Mechanics Exam - AP Physics C Electricity & Magnetism

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Question

With what minimum velocity must a rocket be launched from the surface of the moon in order to not fall back down due to the moon's gravity?

The mass of the moon is $7.210^{22}kg$ and its radius is .

$\small G=6.6710^{-11}\frac{m^2N}{kg^2}$

Answer

Relevant equations:

$E_{total}=K+U_g$

$K = \frac{1}{2}mv^2$

$U_g=\frac{-GMm}{r}$

For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away.

For the minimum energy case as the rocket leaves the surface:

$0 = K + U_g = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{r}$

Rearrange energy equation to isolate the velocity term.

$v_{esc}^2 = \frac{GMm}{\frac{1}{2}mr}$

$v_{esc}=\sqrt{\frac{2GM}{r}}$

Substitute in the given values to solve for the velocity.

$v_{esc}=\sqrt{\frac{2(6.6710^{-11}\frac{m^2N}{kg^2})(7.210^{22}kg)}{(1.8*10^6m)}}=2310\frac{m}{s}\approx 2.3\frac{km}{s}$

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Question

What is the gravitational force of the sun on a $\small 5.0kg$ book on the earth's surface if the sun's mass is $2.010^{30}kg$ and the earth-sun distance is $1.510^{11}m$ ?

$\small G=6.67*10^{-11}\frac{m^2N}{kg^2}$

Answer

Relevant equations:

$|F_g| = \frac{GMm}{r^2}$

Use the given values to solve for the force.

$|F_g| = \frac{(6.6710^{-11}\frac{m^2N}{kg^2})(2.010^{30}kg)(5kg)}{(1.510^{11}m)^2}=0.0296N\approx 3.010^{-2}N$

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Question

Two spheres of equal mass are isolated in space, and are separated by a distance . If that distance is doubled, by what factor does the gravitational force between the two spheres change?

Answer

Newton's law of universal gravitation states:

$F_g=\frac{Gm_1 m_2}{r^2}$

We can write two equations for the gravity experienced before and after the doubling:

$F_{g,1} =\frac{Gm_1 m_2}{d^2}$

$F_{g,2}=\frac{Gm_1 m_2}{(2d)^2}$

The equation for gravity after the doubling can be simplified:

$F_{g,2}=\frac{Gm_1 m_2}{4d^2}$

$F_{g,2}=\frac{1}{4}\frac{Gm_1 m_2}{d^2}$

Because the masses of the spheres remain the same, as does the universal gravitation constant, we can substitute the definition of Fg1 into that equation:

$F_{g,2} = \frac{1}{4}F_{g,1}$

The the gravitational force decreases by a factor of 4 when the distance between the two spheres is doubled.

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Question

Two spheres of equal mass are isolated in space. If the mass of one sphere is doubled, by what factor does the gravitational force experienced by the two spheres change?

Answer

Newton's law of universal gravitation states that:

$F_g=\frac{Gm_1m_2}{r^2}$

We can write two equations representing the force of gravity before and after the doubling of the mass:

$F_{g,1}=\frac{Gm_{a,1}m_{b,1}}{r^2}$

$F_{g,2}=\frac{Gm_{a,2}m_{b,2}}{r^2}$

The problem gives us $m_{a,2}=2m_{a,1}$ and we can assume that all other variables stay constant.

Substituting these defintions into the second equation:

$F_{g,2}=\frac{G(2m_{a,1})m_{b,1}}{r^2}$

This equation simplifies to:

$F_{g,2}=2 \frac{Gm_{a,1}m_{b,1}}{r^2}$

Substituting the definition of Fg1, we see:

$F_{g,2}=2F_{g,1}$

Thus the gravitational forces doubles when the mass of one object doubles.

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Question

You are riding in an elevator that is accelerating upwards at , when you note that a block suspended vertically from a spring scale gives a reading of .

What does the spring scale read when the elevator is descending at constant speed?

Answer

When the elevator accelerates upward, we know that an object would appear heavier. The normal force is the sum of all the forces added up, and in this case it is . We know that the normal force has two components, a component from gravity, and a component from the acceleration of the elevator. Using this equation, we can determine the mass of the block, which doesn't change:

is acceleration due to gravity and is the acceleration of the elevator.

When substituting in the values, we get

Solving for , we get

Since the elevator is descending at constant speed, no additional force is applied, therefore the force that the spring scale reads is only due to gravity, which is calculated by:

$3.097 \cdot 9.8 = 30.35:N$

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Question

The mass and radius of a planet’s moon are $5.35\times10^{22} :kg$ and respectively.

With what minimum speed would a bullet have to be fired horizontally near the surface of this moon in order for it to never hit the ground?

(Note: You can treat the moon as a smooth sphere, and assume there’s no atmosphere.)

Answer

To do this problem we have to realize that the force of gravity acting on the bullet is equal to the centripetal force. The equations for gravitational force and centripetal force are as follows:

$F_g = \frac{GmM}{r^2}, F_c = \frac{mv^2}{r}$

If we set the two equations equal to each other, the small (mass of the bullet) will cancel out and will disappear from the right side of the equation.

$\frac{GM}{r}= v^2 \rightarrow v=\sqrt{\frac{GM}{r}}$

is the universal gravitational constant $6.67384 \times10^{-11}: m^3 kg^{-1} s^{-2}$

is given to be $5.35\times10^{22} kg$ and to be $1740:km \rightarrow 1,740,000:m$ .

