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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

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7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Universal Gravitation

Pluto radius: 1.186*10^6 m

Pluto mass: $1.3*10^{22} kg$

Determine the gravity constant, $g_{pluto}$ on the surface of Pluto.

Possible Answers:

$g_{pluto}=-2.4\frac{m}{s^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

$g_{pluto}=-9.8\frac{m}{s^2}$

$g_{pluto}=-3.1\frac{m}{s^2}$

$g_{pluto}=-14.8\frac{m}{s^2}$

Correct answer:

$g_{pluto}=-.616\frac{m}{s^2}$

Explanation:

Set both forms of gravitational force equal to each other.

$m_{object}g_{pluto}=-G\frac{m_{pluto} m_{object}}{r^2}$

Simplify:

$g_{pluto}=-G\frac{m_{pluto}}{r^2}$

Plug in values:

$g_{pluto}=-6.674*10^{-11}\frac{1.3*10^{22}}{(1.186*10^6)^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

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Example Question #81 : Forces

How would the linear velocity of the Moon be different if it's mass was doubled? Assume that the distance to the Earth stayed the same.

Possible Answers:

Unchanged

Quadrupled

Quartered

Doubled

Halved

Correct answer:

Unchanged

Explanation:

Set centripetal force equal to gravitational force:

$m_M\frac{v^2}{r}=G\frac{m_M m_E}{r^2}$

$\frac{v^2}{r}=G\frac{m_E}{r^2}$

The mass of the Moon, m_M cancels out, thus, there is no effect.

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Example Question #43 : Universal Gravitation

If on earth you have a weight, , what would your new weight be if you were standing on a planet with the same mass as earth, but with half the radius?

Possible Answers:

$\frac{1}{2}$ your earth weight.

The same as your earth weight.

times your earth weight

$\frac{1}{2}$ your earth weight.

times your earth weight.

Correct answer:

times your earth weight.

Explanation:

Your weight is a function of how much force is pulling you down towards the planet, not just your mass. To find force when your standing on a different planet, you would need to use Newton's Law of Universal Gravitation.

$F = - \frac{G M}{r^{2}}$

To compare the forces from each planet, you would set this equation equal to itself.

$- \frac{G M_{earth}}{r_{earth}^{2}} = - \frac{G M_{planet}}{r_{planet}^{2}}$

We then cancel the common terms, in this formula that's the negative sign, (the gravitational constant), and the mass of the earth/planet (because they're the same). After that we can substitute the radius of the new planet for half of the earth radius.

$r_{planet}=\frac{1}{2}{}r_{earth}$

We need to remember that when we square the radius in the Law of Universal Gravitation, we also need to square the $\frac{1}{2}$ , making it $\frac{1}{4}$ . Because the $\frac{1}{4}$ is in the denominator, we take the inverse of it, so you would feel times more force standing on the new planet. Therefore you would have times as much weight.

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Example Question #82 : Forces

An electronic scale is used to find the mass of a lead cube at sea level. The scale and lead cube are then transported to the top of a mountain, over $20000\textup{ ft}$ above sea level. How does the weight reading compare to the weight given at sea level?

Possible Answers:

None of these

It is impossible to determine

It will be larger

It will be the same

It will be smaller

Correct answer:

It will be smaller

Explanation:

The electronic scale measures based on the normal force the scale provides to the object. This in turn is based on the force of gravity on the object by the earth.

$F_G=-G\frac{m_1 m_2}{r^2}$

As height above sea level increases, as does , the distance to the center of the Earth. As increases, F_G decreases. This would decrease the normal force which would decrease the reading on the scale.

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Example Question #83 : Specific Forces

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

Mass of moon: $7.3*10^{22}kg$

Radius of moon: 1737km

A spring of rest length 15cm is placed upright on the moon. A mass of 15kg is gently placed on top and the spring contracts by 1cm . Determine the spring constant.

Possible Answers:

None of these

$\frac{2415N}{m}$

$\frac{3388N}{m}$

$\frac{4449N}{m}$

$\frac{3751N}{m}$

Correct answer:

$\frac{2415N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to the moon's surface:

F=ma

$F=-G\frac{m_1m_2}{r^2}$

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{moon}}{r_{moon}^2}$

Converting to and pugging in values:

$a_{mass}=-6.67*10^{-11}\frac{7.3*10^{22}}{(1737*10^3)^2}$

$a_{mass}=-1.61\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-15*-1.61}{.01}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{2415N}{m}$

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Example Question #53 : Universal Gravitation

A rope is used to accelerate a 5kg mass upwards at $2\frac{m}{s^2}$ . Determine the tension in the rope.

