AP Physics C: Mechanics : Mechanics Exam
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All AP Physics C: Mechanics Resources
Example Questions
Example Question #1 : Interpreting Collision Diagrams
Two identical atoms, A and B, are at rest some distance apart in a vacuum.
If atom A moves at a constant velocity, 
The velocity of atom A is 
Atom A moves with velocity 
The velocities of the identical atoms are the same, but in opposite directions after the collision
The velocity of atom B after the collision is
The speed of both atoms is the same after collision
The velocity of atom A is
Assuming the atoms have the same mass, a collision between atoms in a vacuum is most nearly an elastic collision. This means the kinetic energy of the system is conserved and when they collide, all of the kinetic energy in atom A will be transferred perfectly to atom B. So just after collision, atom A will have velocity 
Example Question #1 : Interpreting Collision Diagrams
A pickup truck with a mass of 
It turns out the truck can only take on an additional 
The equation for conservation of momentum is:
Given:
Plug in given values into the conservation of momentum equation.
Simplify.
Subtract the final velocity from the initial velocity to find change in speed (since these velocities are in the same direction).
Example Question #1 : Interpreting Collision Diagrams
A pickup truck with a mass of 
Assuming no outside forces are acting on the system, how many 
Because of conservation of momentum, the product of the initial mass and velocity should equal the final mass and velocity. The equation for conservation of momentum is:
The initial mass of the truck is
The initial velocty is
The final velocity is 
Plug in known values to the conservation of momentum equation.
Simplify.
Note that while necessary, the final mass is not our final answer. The final mass, 
The mass of the truck is known so set up an equation and solve for the mass of sand added to the truck.
Example Question #1 : Circular Motion
A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?
The ball is experiencing centripetal force so that it can travel in a circular path. This centripetal force is written as the equation below.
Remember that centripetal acceleration is given by the following equation.
Since the centripetal force is coming from the tension of the string, set the tension force equal to the centripetal force.
Since we're trying to find the speed of the ball, we solve for v.
We know the following information from the question.
We can use this information in our equation to solve for the speed of the ball.
Example Question #1 : Circular Motion Concepts And Equations
In uniform circular motion, the net force is always directed ___________.
away from the center of the circle
toward the center of the circle
nowhere; the net force will be zero
in the same direction as the tangential velocity
toward the center of the circle
The correct answer is "toward the center of the circle." Newton's second law tells us that the direction of the net force will be the same as the direction of the acceleration of the object.
In uniform circular motion, the object accelerates towards the center of the circle (centripetal acceleration); the net force acts in the same direction.
Example Question #1 : Circular Motion
A car moves around a circular path of radius 100m at a velocity of 
The force of friction is what keeps the car in circular motion, preventing it from flying off the track. In other words, the frictional force will be equal to the centripetal force.
We can cancel mass from either side of the equation and rearrange to solve for the coefficient of friction:
We can use our given values to solve:
Example Question #1 : Circular Motion
An object of mass 10kg undergoes uniform circular motion with a constant velocity of 
The time for one full revolution can be calculated simply by manipulating the defintion of velocity, where the distance is just the circumference of the circlular path. The time it takes is modeled by the following equation:
Use the given radius and velocity to solve for the time per revolution:
Example Question #1 : Kinetic Energy
A train car with a mass of 2400 kg starts from rest at the top of a 150 meter-high hill. What will its velocity be when it reaches the bottom of the hill, assuming that the bottom of the hill is the reference level.
The law of conservation of energy states:
If the car starts at rest, then the initial kinetic energy = 0 J.
If the car ends at the reference height, the final potential energy = 0 J.
Subsituting these values, the equation becomes:
The initial potential energy can be determined by:
The final kinetic energy equation is:
Substituting the initial potential energy and final kinetic energy into our modified conservation of energy equation, we get:
Example Question #1 : Energy
An object has a mass of 5kg and has a position described by the given function:
What is the object's kinetic energy after two seconds?
Kinetic energy is defined by the equation:
Taking the derivative of the position function allows us to obtain the velocity function:
We can now determine the velocity after two seconds:
Now that we know our velocity, we can solve for the kinetic energy.
Example Question #21 : Work, Energy, And Power
An object starts from rest and accelerates at a rate of 
Kinetic energy is given by the equation:
We can find the velocity using the given acceleration and time:
Use this velocity to find the kinetic energy after three seconds:
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