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AP Physics 1 : Linear Motion and Momentum

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Example Questions

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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Example Question #153 : Linear Motion And Momentum

A tall man standing on a cliff that is 35m high throws a baseball completely horizontally off the cliff at a velocity of $36\frac{m}{s}$ , how far from the base of the cliff will the ball travel before hitting the ground?

Possible Answers:

99m

648m

67m

38m

512m

Correct answer:

99m

Explanation:

In order to find the distance that the ball travels horizontally one must multiply the velocity of the ball by the amount of time that the ball is traveling. Time in the air can be found by figuring out the amount of time that it takes for the ball to hit the ground relative to the vertical distance. Using the equation:

$d=V_{0}t+\frac{1}{2}at^{^{2}}$

where d is 37m (height of the cliff plus height of the man), $V_{0}$ is equal to zero because there is no initial vertical velocity, and acceleration is equal to the gravity constant, time is found to be 2.75s . In order to find distance traveled horizontally, one can use the equation:

$d=V_0t+\frac{1}{2}at^2$

Because the vertical and horizontal axis are completely independent of each other there is no acceleration in the horizontal direction, the equation then becomes simplified to:

$d=V_{0}t$

After plugging in numbers

d=36*2.75

d=99m

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Example Question #154 : Linear Motion And Momentum

Two balls are thrown off the edge of a cliff of height from the same location. Both balls are thrown with an initial velocity of $\small v$ . The first ball is thrown horizontally and the second ball is thrown at and angle of $\small 45^{\circ}$ above the horizontal. When the cliff is a particular height, let's call it $h_{equal}$ , the two balls will land at the same spot on the ground.

Which of the following is true?

Possible Answers:

The ball thrown $45^{\circ}$ above the horizontal will always go further besides when $h=h_{equal}$

There are multiple cliff heights where the two balls will land at the same location

The ball thrown $45^{\circ}$ above the horizontal will only go further than the horizontally thrown ball if $h>h_{equal}$

The ball thrown $45^{\circ}$ above the horizontal will only go further than the horizontally thrown ball if $h<h_{equal}$

The horizontally thrown ball will always go further besides when $h=h_{equal}$

Correct answer:

The ball thrown $45^{\circ}$ above the horizontal will only go further than the horizontally thrown ball if $h<h_{equal}$

Explanation:

If h=0 , the horizontally thrown ball will travel no distance. As the height of the cliff increases, the horizontally thrown ball will have more and more air time until it begins to travel further than the ball thrown $45^{\circ}$ above the horizontal. This is because the distance each ball will go equals the time it's in the air multiplied by its horizontal velocity (since it's not accelerating in the x-direction). The time the ball thrown $45^{\circ}$ above the horizontal is always the same amount greater than the time the ball thrown horizontally is in the air. The ball thrown horizontally has a horizontal velocity greater than the ball thrown $45^{\circ}$ above the horizontal (by a factor of $\sqrt{2}$ ). This means the ball thrown horizontally travels a distance of v*t while the ball thrown $45^{\circ}$ above the horizontal travels a distance of $\frac{\sqrt2v(t+k)}{2}$ . If we equate these two expressions, we get $t=(1+\sqrt{2})k$ . When is less than this, the ball thrown $45^{\circ}$ above the horizontal travels further and when is greater the ball thrown horizontally travels further. $t=(1+\sqrt{2})k$ represents the time is takes for the ball thrown horizontally to hit the ground when the height of this cliff is $h_{equal}$ .

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Example Question #433 : Ap Physics 1

Pulley

2 masses are connected to a pulley system as demonstrated by the image above. Mass 1 (m1) weighs $4.0\ kg$ and mass 2 weighs $6.5\ kg$ .

Assuming there is no friction in the system, how fast would mass 1 be accelerating?

Possible Answers:

$10\ \frac{m}{s^{2}}$

$16\ \frac{m}{s^{2}}$

$20\ \frac{m}{s^{2}}$

$12\ \frac{m}{s^{2}}$

$14\ \frac{m}{s^{2}}$

Correct answer:

$16\ \frac{m}{s^{2}}$

Explanation:

Because the mass of object 1 is given, the force acting on it must be found in order to figure out how fast it is accelerating. These 3 variables are related by the equation:

F=ma

Because there is no friction in the system, the only force acting on the object is the tension in the rope. The magnitude of the force of tension is the magnitude of the gravitational force of mass 2. Gravitational force is equal to:

F=mg

where g is the gravity constant which is equal to $9.81\frac{m}{s^{2}}$ which is also the acceleration of this object. Since this force is the only force acting on the object, accelerating can be found relating the 2 equations above as follows:

$m_{2}g=m_{1}a_{1}$

After plugging in values, the acceleration of mass 1 is found to be:

$6.5*9.81=4.0a_{1}$

$a_{1}\approx 16\ \frac{m}{s^{2}}$

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Example Question #434 : Ap Physics 1

If an object is launched $50\: \frac{m}{s}$ at an angle of $30^{o}$ with respect to the ground, what is its vertical speed after $2\: s$ ?