Possible Answers:

202N

90N

None of these

59N

18N

Correct answer:

59N

Explanation:

Using

F=ma

$F_{total}=F_1+F_2+...$

Combining equations:

$ma=F_{tension}+mg$

Solving for $F_{tension}$ and plugging in values:

$F_{tension}=5*2-5*(-9.8)$

$F_{tension}=59N$

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Example Question #52 : Universal Gravitation

Mass of Earth: $5.972*10^{24} kg$

Universal gravitation constant: $6.67*10^{-11}\frac{N\cdot m^2}{kg^2}$

Radius of earth: 6371km

Determine the magnitude of gravitational force by the earth on a 110kg astronaut 200km above the surface of the earth.

Possible Answers:

1455N

987N

1110N

1015N

1561N

Correct answer:

1015N

Explanation:

Finding total distance from center of Earth to astronaut:

200km+6371km=6571km

Converting to meters

6.571*10^6m

Using Universal Gravitation equation:

$F_G=-G\frac{m_1m_2}{r^2}$

Plugging in values:

$F_G=-6.67*10^{-11}\frac{5.972*10^{24}*110}{(6.571*10^6)^2}$

|F_G|=1015N

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Example Question #53 : Universal Gravitation

Mass of Jupiter: $1.89*10^{27}kg$

Universal gravitation constant: $6.67*10^{-11}\frac{N*m^2}{kg^2}$

Radius of Jupiter: 69911km

A marble is placed 100000km from the surface of Jupiter. Determine the acceleration due to the gravity of Jupiter.

Possible Answers:

$6.11*10^{-2}\frac{m}{s^2}$

$3.55*10^{-2}\frac{m}{s^2}$

None of these

$4.52*10^{2}\frac{m}{s^2}$

$4.52*10^{-2}\frac{m}{s^2}$

Correct answer:

$4.52*10^{-2}\frac{m}{s^2}$

Explanation:

Using

F=ma

and

$|F_G|=G\frac{m_1 m_2}{r^2}$

Combining equations

$m_1a=G\frac{m_1m_2}{r^2}$

Solving for

$a=G\frac{m_2}{r^2}$

Plugging in values:

$a=6.67*10^{-11}\frac{1.89*10^{27}}{(1670*10^6)^2}$

$a=4.52*10^{-2}\frac{m}{s^2}$

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Example Question #52 : Universal Gravitation

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

Mass of Pluto: $1.31*10^{22}kg$

Radius of Pluto: $1.19*10^{6}m$

A spring of rest length 27cm is placed upright on Pluto. A mass of 27kg is gently placed on top and the spring contracts by 4cm . Determine the spring constant.

Possible Answers:

$\frac{557N}{m}$

$\frac{844N}{m}$

None of these

$\frac{416N}{m}$

$\frac{611N}{m}$

Correct answer:

$\frac{416N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to Pluto's surface:

F=ma

$F=-G\frac{m_1m_2}{r^2}$

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{pluto}}{r_{pluto}^2}$

Plugging in values:

$a_{mass}=-6.67*10^{-11}\frac{1.31*10^{22}}{(1.19*10^6)^2}$

$a_{mass}=-.617\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-27*(-.617)}{.04}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{416N}{m}$

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Example Question #54 : Universal Gravitation

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

Mass of Mars: $639*10^{21}kg$

Radius of Mars: $3.39*10^{6}m$

A spring of rest length 18cm is placed upright on Mars. A mass of 65kg is gently placed on top and the spring contracts by 5.5cm . Determine the spring constant.

Possible Answers:

$k=\frac{4385N}{m}$

$k=\frac{3985N}{m}$

None of these

$k=\frac{3751N}{m}$

$k=\frac{4014N}{m}$

Correct answer:

$k=\frac{4385N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to the martian surface:

F=ma

$F=-G\frac{m_1m_2}{r^2}$

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{mars}}{r_{mars}^2}$

Plugging in values:

$a_{mass}=-6.67*10^{-11}\frac{639*10^{21}}{(3.39*10^6)^2}$

$a_{mass}=-3.71\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-65*(-3.71)}{.055}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{4385N}{m}$

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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 #41 : Universal Gravitation

Example Question #81 : Forces

Example Question #43 : Universal Gravitation

Example Question #82 : Forces

Example Question #83 : Specific Forces

Example Question #53 : Universal Gravitation

Example Question #52 : Universal Gravitation

Example Question #53 : Universal Gravitation

Example Question #52 : Universal Gravitation

Example Question #54 : Universal Gravitation

All AP Physics 1 Resources

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