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11:29 MasteringPhysics
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Homework assignment 3 (covers Sections 3.1, 3.2, 3.3, 3.4, 3.5)
Due: 2:00am on Monday, April 19, 2010
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
First review Sections 3.1, 3.2, 3.3, 3.4, and 3.5 of Young and Freedman, including the worked examples. You should then be able to solve the problems given below. Note that you are allowed only 6 answer attempts per problem.
Conceptual Problem about Projectile Motion
Description: Conceptual questions about speed, peak characteristics, flight time, and range for an object undergoing projectile motion. Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: 1. An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, , is constant. 2. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , changes continuously. 3. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time the projectile is in the air.
The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time corresponds to the moment just after the ball is launched from position and . . Its launch velocity, also
called the initial velocity, is
Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time with velocity . Its position at this moment is denoted by its maximum height. The other point, at time or with velocity since it is at , corresponds
to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( since it is at its maximum horizontal range.
Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence .
Part A How do the speeds ANSWER: , = = = , and = > > (at times >0 =0 >0 , , and ) compare?
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> >
> >
>0 =0 . This is because and include vertical speed as well as the
Here equals by symmetry and both exceed constant horizontal speed. Consider a diagram of the ball at time . Recall that
refers to
the instant just after the ball has been launched, so it is still at ground level ( ). However, it is already moving with initial velocity direction is positive x direction. , whose magnitude is counterclockwise from the and
Part B What are the values of the intial velocity vector components components Hint B.1 and (both in and (both in ." ) as well as the acceleration vector
)? Here the subscript 0 means "at time
Determining components of a vector that is aligned with an axis
If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint B.2 Calculating the components of the initial velocity points up and to the right. Since "up" is the positive y axis direction and "to the right" is the positive will both be positive. , , and are three sides of a right triangle, one angle of which is . Thus and and can
Notice that the vector x axis direction, and
As shown in the figure,
be found using the definition of the sine and cosine functions given below. Recall that and note that ,
What are the values of
and
?
Enter your answers numerically in meters per second separated by a comma. ANSWER:
ANSWER:
15.0, 26.0, 0, -9.80
Also notice that at time and components (0, -9.80
, just before the ball lands, its velocity components are (the same size but opposite sign from ), exactly the same as at
(the same as always) will have
by symmetry). The acceleration at time
, as required by Rule 2.
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The peak of the trajectory occurs at time
. This is the point where the ball reaches its maximum height
. At the peak the
ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components components ANSWER: and (both in and (both in ) as well as the acceleration vector .
)? Here the subscript 1 means that these are all at time
15.0, 0, 0, -9.80
At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched ( ) until just before it lands ( ). Hence the flight time can be calculated as , or just in this particular situation since . Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height Hint D.1 Kicking a ball of cliff; a related problem , how long would it take to reach the ground? Ignore air resistance.
Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint D.1.1 Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity ( ) as well as the same acceleration (
downward). They differ only in their x velocity (one is zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time.
The fact that one ball moves horizontally as it falls does not influence its vertical motion. Hence both balls are at the same height at all moments in time and thus they strike the ground at the same instant. Now return to the original question, in which you are asked to compare the flight time for a ball that rises from the ground to a peak and then falls back down to the ground with the flight time for a second ball that only needs to fall from the peak height to the ground.
Check all that apply. ANSWER:
In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range is defined as of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range , or just in this particular situation since . (which is the same at all
Range can be calculated as the product of the flight time
and the x component of the velocity
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times, so
). The value of
can be found from the launch speed
and the launch angle
using trigonometric
functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from and using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position ( ) at time with initial speed and launch angle measured from the horizontal. As was the case above, refers to the flight time and flight time: refers to the range of the projectile.
range:
In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees.
Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Increase Reduce Reduce Reduce Increase above 30 below 30 from 60 from 60 from 60 . to 45 . . . .
to less than 30 up toward 90
A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable.
