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AP Physics 1 : Motion in One Dimension

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Example Question #41 : Motion In One Dimension

Suppose that two people are ice skating. Person A has a mass of $100\:kg$ and person B has a mass of $50\:kg$ . 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.

$(m_{A}+m_{B})v_{i}=m_{A}v_{A}+m_{B}$

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

$0=m_{A}v_{A}+m_{B}v_{B}$

$-m_{A}v_{A}=m_{B}v_{B}$

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

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

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

$-2v_{A}=v_{B}$

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.

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Example Question #41 : Linear Motion And Momentum

When striking, a mosquito can accelerate from rest to a max speed of $12.1\frac{m}{s}$ in 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:

$107\frac{m}{s}$

$111\frac{m}{s}$

$102\frac{m}{s}$

$109\frac{m}{s}$

$105\frac{m}{s}$

Correct answer:

$102\frac{m}{s}$

Explanation:

Use the equation for acceleration.

$a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}$

Plug in the known values and solve for the acceleration.

$a=\frac{12.1\frac{m}{s}-o\frac{m}{s}}{0.119s-0s}$

$a=101.6\frac{m}{s^{2}}$

The mosquito has an incredible acceleration of $102m/s^{2}$ .

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Example Question #43 : Motion In One Dimension

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

Possible Answers:

2.6 s

4.2 s

3.2 s

2.2 s

2.4 s

Correct answer:

3.2 s

Explanation:

Use a kinematic equation to solve for the final time.

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

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

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

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

$t_{f}=3.2 s$

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

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Example Question #41 : Motion In One Dimension

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

Possible Answers:

$15\frac{m}{s}$

$25\frac{m}{s}$

$20\frac{m}{s}$

$10\frac{m}{s}$

$5\frac{m}{s}$

Correct answer:

$15\frac{m}{s}$

Explanation:

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

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

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

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

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

$t_{f}=1.42 s$

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

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

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

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

$V_{f}=13.9\frac{m}{s}$

When rounding and using significant figures the final answer would be 15 s .

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Example Question #321 : Ap Physics 1

A basketball player jumps directly upward with an initial velocity of $18\frac{m}{s}$ . Neglecting air resistance, how long does he spend in the air?

Possible Answers:

3.6 s

3.8 s

4.2 s

4.4 s

4.0 s

Correct answer:

3.6 s

Explanation:

Use a kinematic equation to solve for the time.

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

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

$0\frac{m}{s}=18\frac{m}{s}+(-9.8\frac{m}{s^{2}})(t_{f}-0s)$

$-18\frac{m}{s}= (-9.8\frac{m}{s^{2}})(t_{f})$

$t_{f}=1.8 s$

Since that describes half of the time he spends in the air you must multiply by 2 for the final answer.

$t_{f}=3.6 s$

The basketball player will be int he air for a total of 3.6 s .

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Example Question #322 : Ap Physics 1

A rubber ball bounces off the floor at $45\frac{m}{s}$ , how high will the ball be at it's apex? For this question neglect air resistance.

Possible Answers:

100 m

112 m

110 m

103 m

106 m

Correct answer:

110 m

Explanation:

Use a kinematic equation to solve for the final time.

$V_f=Vo+a(\Delta t)$

Plug in the known values and solve for time final.

$0\frac{m}{s}=45\frac{m}{s}+(-9.8\frac{m}{s^{2}})(\Delta t)$

$-45\frac{m}{s}=(-9.8\frac{m}{s^{2}})(t_{f})$

$t_{f}=4.59 s$

Use another kinematic equation and solve for the final distance.

$X_f=X_i+V_o(\Delta t)+0.5(a)(\Delta t)^{2}$

Plug in and solve for the final distance

$X_{f}=0m+45\frac{m}{s}(4.59s)+\frac{1}{2}(-9.8\frac{m}{s^{2}})(4.59s)^{2}$

$X_{f}=206.55m+(-4.9\frac{m}{s^{2}})(21.07s^{2})$

$X_{f}=206.55m+(-96.7m)$

$X_{f}=109.9m$

The rubber ball will be 110 m off the ground at it's highest.

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Example Question #323 : Ap Physics 1

A train starting in Denver, proceeds down a straight track from rest at a constant acceleration of $1.25\frac{m}{s^2}$ for exactly 1 min . The train then goes at that speed for another 5 min . Neglecting both the effects of friction and air resistance how far did the train get in kilometers away from Denver?

