### All AP Physics 1 Resources

## Example Questions

### Example Question #121 : Linear Motion And Momentum

A rollercoaster car is traveling along an incline at a velocity . What is the direction of the rollercoaster car?

**Possible Answers:**

To the right at an angle above the horizon

To the right at an angle below the horizon

To the left at an angle above the horizon

To the left at an angle below the horizon

**Correct answer:**

To the right at an angle below the horizon

A two-dimensional vector comes in the form . The direction of the vector is found using the equation:

For this problem:

The rollercoaster car is traveling backwards at an angle below the horizon.

### Example Question #31 : Motion In Two Dimensions

Two cyclist leave from the same starting point. One cyclist is traveling east at . The other cyclist is traveling south at . At what speed are the cyclists traveling away from each other?

**Possible Answers:**

**Correct answer:**

To find the speed at which the cyclists are traveling away from each other, we must represent the velocity of each cyclists as a vector. We will then add the two vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

One cyclist is traveling east at , represented as a vector:

The other cyclist is traveling south at , represented as a vector:

Adding the two vectors together, gives us the velocity at which the cyclists are traveling away from each other:

The speed at which the cyclists are traveling away from each other is the magnitude of the velocity.

If a vector is in the form , its magnitude, ,is found by using the equation

Given the velocity vector:

The speed at which the cyclists are traveling away from each other is:

### Example Question #401 : Ap Physics 1

A boat crossing a river is traveling east at . The current of the river is traveling north at . What is the resultant speed of the boat?

**Possible Answers:**

**Correct answer:**

To find the resultant speed of the boat, we must represent the velocity of the boat and the current as vectors. We will then add the two vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

The boat is traveling east at , represented as a vector

The current is traveling north at , represented as a vector

Adding the two vectors together, gives us the resultant velocity of the boat:

The resultant speed of the boat is the magnitude of the velocity.

If a vector is in the form , its magnitude, ,its found by using the equation

Given the velocity vector:

The resultant speed of the boat is:

### Example Question #401 : Ap Physics 1

Object 1 has velocity and mass . Object 2 has velocity and mass . Determine the total momentum of the system.

**Possible Answers:**

**Correct answer:**

Definition of momentum:

Combine equations:

Plug in values:

### Example Question #33 : Motion In Two Dimensions

There is an annual contest around halloween where contestants build cannons to see who can launch a pumpkin the farthest. One contestant launches his pumpkin out of the cannon at an angle of to the horizontal at a speed of . The height of the cannon head where the pumpkin emerges is off the ground.

Assuming the terrain around the area is flat and there is no air resistance on the pumpkin, how far away is the pumpkin when it is at the same height as when it was fired?

**Possible Answers:**

**Correct answer:**

A picture of the pumpkins motion is shown above. We wish to find the range that the pumpkin travels. We align our axes so that the x-axis points in the direction of the horizontal motion of the pumpkin, with x=0 being the x position of the cannon head and the y-axis points vertically off the ground. Therefore the pumpkin's intial position is and . Our pumpkin's initial velocity is at to the horizontal. We want the x and y components of the velocity, and that is found by resolving the vector into x and y-components given by

and

Since we want the final x position of the pumpkin, we write out our kinematics equation for position in the x-coordinate:

Since there is no air resistance, , and substituting in for and for gives us

However, we still don't know how long the pumpkin is in the air. But we do know that it's final y-position . Therefore we will use our kinematics equation for position in the y-coordinate to find the time. We have:

Since this is a projectile motion problem, we know that , and we already know that , , and . Plugging these in to the y-position equation gives us:

This is a quadratic equation that we can solve for time by using the quadratic formula. Doing so will give us two values:

and

Since time is always positive, we use the value . Thus the pumpkin flies in the air for this long. Now we can plug this into our x-position equation to find . Doing so gives:

. Therefore the range of the projectile is .

### Example Question #402 : Ap Physics 1

A meteor is traveling at and has a mass of . The meteor then spontaneously splits apart into two chunks of equal mass. One chunk splits off at an angle of to the original trajectory, and the other chunk splits off at an angle of to the original trajectory.

Assuming no forces act on the meteor during the split, what is the speed of the chunk that splits off at ?