Position, Velocity, and Acceleration
Description: Conceptual questions about when the position, velocity and acceleration (or their magnitudes) change in a collision or in uniform circular motion. Learning Goal: To identify situations when position, velocity, and /or acceleration change, realizing that change can be in direction or magnitude. If an object's position is described by a function of time, (measured from a nonaccelerating reference frame), then the , and the object's acceleration is described by
object's velocity is described by the time derivative of the position, the time derivative of the velocity, .
It is often convenient to discuss the average of the latter two quantities between times
and
:
and
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.
Part A You throw a ball. Air resistance on the ball is negligible. Which of the following functions change with time as the ball flies through the air? Hint A.1 Newton's 2nd Law ) says that every acceleration is caused by a force. What force acts on the ball after it leaves
Newton's 2nd Law (
your hand? Does this force change during the flight of the ball, or is it constant through time? ANSWER: only the position of the ball only the velocity of the ball only the acceleration of the ball the position and velocity of the ball the position and the velocity and acceleration of the ball
Part B You are driving a car at 65 mph. You are traveling north along a straight highway. What could you do to give the car a nonzero acceleration? Hint B.1 What constitutes a nonzero acceleration?
The velocity of the car is described by a vector function, meaning it has both magnitude (65 mph) and direction (north). The car experiences a nonzero acceleration if you change either the magnitude of the velocity or the direction of the velocity. ANSWER: Press the brake pedal. Turn the steering wheel. Either press the gas or the brake pedal. Either press the gas or the brake pedal or turn the steering wheel.
Part C A ball is lodged in a hole in the floor near the outside edge of a merry-go-round that is turning at constant speed. Which kinematic variable or variables change with time, assuming that the position is measured from an origin at the center of the merry-go-round? Hint C.1 Change of a vector
A vector quantity has both magnitude and direction. The vector changes with time if either of these quantities changes with time. ANSWER: the position of the ball only the velocity of the ball only the acceleration of the ball only both the position and velocity of the ball the position and velocity and acceleration of the ball
Part D For the merry-go-round problem, do the magnitudes of the position, velocity, and acceleration vectors change with time? Hint D.1 Change of magnitude of a vector
A vector quantity has both magnitude and direction. The magnitude of a vector changes with time only if the length changes with time. ANSWER: yes no
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Introduction to Projectile Motion
Description: Conceptual questions about projectile motion and some easy calculations. (uses applets) Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . Consider a particle with initial velocity axis. Part A What is the x component of ? that has magnitude 12.0 and is directed 60.0 above the negative x
Express your answer in meters per second. ANSWER: =
Part B What is the y component of ?
Express your answer in meters per second. ANSWER: =
Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The xcomponent motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle's shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity.
As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is
no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation .
Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 Part D
.
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How long Hint D.1
does it take for the balls to reach the ground? Use 10 How to approach the problem at time
for the magnitude of the acceleration due to gravity.
The balls are released from rest at a height of 5.0
. Using these numbers and basic kinematics, you can
determine the amount of time it takes for the balls to reach the ground. Express your answer in seconds to two significant figures. ANSWER: =
This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Hint E.1 How to approach the problem
Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is .
Express your answer in meters per second to two significant figures. ANSWER: =
You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time).
A Wild Ride
Description: Find the speed of a roller coaster car given its trajectory. A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from to s, the trajectory of the car along a certain section of the track is given by , where Part A At Hint A.1 is the roller coaster car ascending or descending? How to approach the problem is a positive dimensionless constant.
The direction of motion of a particle is given by the direction of its velocity. In this particular case, you need to establish if velocity of the car points upward or downward. That can be easily determined by simply looking at the sign of the vertical component of the velocity of the car. Hint A.2 Find the vertical component of the velocity of the car , at .
Find the y component of the velocity of the car, Hint A.2.1 Velocity components
Consider a particle moving in the xy plane with position vector defined as the time derivative of , that is,
. The instantaneous velocity
of the particle is
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.
Then, the components of the velocity vector are .
Express your answer in meters per second in terms of ANSWER:
.
=
What does a negative y component of velocity tell you about the direction of the motion of the car? ANSWER: ascending descending
Part B Derive a general expression for the speed Hint B.1 of the car.
