All AP Physics B Resources
Example Questions
Example Question #2 : Motion In Two Dimensions
A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
What is the velocity of the box just before it hits the ground?
We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation .
We can plug these values into the following distance equation and solve for time.
Now that we know the acceleration on the box and the time of travel, we can use the equation to solve for the velocity.
Example Question #21 : Ap Physics B
A ball gets pushed off a high table with a horizontal speed of . How far does the ball travel horizontally before hitting the ground?
This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion.
Choosing the ground to be the zero height, we have and .
Also, knowing that the initial vertical velocity is zero, we know that .
The kinematic equation simplifies using these values.
Rearrange the equation to isolate time.
We know that is the acceleration due to gravity: . Plug in the values to solve for time.
We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation .
We know that and that . Using these values and the time, we can solve for the horizontal distance travelled.
Example Question #2 : Motion In Two Dimensions
A car drives north at for , then turns east and drives at for . What is the magnitude and direction of the average velocity for the trip?
First, determine how far the car travels in each direction:
Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem.
Find the average velocity by dividing the total displacement by the total time.
Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry.
Use the north and the east directional displacements to find the angle.
Our final answer will be:
Example Question #3 : Motion In Two Dimensions
A ball is launched at an angle of above the horizontal with an initial velocity of . At what time is its vertical velocity ?
Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time.
We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity.
Using this in our kinematics formula, we solve for the time.
Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity.
Example Question #3 : Motion In Two Dimensions
A 2kg box is at the top of a frictionless ramp at an angle of . The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
When the box is released, how long will it take the box to reach the ground?
We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation .
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of . Since the angle is 60o, the acceleration on the box is
Finally, we can plug these values into the following distance equation and solve for time.
Example Question #1 : Ap Physics 1
A spring is compressed from equilibrium by a force of . What is the spring constant for this spring?
We use Hooke's law equation to relate the force, displacement, and spring constant:
We are given the force and the displacement, allowing us to solve for the spring constant.
Note that the displacement is negative, since the spring is compressed. For springs, compressions represents a negative displacement, while stretching represents a positive displacement.
Example Question #1 : Using Spring Equations
A mass of weight is suspended vertically from a spring and stretches it from equilibrium. What is the energy stored in the spring?
First, we need to solve for the spring constant by using the force on the spring. We can use Hooke's Law:
The magnitude of the force on the spring will be equal to the force of gravity on the mass, which is given to be . The distance the spring it stretched is in the downward direction, so we must use a negative displacement. Use these values to calculate the spring constant.
Next, use the spring energy equation with the displacement and spring constant to solve for the energy stored in the spring.
Example Question #22 : Ap Physics B
A block is sliding on a horizontal frictionless floor at a speed of and runs into a horizontal spring. The spring has a spring constant of . What is the maximum compression of the spring after the collision?
First, we calculate the kinetic energy of the block as it slides:
We know that all of the kinetic energy is converted to spring potential energy during the collision. We can use the equation for the potential energy of a spring and set it equal to the kinetic energy. Then, solve for the compression distance:
Example Question #1 : Other Circular And Rotational Motion Concepts
A wheel of radius rolls along a flat floor and makes rotations over a period of time. What distance has the wheel traveled?
In order to find the distance the wheel travels, we need a way to convert angular displacement to linear displacement. We know that the circumference of a circle (or a wheel, in this case) is . This means that in one rotation, the wheel would travel a distance equal to its circumference.
Distance of one rotation equals: .
Since the wheel travels rotations, the total distance that the wheel travels will be equal to the distance traveled by one rotation multiplied by the number of rotations.
Distance of rotations equals:
Example Question #4 : Ap Physics 1
A ball of mass on a string is rotating about the center axis with a velocity . If the tension force in the string is , what is the radius of the circular path of the ball?
We use the equation for centripetal force to find the radius:
Since the string ties the ball to the axis, the force of tension will be equal to the centripetal force.
Use the given mass and velocity to solve for the radius.
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