### All AP Physics B Resources

## Example Questions

### Example Question #1 : Ap Physics 1

A ball is thrown at a velocity of at an angle of from the horizontal. What are the ball's horizontal and vertical velocities?

**Possible Answers:**

There is not enough information to solve this problem

**Correct answer:**

The velocity of can be broken into horizontal and vertical components by using trigonometry. Think of the figure below, where x and y velocity components of the total velocity are shown.

Use the total velocity, the x-component, and the y-component to form a right triangle below.

Treating as the hypotenuse, x-component as the leg adjacent, and y-component as the leg opposite, you can conclude that the velocities are related through trigonometric identities.

Plugging in the given values, we can solve for the x and y velocity components.

### Example Question #1 : Understanding Vector And Scalar Quantities

Which answer choice below includes only scalar quantities?

**Possible Answers:**

Force, time, velocity

Distance, speed, time

Acceleration, energy, displacement

Displacement, time, acceleration

Velocity, displacement, force

**Correct answer:**

Distance, speed, time

Scalar quantities are those that can be described with magnitude only, as opposed to vectors, which include both magnitude and direction components. Distance, speed, and time are all scalars. Displacement is not a scalar, as it involves both the distance and the direction moved from a starting point. Velocity also includes a direction component, and is therefore a vector quantity.

### Example Question #1 : Ap Physics 1

If a 15kg object is subjected to a force of 175N to the east and another force of 1020N to the west, what is the object’s acceleration?

**Possible Answers:**

1.4m/s^{2}

5.6m/s^{2}

113m/s^{2}

56m/s^{2}

**Correct answer:**

56m/s^{2}

First, find the net force by subtracting the opposing forces.

F = 1020N – 175N = 845 N to the west

Next, find acceleration using Newton's second law, .

### Example Question #1 : Newtonian Mechanics

Three blocks (in left-to-right order: block A, block B, block C) are pushed along a frictionless horizontal surface by a 50N force to the right, which acts on block A.

If the masses of the blocks are given by , , and , which of the following best describes the relationship between the accelerations of the blocks?

**Possible Answers:**

**Correct answer:**

The three blocks must remain in contact as they move, so they will each have the same velocity and acceleration regardless of their different masses. So, .

### Example Question #1 : Newtonian Mechanics

A block slides down a incline. If the coefficient of friction between the block and the incline is , determine the acceleration of the block as it slides down the incline.

**Possible Answers:**

**Correct answer:**

Know what forces are involved by drawing a force diagram.

The gravitational force is broken into the x and y components. The net force on the block in the y-direction is the normal force minus the y component of the gravitational acceleration ().

Notice that the net y-force is equal to zero to show that the block is not moving anywhere in the y-direction. Now, we can isolate the normal force.

The net force in the x-direction is . We know that the block is accelerating in the x-direction; therefore the net force is equal to .

We can use the friction equations to substitute for the x-direction forces.

We can isolate the acceleration and solve using the provided values.

### Example Question #1 : Newtonian Mechanics

A car of mass is initially at rest, and then accelerates at for . What is the kinetic energy of the car at time ?

**Possible Answers:**

**Correct answer:**

The first step will be to find the final velocity of the car. We know the acceleration and time, so we can find the final velocity using kinematics. The initial velocity is zero, since the car starts at rest.

Use this velocity and the mass of the car to solve for the final kinetic energy.

### Example Question #1 : Linear Motion

A man throws a ball straight up in the air at a velocity of . If there is a constant air resistance force of against the motion of the ball, what is the maximum height of the ball?

**Possible Answers:**

**Correct answer:**

We first need to find the net force acting on the ball during flight. We can then use the net force and Newton's second law to find the total acceleration on the ball.

Use this net force to find the acceleration.

From here, there are two ways to solve. One way uses kinematic equations, and the other uses energy. We will solve using energy.

Total energy must be conserved during the throw, so the initial kinetic energy must equal the final potential energy (since velocity is zero at the maximum height).

Use the given initial velocity to find the final height.

### Example Question #1 : Newton's Second Law

What is the acceleration of the system shown above? (Assume the table is frictionless and the mass of the rope connecting blocks is negligible).

**Possible Answers:**

**Correct answer:**

The force that translates to the entire system is that of gravity acting on the mass hanging over the ledge.

140N is the total force acting on the system, which has a mass equal to both blocks combined (65 kg + 14 kg = **79 kg**). We can find the acceleration using Newton's second law.

### Example Question #1 : Linear Motion

A car moving at 40m/s suddenly applies a braking force and comes to rest in 20s, with constant deceleration. If the car has a mass of 2000kg, what is the braking force?

**Possible Answers:**

–4000N

–8000N

4000N

40000N

–2000N

**Correct answer:**

–4000N

First we can calculate the acceleration.

Using F = ma with the magnitude of the acceleration we can find the force.

### Example Question #1 : Newtonian Mechanics

A box is placed on a 30^{o} frictionless incline. What is the acceleration of the box as it slides down the incline?

**Possible Answers:**

**Correct answer:**

To find the acceleration of the box traveling down the incline, the mass is not needed. Using the incline of the plane as the x-direction, we can see that there is no movement in the y-direction; therefore, we can use Newton's second, F = ma, in the x-direction.

There is only one force in the x-direction (gravity), however gravity is not just equal to “mg” in this case. Since the box is on an incline, the gravitational force will be equal to mgsin(30^{o}). Substituting force into F =ma we find that mgsin(30^{o}) = ma. We can now cancel out masses and solve for acceleration.

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