How to approach the problem
The speed of a particle is the magnitude of the velocity vector of the particle. Since the magnitude of a vector depends on its components, to find the speed of the car you need to know the components of the car's velocity. Hint B.2 Magnitude of a vector , whose components are and , is given by .
The magnitude of a vector
Hint B.3
Find the components of the velocity of the car and , that is, the x and y components of the velocity of the car.
Find a general expression for
Hint B.3.1 The velocity vector Consider a particle moving in the xy plane with position vector defined as the time derivative of , that is, . . The instantaneous velocity of the particle is
Then, the components of the velocity vector are .
Express your answers in meters per second in terms of ANSWER: , =
and . Separate the velocity components with a comma.
Express your answer in meters per second in terms of ANSWER: =
and .
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Part C The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding the maximum value of Hint C.1 allowed by these regulations. . Find
How to approach the problem
To comply with the regulations, the speed of the car cannot exceed the given safety limit at any time. Thus, you need to determine what the maximum value of the speed is and impose the condition that such a value cannot be greater than the safety limit. Hint C.2 Find the maximum value of the speed of the car in terms of .
Given the expression found in Part B, find the maximum speed Hint C.2.1 Using the calculus Recall that a function has a local maximum at if
and
, where
and
.
If you are trying to find the maximum value of a function over a fixed interval, you must also check whether the function is maximized at one of the endpoints of the interval. It might help to sketch the velocity versus time. Hint C.2.2 Find the first derivative of the speed As you found in Part B, the speed of the car is described by the following function:
. Find an expression for the first derivative of the speed with respect to time. Express your answer as a function of ANSWER: = and .
Note that the first derivative of the speed is not necessarily equal to the magnitude of the car's acceleration. In two dimensions, to find the magnitude of the acceleration, you would first need to find the x and y components of the acceleration, then compute the magnitude of the acceleration vector. Hint C.2.3 Find the time at which the speed reaches its maximum value At what time does the speed reach its maximum value?
Remember to check not only the points where the derivative of the function is zero but also the endpoints of the interval to . Express your answer in seconds. ANSWER: =
Now calculate the value of the speed of the car at Express your answer in meters per second. ANSWER: =
.
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Now impose the condition that the maximum value of the speed of the car cannot be greater than the safety limit and solve for . Express your answer using two significant figures. ANSWER: =
Accelerating along a Racetrack
Description: Determine the direction of acceleration of cars at various points along a race track A road race is taking place along the track shown in the figure . All of the cars are moving at constant speeds. The car at point F is traveling along a straight section of the track, whereas all the other cars are moving along curved segments of the track.
Part A Let be the velocity of the car at point A. What can you say about the acceleration of the car at that point? Acceleration along a curved path
Hint A.1
Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector along the curved path and is directed toward the center of curvature of the path. ANSWER: The acceleration is parallel to . and directed toward the inside of the track. and directed toward the outside of the track. . at each point
The acceleration is perpendicular to The acceleration is perpendicular to
The acceleration is neither parallel nor perpendicular to The acceleration is zero.
Part B Let be the velocity of the car at point C. What can you say about the acceleration of the car at that point? Acceleration along a curved path
Hint B.1
Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector along the curved path and is directed toward the center of curvature of the path. ANSWER: The acceleration is parallel to . and pointed toward the inside of the track.
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at each point
The acceleration is perpendicular to
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The acceleration is perpendicular to
and pointed toward the outside of the track. .
The acceleration is neither parallel nor perpendicular to The acceleration is zero.
Part C Let be the velocity of the car at point D. What can you say about the acceleration of the car at that point? Acceleration along a curved path
Hint C.1
Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector along the curved path and is directed toward the center of curvature of the path. ANSWER: The acceleration is parallel to . and pointed toward the inside of the track. and pointed toward the outside of the track. . at each point
The acceleration is perpendicular to The acceleration is perpendicular to
The acceleration is neither parallel nor perpendicular to The acceleration is zero.