If we plug everything in, we get

or

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Question

With what minimum velocity must a rocket be launched from the surface of the moon in order to not fall back down due to the moon's gravity?

The mass of the moon is $7.210^{22}kg$ and its radius is .

$\small G=6.6710^{-11}\frac{m^2N}{kg^2}$

Answer

Relevant equations:

$E_{total}=K+U_g$

$K = \frac{1}{2}mv^2$

$U_g=\frac{-GMm}{r}$

For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away.

For the minimum energy case as the rocket leaves the surface:

$0 = K + U_g = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{r}$

Rearrange energy equation to isolate the velocity term.

$v_{esc}^2 = \frac{GMm}{\frac{1}{2}mr}$

$v_{esc}=\sqrt{\frac{2GM}{r}}$

Substitute in the given values to solve for the velocity.

$v_{esc}=\sqrt{\frac{2(6.6710^{-11}\frac{m^2N}{kg^2})(7.210^{22}kg)}{(1.8*10^6m)}}=2310\frac{m}{s}\approx 2.3\frac{km}{s}$

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Question

What is the gravitational force of the sun on a $\small 5.0kg$ book on the earth's surface if the sun's mass is $2.010^{30}kg$ and the earth-sun distance is $1.510^{11}m$ ?

$\small G=6.67*10^{-11}\frac{m^2N}{kg^2}$

Answer

Relevant equations:

$|F_g| = \frac{GMm}{r^2}$

Use the given values to solve for the force.

$|F_g| = \frac{(6.6710^{-11}\frac{m^2N}{kg^2})(2.010^{30}kg)(5kg)}{(1.510^{11}m)^2}=0.0296N\approx 3.010^{-2}N$

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Question

Two spheres of equal mass are isolated in space, and are separated by a distance . If that distance is doubled, by what factor does the gravitational force between the two spheres change?

Answer

Newton's law of universal gravitation states:

$F_g=\frac{Gm_1 m_2}{r^2}$

We can write two equations for the gravity experienced before and after the doubling:

$F_{g,1} =\frac{Gm_1 m_2}{d^2}$

$F_{g,2}=\frac{Gm_1 m_2}{(2d)^2}$

The equation for gravity after the doubling can be simplified:

$F_{g,2}=\frac{Gm_1 m_2}{4d^2}$

$F_{g,2}=\frac{1}{4}\frac{Gm_1 m_2}{d^2}$

Because the masses of the spheres remain the same, as does the universal gravitation constant, we can substitute the definition of Fg1 into that equation:

$F_{g,2} = \frac{1}{4}F_{g,1}$

The the gravitational force decreases by a factor of 4 when the distance between the two spheres is doubled.

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Question

Two spheres of equal mass are isolated in space. If the mass of one sphere is doubled, by what factor does the gravitational force experienced by the two spheres change?

Answer

Newton's law of universal gravitation states that:

$F_g=\frac{Gm_1m_2}{r^2}$

We can write two equations representing the force of gravity before and after the doubling of the mass:

$F_{g,1}=\frac{Gm_{a,1}m_{b,1}}{r^2}$

$F_{g,2}=\frac{Gm_{a,2}m_{b,2}}{r^2}$

The problem gives us $m_{a,2}=2m_{a,1}$ and we can assume that all other variables stay constant.

Substituting these defintions into the second equation:

$F_{g,2}=\frac{G(2m_{a,1})m_{b,1}}{r^2}$

This equation simplifies to:

$F_{g,2}=2 \frac{Gm_{a,1}m_{b,1}}{r^2}$

Substituting the definition of Fg1, we see:

$F_{g,2}=2F_{g,1}$

Thus the gravitational forces doubles when the mass of one object doubles.

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Question

You are riding in an elevator that is accelerating upwards at , when you note that a block suspended vertically from a spring scale gives a reading of .

What does the spring scale read when the elevator is descending at constant speed?

Answer

When the elevator accelerates upward, we know that an object would appear heavier. The normal force is the sum of all the forces added up, and in this case it is . We know that the normal force has two components, a component from gravity, and a component from the acceleration of the elevator. Using this equation, we can determine the mass of the block, which doesn't change:

is acceleration due to gravity and is the acceleration of the elevator.

When substituting in the values, we get

Solving for , we get

Since the elevator is descending at constant speed, no additional force is applied, therefore the force that the spring scale reads is only due to gravity, which is calculated by:

$3.097 \cdot 9.8 = 30.35:N$

Compare your answer with the correct one above

Question

The mass and radius of a planet’s moon are $5.35\times10^{22} :kg$ and respectively.

With what minimum speed would a bullet have to be fired horizontally near the surface of this moon in order for it to never hit the ground?

(Note: You can treat the moon as a smooth sphere, and assume there’s no atmosphere.)

Answer

To do this problem we have to realize that the force of gravity acting on the bullet is equal to the centripetal force. The equations for gravitational force and centripetal force are as follows:

$F_g = \frac{GmM}{r^2}, F_c = \frac{mv^2}{r}$

If we set the two equations equal to each other, the small (mass of the bullet) will cancel out and will disappear from the right side of the equation.

$\frac{GM}{r}= v^2 \rightarrow v=\sqrt{\frac{GM}{r}}$

is the universal gravitational constant $6.67384 \times10^{-11}: m^3 kg^{-1} s^{-2}$

is given to be $5.35\times10^{22} kg$ and to be $1740:km \rightarrow 1,740,000:m$ .

If we plug everything in, we get

or

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

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Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

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Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

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