Possible Answers:

$5.4\: \frac{m}{s}$

$25\: \frac{m}{s}$

$16.2\: \frac{m}{s}$

$35.8\: \frac{m}{s}$

$30.4\: \frac{m}{s}$

Correct answer:

$5.4\: \frac{m}{s}$

Explanation:

To solve this question, we'll first need to break up the velocity into its component parts. For this question, we're only concerned about motion in the vertical direction, so we'll first need to find the initial velocity in the vertical direction.

$v_{yi}=v_{i}sin(\Theta )$

$v_{yi}=(50\: \frac{m}{s})sin(30^{o})=25\: \frac{m}{s}$

Now that we know what the initial velocity is, we can determine what the final velocity will be after $2\: s$ .

$\frac{\Delta v}{t}=a$

$\frac{v_{yf}-v_{yi}}{t}=a$

$v_{yf}=v_{yi}+at$

$v_{yf}=25\: \frac{m}{s}+(-9.8\: \frac{m}{s^{2}})(2\:sec)$

$v_{yf}=25\: \frac{m}{s}-19.6\: \frac{m}{s}=5.4\: \frac{m}{s}$

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Example Question #155 : Linear Motion And Momentum

A baseball hits a perfectly horizontal ball at a speed of $17\frac{m}{s}$ from a height of 1.2m . How far has the ball traveled before it hits the ground?

$g = 10\frac{m}{s^2}$

Possible Answers:

8.3m

7.6m

6.1m

4.2m

1.9m

Correct answer:

8.3m

Explanation:

First, we can calculate how long it takes for the ball to hit the ground using the following expression:

$\Delta y = v_{i}+\frac{1}{2}at^2$

Since the original vertical velocity is 0, we can rearrange to get:

$t = \sqrt{\frac{2\Delta y}{a}}$

Plugging in our values, we get:

$t = \sqrt{\frac{2(1.2m)}{10\frac{m}{s^2}}}$

t = 0.49s

Multiplying this by the velocity given in the problem statement (horizontal velocity), we get:

$d = \left ( 17\frac{m}{s}\right )(0.49s) = 8.3m$

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Example Question #436 : Ap Physics 1

A baseball player hits a ball from a height of at a velocity of $30\frac{m}{s}$ and an angle of $15^{\circ}$ above the horizontal. The ball is hit toward another player that is 25m away from the batter. At what height is the ball when it reaches the other player?

$g = 10\frac{m}{s^2}$

Possible Answers:

2.41m

1.77m

0.16m

3.98m

0.99m

Correct answer:

3.98m

Explanation:

First, let's separate the given velocity into its vertical and horizontal components:

$v_x = vcos(\theta)$

$v_y = vsin(\theta)$

$v_x= \left ( 30\frac{m}{s}\right )cos(15^{\circ})$

$v_y= \left ( 30\frac{m}{s}\right )sin(15^{\circ})$

$v_ x= 28.97777\frac{m}{s}$

$v_y = 7.76457\frac{m}{s}$

Since we are neglecting air resistance, we can easily calculate how long it takes for the ball to reach the second player:

$t = \frac{d}{v_x}=\frac{25m}{28.97777\frac{m}{s}} = 0.86273s$

Now we can use this time to calculate the height of the ball when it reaches the second player:

$y_f = y_i+v_{y_i}t+\frac{1}{2}at^2$

Plugging in our values, we get:

$y_f = 1m+\left ( 7.76457\frac{m}{s}\right )(0.86273s)+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )(0.86273s)^2$

y_f=3.98m

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AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

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

Example Question #153 : Linear Motion And Momentum

Example Question #154 : Linear Motion And Momentum

Example Question #433 : Ap Physics 1

Example Question #434 : Ap Physics 1

Example Question #155 : Linear Motion And Momentum

Example Question #436 : Ap Physics 1

All AP Physics 1 Resources

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