Part D Let be the velocity of the car at point F. What can you say about the acceleration of the car at that point? Acceleration along a straight path
Hint D.1
The velocity of an object that moves along a straight path is always parallel to the direction of the path, and the object has a nonzero acceleration only if the magnitude of its velocity changes in time. ANSWER: The acceleration is parallel to . and pointed toward the inside of the track. and pointed toward the outside of the track. .
The acceleration is perpendicular to The acceleration is perpendicular to
The acceleration is neither parallel nor perpendicular to The acceleration is zero.
Part E Assuming that all cars have equal speeds, which car has the acceleration of the greatest magnitude, and which one has the acceleration of the least magnitude? Hint E.1 How to approach the problem
Recall that the magnitude of the acceleration of an object that moves at constant speed along a curved path is inversely proportional to the radius of curvature of the path. Use A for the car at point A, B for the car at point B, and so on. Express your answer as the name the car that has the greatest magnitude of acceleration followed by the car with the least magnitude of accelation, and separate your answers with a comma.
F
Part F Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car at point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E. Hint F.1 Let Find the acceleration of the car at point E of the
be the radius of the two curves along which the cars at points A and E are traveling. What is the magnitude
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acceleration of the car at point E? Hint F.1.1 Uniform circular motion The magnitude given by . of the acceleration of an object that moves with constant speed along a circular path of radius is
Express your answer in terms of the radius of curvature ANSWER: =
and the speed
of car E.
Hint F.2 If
Find the acceleration of the car at point A , what is the acceleration of the car at point A? Let be the radius of the two curves along which the cars at
points A and E are traveling. Hint F.2.1 Uniform circular motion The magnitude of the acceleration of an object that moves with constant speed by . along a circular path of radius is given
Express your answer in terms of the speed ANSWER: =
of the car at E and the radius .
Since the magnitude of the acceleration of an object that moves with constant speed along a circular path is proportional to the square of the speed, the acceleration of the car at point A is proportional to .
ANSWER:
The magnitude of the acceleration of the car at point A is twice that of the car at point E. The magnitude of the acceleration of the car at point A is the same as that of the car at point E. The magnitude of the acceleration of the car at point A is half that of the car at point E. The magnitude of the acceleration of the car at point A is four times that of the car at point E.
Arrow Hits Apple
Description: A variation of trajectory problem; good test of understanding An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at
the same height above the ground as it was shot. Use Part A Find , the time that the arrow spends in the air.
for the magnitude of the acceleration due to gravity.
Hint A.1
Find the initial upward component of velocity in terms of D. for the initial components of velocity. Then use kinematics to relate them and of the initial velocity?
Introduce the (unknown) variables and solve for . What is the vertical component Hint A.1.1 Find
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Find the horizontal component
of the initial velocity. and given symbolic quantities.
Express your answer symbolically in terms of ANSWER: =
Hint A.1.2
Find of the initial velocity? .
What is the vertical component
Express your answer symbolically in terms of ANSWER: =
Express your answer symbolically in terms of ANSWER: =
and
.
Hint A.2
Find the time of flight in terms of the initial vertical component of velocity. in terms of and .
From the change in the vertical component of velocity, you should be able to find Hint A.2.1 Find
When applied to the y-component of velocity, in this problem the formula for
with constant acceleration
is
What is
, the vertical component of velocity when the arrow hits the tree? only.
Answer symbolically in terms of ANSWER: =
Give your answer in terms of ANSWER: =
and .
Hint A.3
Put the algebra together to find
symbolically. and , and another in terms of
If you have an expression for the initial vertical velocity component in terms in terms of and , you should be able to eliminate this initial component to find an expression for
Express your answer symbolically in terms of given variables. ANSWER: =
Answer numerically in seconds, to two significant figures.
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ANSWER:
=
Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Hint B.1 When should the apple be dropped
The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint B.2 Find the time it takes for the apple to fall 6.0 meters
How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: =
Express your answer numerically in seconds, to two significant figures. ANSWER: =
Circular Launch
Description: Simple combination of centripetal acceleration and projectile motion. Short and quantitative. A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 .
