All AP Physics 1 Resources
Example Questions
Example Question #61 : Motion In Two Dimensions
Two space ships are racing deep in space. At , Spaceship A is at the origin while traveling at
. It is coasting, with it’s rockets off. Spaceship B is at
. It’s traveling at
. It’s rockets are firing with a constant force of
. Both rockets have a mass of
.
What magnitude velocity will spaceship have at the origin?
None of these
Using
Solving for
Plugging in values:
Example Question #153 : Linear Motion And Momentum
A tall man standing on a cliff that is
high throws a baseball completely horizontally off the cliff at a velocity of
, how far from the base of the cliff will the ball travel before hitting the ground?
In order to find the distance that the ball travels horizontally one must multiply the velocity of the ball by the amount of time that the ball is traveling. Time in the air can be found by figuring out the amount of time that it takes for the ball to hit the ground relative to the vertical distance. Using the equation:
where d is (height of the cliff plus height of the man),
is equal to zero because there is no initial vertical velocity, and acceleration is equal to the gravity constant, time is found to be
. In order to find distance traveled horizontally, one can use the equation:
Because the vertical and horizontal axis are completely independent of each other there is no acceleration in the horizontal direction, the equation then becomes simplified to:
After plugging in numbers
Example Question #154 : Linear Motion And Momentum
Two balls are thrown off the edge of a cliff of height from the same location. Both balls are thrown with an initial velocity of
. The first ball is thrown horizontally and the second ball is thrown at and angle of
above the horizontal. When the cliff is a particular height, let's call it
, the two balls will land at the same spot on the ground.
Which of the following is true?
The ball thrown above the horizontal will always go further besides when
There are multiple cliff heights where the two balls will land at the same location
The ball thrown above the horizontal will only go further than the horizontally thrown ball if
The ball thrown above the horizontal will only go further than the horizontally thrown ball if
The horizontally thrown ball will always go further besides when
The ball thrown above the horizontal will only go further than the horizontally thrown ball if
If , the horizontally thrown ball will travel no distance. As the height of the cliff increases, the horizontally thrown ball will have more and more air time until it begins to travel further than the ball thrown
above the horizontal. This is because the distance each ball will go equals the time it's in the air multiplied by its horizontal velocity (since it's not accelerating in the x-direction). The time the ball thrown
above the horizontal is always the same amount greater than the time the ball thrown horizontally is in the air. The ball thrown horizontally has a horizontal velocity greater than the ball thrown
above the horizontal (by a factor of
). This means the ball thrown horizontally travels a distance of
while the ball thrown
above the horizontal travels a distance of
. If we equate these two expressions, we get
. When
is less than this, the ball thrown
above the horizontal travels further and when
is greater the ball thrown horizontally travels further.
represents the time is takes for the ball thrown horizontally to hit the ground when the height of this cliff is
.
Example Question #151 : Linear Motion And Momentum
2 masses are connected to a pulley system as demonstrated by the image above. Mass 1 (m1) weighs and mass 2 weighs
.
Assuming there is no friction in the system, how fast would mass 1 be accelerating?
Because the mass of object 1 is given, the force acting on it must be found in order to figure out how fast it is accelerating. These 3 variables are related by the equation:
Because there is no friction in the system, the only force acting on the object is the tension in the rope. The magnitude of the force of tension is the magnitude of the gravitational force of mass 2. Gravitational force is equal to:
where g is the gravity constant which is equal to which is also the acceleration of this object. Since this force is the only force acting on the object, accelerating can be found relating the 2 equations above as follows:
After plugging in values, the acceleration of mass 1 is found to be:
Example Question #434 : Ap Physics 1
If an object is launched at an angle of
with respect to the ground, what is its vertical speed after
?
To solve this question, we'll first need to break up the velocity into its component parts. For this question, we're only concerned about motion in the vertical direction, so we'll first need to find the initial velocity in the vertical direction.
Now that we know what the initial velocity is, we can determine what the final velocity will be after .
Example Question #155 : Linear Motion And Momentum
A baseball hits a perfectly horizontal ball at a speed of from a height of
. How far has the ball traveled before it hits the ground?
First, we can calculate how long it takes for the ball to hit the ground using the following expression:
Since the original vertical velocity is 0, we can rearrange to get:
Plugging in our values, we get:
Multiplying this by the velocity given in the problem statement (horizontal velocity), we get:
Example Question #436 : Ap Physics 1
A baseball player hits a ball from a height of at a velocity of
and an angle of
above the horizontal. The ball is hit toward another player that is
away from the batter. At what height is the ball when it reaches the other player?
First, let's separate the given velocity into its vertical and horizontal components:
Since we are neglecting air resistance, we can easily calculate how long it takes for the ball to reach the second player:
Now we can use this time to calculate the height of the ball when it reaches the second player:
Plugging in our values, we get:
Example Question #61 : Motion In Two Dimensions
A batter hits a ball at an angle of to the horizontal. If the home run fence is
away with a height the same as when the batter hits the ball, what is the minimal initial velocity that will result in a home run?
This problem is greatly simplified since the home run fence has the same height as the initial height of the ball. Therefore, we can use this range equation to solve this problem in one calculation:
Rearranging for velocity, we get:
Plugging in our values, we get:
Example Question #62 : Motion In Two Dimensions
A batter hits a ball at an angle of to the horizontal. If the home run fence is
away with a height of
, and the ball is hit from an initial height of
, what is the minimal initial velocity that will result in a home run?
There are multiple ways to attack this problem, so don't worry if you took a different route. However, no matter which way you go, you'll be substituting small expressions into larger ones.
First, let's develop an expression of how long it takes the ball to reach the home run fence. We know that:
Rearranging for time:
Then we can substitute the following expression in for horizontal velocity:
Now that we have an expression for time, we can use the following expression:
Plugging in an expression for initial vertical velocity, we get:
Plugging in our expression for time, we get:
Now let's simplify this a bit:
Rearranging:
And some more:
One last time:
It may look a bit nasty, but we know all of these values, so time to plug and chug:
Example Question #71 : Motion In Two Dimensions
A home run derby is being held in a stadium where the home run wall is from home plate. The wall itself has a height of
. The stands begin at the wall and with equal height, and rise at an angle of
to the horizontal.
If a home run is hit a total distance of from an initial height of
and an angle of
to the horizontal, what is the ball's initial velocity?
From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:
If then
If then
Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of , a height of
, and rises at an angle of
.
Using the total horizontal distance traveled, we can calculate the ball's final height using the above expression:
Now we can use the following two kinematics expression:
Where d is the horizontal distance traveled. Replacing out component velocities, we get:
Substituting our expression for time into the other expression, we get:
Simplifying, we get:
Rearranging:
And again:
One last time:
It's nice and messy, but we have all of these values, so time to plug and chug:
Certified Tutor
All AP Physics 1 Resources
![Learning Tools by Varsity Tutors](https://vt-vtwa-app-assets.varsitytutors.com/assets/problems/og_image_practice_problems-9cd7cd1b01009043c4576617bc620d0d5f9d58294f59b6d6556fd8365f7440cf.jpg)