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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #81 : Linear Motion And Momentum

The tallest Pine tree in a nearby forest is 150 meters tall. If a pine cone situated at the very top of the tree fell, how long would it take to hit the ground? $g=10\frac{m}{s^2}$

Possible Answers:

6.2 seconds

4.0 seconds

12 seconds

15 seconds

5.5 seconds

Correct answer:

5.5 seconds

Explanation:

Since the pine cone starts from rest and we are given a distance we will use:

$\Delta x=V_{0}t+.5at^2$ .

$-150m=.5(-10\frac{m}{s^2})t^2$

Solving for t:

$\frac{-150m}{-5\frac{m}{s^2}}=t^2$

$\sqrt{t^2}=\sqrt{30s} \rightarrow \boldsymbol{t=5.5s}$

*Note that the distance is negative since the object is falling and gravitational acceleration is downward. One could also just ignore the signs since time cannot be negative.

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Example Question #82 : Linear Motion And Momentum

The bottom of a mine shaft is filled with water to a depth of 5 feet. If a rock is dropped from the surface and is heard hitting the water 5 seconds later, how deep is the mine shaft? Use $g=10\frac{m}{s^2}$ and consider the travel time of the sound waves negligible.

Possible Answers:

135 meters

130 meters

125 meters

100 meters

150 meters

Correct answer:

130 meters

Explanation:

Since the stone is dropped from rest and we are searching for a distance, we can use the kinematic equation : $\Delta x=V_{0}t+.5at^2$ .

We can eliminate the second term, initial velocity times time because the stone is dropped.

$\Delta x=.5(10\frac{m}{s^2})5^2$

$\Delta x=125m$

But since the mine shaft is filled with 5 meters of water its actual depth would 130 meters.

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Example Question #86 : Motion In One Dimension

A high powered cannon projectile leaves the barrel with a muzzle velocity of $612 \frac{m}{s}$ . If the barrel is 1.2 meters long, what is the acceleration of the bullet in the barrel? Assume uniform acceleration.

Possible Answers:

$155,098\frac{m}{s^2}$

None of these

$130,460 \frac{m}{s^2}$

$149,000 \frac{m}{s^2}$

$156,060 \frac{m}{s^2}$

Correct answer:

$156,060 \frac{m}{s^2}$

Explanation:

The projectile started from rest and we are given its initial velocity, the muzzle velocity that is. Use the kinematic equation: $V^2=V_{0}^2+2ad$ .

The initial velocity is canceled out:

$(612\frac{m}{s})^2=2a(1.2m)$

Solve for a:

$(612\frac{m}{s})^2=a(2.4m)$

$\frac{(612\frac{m}{s})^2}{(2.4m)}=a$

$\boldsymbol{a=156,060\frac{m}{s^2}}$

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Example Question #363 : Ap Physics 1

Some grasshoppers can jump up to 1.7 meters in the air. What is this insects take off speed from the ground?

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

Possible Answers:

$V_{0}=4.76\frac{m}{s}$

$V_{0}=3.58\frac{m}{s}$

None of these

$V_{0}=7.46\frac{m}{s}$

$V_{0}=5.83\frac{m}{s}$

Correct answer:

$V_{0}=5.83\frac{m}{s}$

Explanation:

Use the kinematic equation $V^2=(V_{0})^2+2ad$ . Note that the grasshoppers final speed will be equal to 0.

$0=(V_{0})^2+2(-10\frac{m}{s^2})(1.7m)$

$0=(V_{0})^2-34\frac{m^2}{s^2}$

$34\frac{m^2}{s^2}=(V_{0})^2$

Square rooting both sides:

$\mathbf{V_{0}=5.83\frac{m}{s}}$

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Example Question #83 : Linear Motion And Momentum

If a person drops his keys on earth from 1.6 meters and an astronaut dropped hers on the moon, which set of keys would hit the ground first? Assume air friction is negligible on Earth and that both sets of sets of keys are the same mass.

Possible Answers:

The keys will hit the ground at the same time on both Earth and the Moon.

The Moon because the pull of gravity is weaker.

The Moon because the pull of gravity is stronger.

The Earth because the pull of gravity is stronger.

The Earth because the pull of gravity is weaker.

Correct answer:

The Earth because the pull of gravity is stronger.

Explanation:

The Earth's gravitational acceleration is approximately $10\frac{m}{s^2}$ while the acceleration of gravity on the Moon is about 1.6 $1.6 \frac{m}{s^2}$ . In other words the Earth's pull is much stronger than the Moon's. Objects will fall faster here on Earth.