Part A How far from the bottom of the chute does the ball land? Hint A.1 Speed of ball upon leaving chute
How fast is the ball moving at the top of the chute? Hint A.1.1 Equation of motion The centripetal acceleration for a particle moving in a circle is radius of rotation. ANSWER:
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, where
is its speed and
is its instantaneous
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ANSWER: =
Hint A.2
Time of free fall
How long is the ball in free fall before it hits the ground? Hint A.2.1 Equation of motion There is constant acceleration due to gravity, so you can use the general expression .
Write the values of , , and (separated by commas) that are appropriate for this situation. Use the standard convention that is the magnitude of the acceleration due to gravity. Take at the ground, and take the positive y direction to be upward. ANSWER: , , =
Hint A.2.2 Equation for the height of the ball To find the time in free fall before the ball hits the ground, the ground. Answer in terms of ANSWER: , , and . , set the general equation for the height equal to the height of
Express the free-fall time in terms of ANSWER: =
and .
Hint A.3
Finding the horizontal distance , where . and were found in Parts i and ii
The horizontal distance follows from respectively.
Your answer for the distance the ball travels from the end of the chute should contain ANSWER:
.
=
Crossing a River
Description: Find the speed necessary to cross a river with a current, swimming at a specific angle. A swimmer wants to cross a river, from point A to point B, as shown in the figure. The distance (from A to C) is 200 , the distance (from C to B) is 150 , and the speed of the current
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in the river is 5
. Suppose that the swimmer's velocity with the
relative to the water makes an angle of line from A to C, as indicated in the figure.
Part A To swim directly from A to B, what speed Hint A.1 , relative to the water, should the swimmer have?
Use the motion in the y direction to cross the river and arrive at B. Find an expression for by considering only (the
Suppose it takes the swimmer time
y component of the swimmer's motion with respect to the shore). Hint A.1.1 The y component of the velocity relative to the shore The y component (see the figure in the problem introduction) of the swimmer's velocity relative to the shore is . Answer in terms of ANSWER: , , and .
Hint A.2
Use the motion in the x direction by considering only the x component of the swimmer's motion.
Now find an expression for
Hint A.2.1 The x component of the velocity relative to the shore The x component (see the figure in the problem introduction) of the swimmer's velocity relative to the shore is . Express your answer in terms of ANSWER: , , , and .
Note that this is the same time that you found when you considered only the motion in the y direction, even though the expressions are different. If you set both expressions equal to each other, you can solve for .
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Hint A.3
Solve for . If you used the previous hints, solve for by eliminating from the two equations you
Find a symbolic expression for derived.
Express your answer in terms of ANSWER: =
,
,
, and .
Express the swimmer's speed numerically, to three significant figures, in kilometers per hour. ANSWER: =
Another way to do this problem is to add the swimmer's and the river's velocities vectorially, and set the angle that the resultant vector makes with AC or the river bank equal to that which AB makes with the same.
Curved Motion Diagram
Description: Motion diagram for pendulum released from 45 degrees: find acceleration (velocity hints), identify possibly physical problem. Hard version. The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a "compass in rose" which directions are labeled with the letters of the alphabet.
Part A What is the direction of the acceleration of the object at moment 5? Hint A.1 How to approach the problem
The acceleration of the object at moment 5 is the acceleration found from the change in velocity between moments 4 and 5 and moments 5 and 6. Hint A.2 Definition of acceleration
Acceleration is defined as the change in velocity per unit time. Mathematically, . Since velocity is a vector, acceleration is a vector that points in the direction of the change in the velocity. Hint A.3 Change of velocity: a graphical interpretation to a final value . Then, the
Let us assume that in a second the velocity of an object changes from an initial value change in velocity that the object undergoes in that interval of time is
. If one represents the velocity of the
object graphically with vectors, then the change of velocity can be evaluated simply by applying the rule of subtraction of vectors, as shown in the picture.
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Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length.
