Both rockets have a mass of 1.5*10^5 kg .

At what time will spaceship B pass Spaceship A if this continues? Assume that they are at different locations in the z-axis and thus will not crash.

Possible Answers:

t=91.3s

None of these

t=98.9s

t=12.5s

t=46.7s

Correct answer:

t=91.3s

Explanation:

Using position update formula:

r=r_i+v_i*t+.5at^2

The positions need to be equal, so:

r_i_B+v_i_B*t+.5a_Bt^2=r_i_A+v_i_A*t+.5a_At^2

Converting into quadratic form:

0=(.5a_B-.5a_A)t^2+(v_i_B-v_i_A)t+(r_i_B-r_i_A)

Plugging in values:

0=(.5<10,6.66>-.5*<0,0>)t^2+(<150,100>-<600,400>)t+(<-600,-400>-<0,0>)

0=<5t^2,3.33t^2>+<-450t,-300t>+<-600,-400>

0=5t^2-450t-600

0=3.33t^2-300t-400

Solving for a quadratic yields:

t=91.3s

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Example Question #152 : Linear Motion And Momentum

What magnitude velocity will spaceship have at the origin?

Possible Answers:

$175\frac{m}{s}$

None of these

$325\frac{m}{s}$

$388\frac{m}{s}$

$229\frac{m}{s}$

Correct answer:

$229\frac{m}{s}$

Explanation:

Using

W=E_f-E_i

F*d=.5mv_f^2-.5mv_i^2

Solving for v_f

$\sqrt{\frac{F*d+.5mv_i^2}{.5m}}=v_f$

Plugging in values:

$\sqrt{\frac{<1.5*10^6,1.5*10^6>*<600,400>+.5*1.5*10^5*180.3^2}{.5(1.5*10^5)}}=v_f$

$v_f=229\frac{m}{s}$

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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 #51 : Motion In Two Dimensions

Example Question #53 : Motion In Two Dimensions

Example Question #51 : Motion In Two Dimensions

Example Question #54 : Motion In Two Dimensions

Example Question #57 : Motion In Two Dimensions

Example Question #55 : Motion In Two Dimensions

Example Question #53 : Motion In Two Dimensions

Example Question #152 : Linear Motion And Momentum

Example Question #151 : Linear Motion And Momentum

Example Question #152 : Linear Motion And Momentum

All AP Physics 1 Resources

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