Possible Answers:

23.6 km

25.2 km

24.8 km

24.2 km

26.0 km

Correct answer:

24.8 km

Explanation:

Use the equation for acceleration to solve for the final velocity.

$a=\frac{(V_f-V_i)}{(t_f-t_i)}$

Plug in and solve for the final velocity.

$1.25\frac{m}{s^{s}}=\frac{V_{f}-0\frac{m}{s}}{60s-0s}$

$V_{f}=75\frac{m}{s}$

use another kinematic equation to solve for the final distance

$X_f=X_i+V_o(\Delta t)+0.5(a)(\Delta t)^{2}$

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

$X_{f}=0m+(75\frac{m}{s})(300s)+\frac{1}{2}(1.25\frac{m}{s^{2}})(60s)^{2}$

$X_{f}=22500m+2250m$

$X_{f}=24750m$

The answer needs to be in kilometers so divide by 1000m

$\frac{24750m}{1000m}=24.75 km$

Using significant figures 24.8km .

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Example Question #281 : Newtonian Mechanics

My roommate locked himself out of the apartment and forgot his keys. He asks me to walk over to the balcony and drop my key off the edge so he can get in. The keys take 6.0 s to reach the ground from the time it is released from my hand. From what height were the keys released, neglect the effects due to air resistance?

Possible Answers:

174 m

178 m

180 m

176 m

172 m

Correct answer:

176 m

Explanation:

Use a kinematic equation to solve for the height of the building.

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

Plug in and solve for the height of the building.

$0m=X_i+0\frac{m}{s}(6.0s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}(6.0s)^{2}$

$0m=X_i+(-4.9\frac{m}{s^{2}})(36s^{2})$

X_i= 176.4m

The building was 176 m tall.

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Example Question #325 : Ap Physics 1

A leaf falls from an Aspen tree at an initial height of 20 m . The leaf maintains a high surface to volume ratio and we can not neglect air resistance. Now, the new free fall acceleration due to gravity is equal to 75% it's usual value. How long does it take the leaf to fall directly to the ground bellow?

Possible Answers:

1 s

5 s

2 s

4 s

3 s

Correct answer:

2 s

Explanation:

Solve for the new acceleration due to gravity.

$a=(-9.80\frac{m}{s^{2}})(0.75)$

$a=-7.35\frac{m}{s^{2}}$

Use a kinematic equation to solve for the final time.

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

Plug in and solve for the final time.

$0m=20m+0\frac{m}{s}(\Delta t)+\frac{1}{2}(-7.35\frac{m}{s^{2}})(\Delta t)^{2}$

$-20m=-3.675\frac{m}{s^{2}}(t_{f}-0s)^{2}$

$5.44s^{2}=t_{f}^{2}$

$t_{f}=2.33 s$

Using significant figures and rounding the answer becomes 2 s .

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Example Question #321 : Ap Physics 1

A rocket ship is launched straight up from rest with a constant acceleration of $36\frac{m}{s^2}$ . After 200 seconds, how fast is the rocket ship traveling? Ignore both the effects of friction and of acceleration due to gravity.

Possible Answers:

$1100\frac{m}{s}$

$9600\frac{m}{s}$

$4800\frac{m}{s}$

$7200 \frac{m}{s}$

$2700\frac{m}{s}$

Correct answer:

$7200 \frac{m}{s}$

Explanation:

Use the following kinematic formula to find the final velocity after 200 seconds:

V_f= V_o + at

Plug in known values and solve.

$V_f=0\frac{m}{s}+36\frac{m}{s^2}*200s$

$V_f=7200\frac{m}{s}$

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AP Physics 1 : Motion in One Dimension

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #41 : Motion In One Dimension

Example Question #41 : Linear Motion And Momentum

Example Question #43 : Motion In One Dimension

Example Question #41 : Motion In One Dimension

Example Question #321 : Ap Physics 1

Example Question #322 : Ap Physics 1

Example Question #323 : Ap Physics 1

Example Question #281 : Newtonian Mechanics

Example Question #325 : Ap Physics 1

Example Question #321 : Ap Physics 1

All AP Physics 1 Resources

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