All AP Physics C: Mechanics Resources
Example Questions
Example Question #101 : Mechanics Exam
What is the gravitational force of the sun on a book on the earth's surface if the sun's mass is and the earth-sun distance is ?
Relevant equations:
Use the given values to solve for the force.
Example Question #3 : Calculating Gravitational Forces
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?
No change
Newton's law of universal gravitation states:
We can write two equations for the gravity experienced before and after the doubling:
The equation for gravity after the doubling can be simplified:
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:
The the gravitational force decreases by a factor of 4 when the distance between the two spheres is doubled.
Example Question #4 : Calculating Gravitational Forces
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?
Newton's law of universal gravitation states that:
We can write two equations representing the force of gravity before and after the doubling of the mass:
The problem gives us and we can assume that all other variables stay constant.
Substituting these defintions into the second equation:
This equation simplifies to:
Substituting the definition of Fg1, we see:
Thus the gravitational forces doubles when the mass of one object doubles.
Example Question #25 : Forces
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?
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:
Example Question #26 : Forces
The mass and radius of a planet’s moon are 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.)
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:
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.
is the universal gravitational constant
is given to be and to be .
If we plug everything in, we get
or
Example Question #1 : Calculating Weight
The force on an object due to gravity on the moon is one-sixth of that found on Earth. What is the acceleration due to gravity on the moon?
We can use Newton's second law:
Set up equations for the force on the moon and the force on Earth:
Now we can use substitution:
From this, we can see that . Using the acceleration due to gravity on Earth, we can find the acceleration due to gravity on the moon.
Example Question #102 : Mechanics Exam
A large planet exerts a gravitational force five times stronger than that experienced on the surface of Earth. What is the weight of a 50kg object on this planet?
The weight of the object on Earth's surface is:
The force on the new planet is five times that on Earth, so we can simply multiply:
Example Question #2 : Calculating Weight
A space woman finds herself in an unkown planet with gravity . If her weight on Earth is 500N, what is her weight on the unkown planet?
We know that the weight of an object is given by:
is the mass of the object and is the gravitational acceleration of whatever planet the object happens to be on.
We know the gravity on the unkown planet, so the weight of the woman is given by:
We need only to find the mass of the woman to solve the problem. Since the mass of the woman is constant, we can use the information about her weight on Earth to figure out her mass.
Use this mass to solve for her weight on the new planet.
Example Question #1 : Understanding Orbits
If is the escape velocity from the surface of a planet of mass and radius . What is the velocity necessary for an object launched from the surface of a planet of the same mass, but with a radius that is , to escape the gravitational pull of this smaller, denser planet?
Not enough information
(the same velocity for both planets)
Escape velocity is the velocity at which the kinetic energy of an object in motion is equal in magnitude to the gravitational potential energy. This allows us to set two equations equal to each other: the equation for kinetic energy of an object in motion and the equation for gravitational potential energy. Therefore:
In this case, . Substitute.
Since the radius of the new planet is or the radius of the original planet, and they have the same masses, we can equate:
Where is the escape velocity of the larger planet and is the escape velocity of the smaller planet.
Example Question #1 : Understanding Orbits
A comet is in an elliptical orbit about the Sun, as diagrammed below:
The mass of the comet is very small compared to the mass of the Sun.
In the diagram above, how does the net torque on the comet due to the Sun's gravitational force compare at the two marked points?
The torque is greater at point b
The torque is zero at both a and b
It cannot be determined from the information given
The torque is the same at points a and b, but it is not zero
The torque is greater at point a
The torque is zero at both a and b
Since the direction of the gravitational force is directly towards the center of the Sun, it lies in the plane of the comet's orbit. Since torque is , the torque is always zero. That is why the comet's angular momentum is conserved in orbit.