Part B What is the direction of the acceleration of the object at moments 0 and 10? Hint B.1 Find the direction of the velocity
What is the direction of the velocity of this object at moments 1 and 9? Enter the letters of the corresponding directions from the compass rose, separated by commas. Type Z if the velocity vector has zero length. directions at time step 1, time step 9 = B
Hint B.2
Definition of acceleration
Acceleration is defined as the change in velocity per unit time. Mathematically, . Since velocity is a vector, acceleration is a vector that points in the direction of the change in the velocity. Hint B.3 Applying the definition of acceleration
To find the acceleration at moment 0, subtract the (vector) velocity at moment 0 from the velocity at moment 1. Similarly, to find the acceleration at moment 10, subtract the (vector) velocity at moment 9 from the velocity at moment 10. Enter the letters corresponding to the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. directions at time step 0, time step 10 = F
The Archerfish
Description: Find the initial speed of a drop of water that an archerfish spits to hit a floating insect, given the position of the insect and the launch angle. The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers. When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating
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a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface. Part A At what speed surface. Hint A.1 How to approach the problem should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle above the water
This problem involves projectile motion. A drop of water is launched at a certain angle above the surface, and you need to calculate the initial speed required to hit the target. Note that the target is an insect floating over the surface of the water. Hint A.2 Find how long it takes the water drop to fall back into the pond above the surface with an initial speed . The time of the drop. Find a formula for it takes the drop to fall and
Assume that the drop is launched at an angle back into the water depends on the initial speed , the acceleration due to gravity. Hint A.2.1 Projectile motion
that depends only on , the angle
The motion of a projectile can be described by two sets of equations: one that describes uniform motion at constant velocity in the horizontal direction and another one describing free-fall motion in the vertical direction. Specifically, the equations for displacement are
, where and are, respectively, the x coordinate and the y coordinate of the initial position of the projectile, and are, respectively, the x and y components of the initial velocity of the projectile, is the acceleration due to gravity, and is time. Note that if the launch angle is known, then the components of the initial velocity are given by
,
where
is the initial speed of the projectile.
Hint A.2.2 Initial and final position of the projectile Since the question asks for the time it takes the drop to fall back into the water, the initial and final vertical position of the drop must be the same. This gives you enough information to find from the y component of the equation for the projectile motion. Express your answer in terms of the angle , the initial speed , and , the acceleration due to gravity. ANSWER: =
Hint A.3 Find how far from the fish the drop falls Now that you have found an expression for the time drop's initial speed , determine at what distance terms of that it takes for the drop to fall back into the water in terms of the from the fish the drop falls. As previously, find an expression for in
using the information given in the problem introduction and your result found in the previous hint.
Express your answer in terms of the angle , the initial speed , and , the acceleration due to gravity. ANSWER: =
Now make the distance
equal to the location of the insect and solve for .
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Express your answer in meters per second. ANSWER: =
Some archerfish can "shoot" as far as 3.5
, hitting their targets with reasonable accuracy as far as 1.2
. They have
binocular vision, which helps them judge distance. Part B Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle insect? Hint B.1 How to approach the problem above the surface and with the same initial speed as before. At what height above the surface was the
Use the kinematics equations that describe projectile motion. However, now you are given the projectile's initial speed, launch angle, and final distance along the x axis and you need to calculate the corresponding height reached by the projectile. Hint B.2 Find the time it takes the water drop to hit the insect
Considering that a water drop launched by the archerfish successfully hit the insect resting on the leaf, how long did it take? Hint B.2.1 Kinematics equation in the horizontal direction Since you know that the insect is located at a horizontal distance 0.600 from the fish, you can calculate the time it takes the water drop to travel this distance by using the x component of the equation for the projectile motion, that is, the equation of motion at constant velocity. Express your answer in seconds. ANSWER:
=
Now calculate the height reached by the drop at this time. Express your answer in meters. ANSWER: =
Experiments have shown that the archerfish can predict the point where the disabled prey will fall and hit the water by simply looking at the initial trajectory of the dislodged insect for only 10 . The archerfish then darts to the place where it has "calculated" the insect will hit the water, planning to get there before another fish does.