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Example Question #84 : Linear Motion And Momentum

For a stone thrown straight up and caught at the same height it was thrown, which of the following are true. Ignore air friction.

Possible Answers:

Acceleration of the stone is zero after being caught.

Two of these

Acceleration and velocity of the stone always point in the same direction.

Acceleration of the stone is zero at the apex when the stone pauses for a split second.

The time going up equals the time going down.

Correct answer:

The time going up equals the time going down.

Explanation:

Gravity is the same at all heights with minor exceptions over great distances such as deep in the Earth and at 30,000 feet; gravity would be stronger that much closer to Earth.

Acceleration of gravity pulls everything downward- all the time and even if an object isn't moving. So acceleration and velocity can't always point in the same direction as the stones initial velocity points up.

The time traveling up will be equal to time falling as the motions just cancel each other. Gravity is slowing the stone as it travels upward until it stops very, very briefly and begins to accelerate and travel downward.

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Example Question #321 : Newtonian Mechanics

A drone must accelerate down a runway at $1.2 \frac{m}{s^2}$ for 20 seconds until it takes off. What is the minimum length of its runway?

Possible Answers:

None of these

240 meters

420 meters

402 meters

204 meters

Correct answer:

240 meters

Explanation:

Since the drone starts from rest we use $\Delta x=V_{o}t+\frac{1}{2}at^2$ and cross out the first term after the equals sign: initial velocity multiplied by time.

$\Delta x=\frac{1}{2}(1.2\frac{m}{s^2})(20s)^2$

$\boldsymbol{\Delta x=240 m}$

*Notice how the time units cancel

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Example Question #322 : Newtonian Mechanics

An astronaut drops his keys while on the moon from a height of 1.6 meters. How long will it take for them to hit the ground?

Possible Answers:

2.1 seconds

1.2 seconds

0.57 seconds

1.4 seconds

2.3 seconds

Correct answer:

1.4 seconds

Explanation:

We are given a distance and since the object is dropped we can cross out initial velocity in the kinematic equation: $\Delta x=V_{0}t+.5at^2$ . Remember to use the acceleration for gravity on the moon. It is $1.62\frac{m}{s^2}$ .

$-1.6m=.5(-1.62\frac{m}{s^2})t^2$

$-1.6m=-.81\frac{m}{s^2}t^2$

Dividing and square rooting the quotient:

$\mathbf{t=1.41 s}$

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Example Question #1 : Motion In Two Dimensions

An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?

Possible Answers:

420m

280m

70m

140m

Correct answer:

140m

Explanation:

The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, v_y, use $v_y = v_o sin\theta$ .

$v_y = (75 \frac{m}{s})sin 45^o = 53 \frac{m}{s}$

Next we find how long it takes to reach the top of its trajectory using v_f = v_o +at .

$0 \frac{m}{s} = 53 \frac{m}{s} + (-10 \frac{m}{s^2})t$

t = 5.3s

Finally, find how high the object goes with $d = v_ot +\frac{1}{2}at^{2}$ .

$d = (53\frac{m}{s})5.3s + \frac{1}{2}(-10\frac{m}{s^2})(5.3s)^{2} = 140 m$

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Example Question #1 : Motion In Two Dimensions

An object is shot from the ground at 125m/s at an angle of 30^o above the horizontal. How far away does the object land?

Possible Answers:

1350m

675m

250m

62.5m

Correct answer:

1350m

Explanation:

First, find the horizontal (x) and vertical (y) components of the velocity

$v_x = v_o cos\theta$

$v_x = (125 \frac{m}{s}) cos(30^o) = 108\frac{m}{s}$

$v_y = v_o sin\theta$

$v_y = (125\frac{m}{s})sin(30^o) = 62.5\frac{m}{s}$

Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.

v_f = v_o + at

$0 \frac{m}{s} = 62.5 \frac{m}{s} + (-10 \frac{m}{s^2})t$

t = 6.25s

Total time in the air is therefore 12.5s (twice this value).

Finally, find distance traveled my multiplying horizontal velocity and time.

d_x = v_x(t)

$d_x = 108 \frac{m}{s} (12.5s) = 1,350 m$

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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 #81 : Linear Motion And Momentum

Example Question #82 : Linear Motion And Momentum

Example Question #86 : Motion In One Dimension

Example Question #363 : Ap Physics 1

Example Question #83 : Linear Motion And Momentum

Example Question #84 : Linear Motion And Momentum

Example Question #321 : Newtonian Mechanics

Example Question #322 : Newtonian Mechanics

Example Question #1 : Motion In Two Dimensions

Example Question #1 : Motion In Two Dimensions

All AP Physics 1 Resources

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