**Possible Answers:**

None of these

**Correct answer:**

The above diagram illustrates the situation. Originally, the meteor (shown on the left) is traveling with some initial momentum. And on the right the two chunks are shown splitting off. Each must have half the original mass as given in the problem, so the masses are each , but the velocities are unknown. Since no forces are acting on the system, momentum is conserved, and we can apply the conservation of momentum for the system. We split the conservation of momentum into two directions, one along the original trajectory which we will call the x direction, and one normal to the original trajectory which we will call the y direction. Then our conservation of momentum equations are and . For our initial momentums, we see that the meteor has momentum in the x-direction but no momentum in the y-direction, therefore:

and

For the final momentum, we see that both meteors have momentum in the positive x-direction, and opposite momenta in the y-direction. Therefore we have:

Therefore, our equations of conservation become:

To make things simpler, we divide through by the mass to get:

Now we have two equations in two unknowns, and . Since we want the velocity of the meteor that split at , we want to find . Therefore, we solve the second equation for to obtain . Now we substitute this value into the first equation to obtain:

Factoring out the and using the trigonometric identity , we get:

Now we divide each side by the expression in the brackets to obtain , and we get:

Therefore the speed of the chunk flying out at is .

### Example Question #35 : Motion In Two Dimensions

Suppose you're at the circus, and you see a clown launched out of a cannon. If you know that the clown's initial velocity when coming out of the cannon is and that the cannon was at an angle of with respect to the ground, what is the maximum height the clown reaches during his path across the circus?

**Possible Answers:**

**Correct answer:**

In this question, we're told that a projectile (the clown) is launched at a certain velocity and a certain angle. We're asked to find the maximum height of the projectile.

We need to recognize that this is a question concerning motion in two dimensions, both x and y. This means that the projectile will be launched such that it travels in a parabolic path. However, for the purposes of this question, we can neglect motion in the x direction and instead direct our attention to the projectile's vertical motion.

Since the projectile travels in a projectile path, we know that at the instant it reaches its maximum height, its velocity will fall to zero. We can take advantage of this fact, combined with one of the kinematic equations (since we know acceleration due to gravity is constant), in order to find our answer.

And since we know that gravity is oriented in the downward direction, we can give it a negative value.

Now, we need to find an expression that gives us the clown's initial velocity in the vertical direction. Remember, the question stem gives us the clown's initial velocity, but this is not the same as its velocity in the y direction. The value that is given in the question stem is a combination of both the x direction velocity as well as the y direction velocity. Thus, we'll need to use the information about the clown's initial velocity provided in the question stem in order to calculate the initial vertical velocity.

Now that we have an expression for the initial vertical velocity, we can go ahead and plug this value into the prior expression that we derived.

Finally, if we plug in the known values, we can calculate our answer.

### Example Question #32 : Motion In Two Dimensions

An arrow is shot with an initial velocity at an angle of above the horizontal at time . How far does the arrow travel laterally between and . Neglect air resistance.

**Possible Answers:**

**Correct answer:**

Since we are asked for the horizontal distance, we need to calculate the horizontal component of the arrow's initial velocity. Also, since we are neglecting air resistance, we know that this velocity is constant. Therefore, we simply need to multiply it by the time interval given in the problem statement to calculate the distance.

Now we can multiply this by the time interval given, which is 0.5 seconds:

### Example Question #33 : Motion In Two Dimensions

An arrow is shot with an initial velocity at an angle of above the horizontal at time . How long does it take for the arrow to reach its maximum height? Neglect air resistance.

**Possible Answers:**

**Correct answer:**

The arrow will reach its maximum height when the vertical component of its velocity is 0. Therefore, we will first calculate the vertical component of the arrow's initial velocity:

Now we can calculate how long it takes for this velocity to equal 0

### Example Question #132 : Linear Motion And Momentum

An arrow is shot with an initial velocity at an angle of above the horizontal from a height of . What is the maximum height reached by the arrow? Neglect air resistance.

**Possible Answers:**

**Correct answer:**

First, we need to calculate the vertical component of the initial velocity of the arrow:

Then we can use the following kinematic equation to calculate the change in height of the arrow:

Rearranging for change in height:

The arrow reaches its maximum height when the vertical component of its velocity is 0. Plugging in values for each variable:

Now adding this to the initial height we get:

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