Delivering a Package by Air
Description: A package is dropped out of a plane. How far from an island should the plane be when the package is dropped in order for it to land on the island? A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 270 at an altitude of 900 . The positive x and y directions
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are defined in the figure. For all parts, assume that the "island" refers to the point at a distance from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is = 9.80 .
Part A After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 270 in the horizontal direction. Hint A.1 Knowns and unknowns: what are the initial conditions?
Take the origin of the coordinate system to be at the point on the surface of the water directly below the point at which the package is released. The directions of the axes are shown in the figure in the problem introduction. In this coordinate system, what are the values of , , , of the package? Hint A.1.1 Initial velocity in the y direction Because the package is ejected horizontally, the vertical component of its initial velocity is zero. Express your answers numerically and enter them, separated by commas, in the order meters and mph for distances and speeds, respectively. ANSWER: , , , = , , , . Use units of
Hint A.2
What are the knowns and unknowns when the package hits the ground?
Take the origin of the coordinate system to be at the point on the surface of the water directly below the point at which the package is released. The directions of the axes are shown in the figure in the problem introduction. Let be the time when the package hits the ground. In this coordinate system, Which of the following values is/are known? Check all that apply. ANSWER: 270 since there is no acceleration in the x direction.
Of course, you also know that
.
Hint A.3
Find the best equation to use when the packet hits the ground?
Which of the equations below could you use to find the time Hint A.3.1 How to determine which equation to use Only one of the quantities variable. ANSWER: , ,
is known. Which one? You have to use the equation that contains this
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Substitute the values into this equation and solve for
.
Express your answer numerically in seconds. Neglect air resistance. ANSWER: =
Part B If the package is to land right on the island, at what horizontal distance released? Hint B.1 How to approach the problem , which is also the change in the x coordinate of the package over the time spent in the air. You . from the plane to the island should the package be
You are asked to find
should have calculated this time interval in Part A. Use it to find Hint B.2 The equation for
Since there is no acceleration in the horizontal direction, the equation for . Express the distance numerically in meters. ANSWER: =
is
Part C What is the speed Hint C.1 of the package when it hits the ground?
How to approach the problem . You already know that 270 , so you need to find .
The speed is the magnitude of the velocity, i.e.,
Hint C.2
The equation for the velocity in the y direction is .
The equation for the velocity in the y direction
If you have completed the earlier parts, you should know all the quantities on the right side. Express your answer numerically in miles per hour. ANSWER: =
Part D The speed at which the package hits the ground is really fast! If a package hits the ground at such a speed, it can be crushed and also cause some serious damage on the ground. Which of the following would help decrease the speed with which the package hits the ground? ANSWER: Increase the plane's speed and height Decrease the plane's speed and height
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This is why it would be nice for rescue teams to have hybrid airplane-helicopters. Of course, then they can just airlift the stranded group.
Graphing Projectile Motion
Description: Graph x and y components of position and velocity vs. time for a given motion diagram. (graphing applet) For the motion diagram given , sketch the shape of the corresponding motion graphs in Parts A to D. Use the indicated coordinate system. One unit of time elapses between consecutive dots in the motion diagram.
Part A Construct a possible graph for x position versus time, Hint A.1 Determine the initial value of .
Is the initial value of the x position positive, negative, or zero? ANSWER: positive negative zero
Hint A.2
Specify the shape of the
graph
Does the x position change at a constant rate or a changing rate? You can determine this by looking at the change in x coordinate from one dot to the next. ANSWER: constant changing
Since the x position changes at a constant rate (implying a constant x velocity), it must be represented by a graph with a constant slope. ANSWER:
View
Part B Construct a possible graph for the y position versus time, Hint B.1 Determine the initial value of .
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Is the initial value of the y position positive, negative, or zero? ANSWER: positive negative zero
Hint B.2
Specify the shape of the
graph
Does the y position change at a constant rate or a changing rate? ANSWER: constant changing
Since the y position changes at a variable rate (implying a changing y velocity), it must be represented by a graph with a changing slope. ANSWER:
View
Part C Construct a possible graph for the x velocity versus time, Hint C.1 Determine the initial value of .
Is the initial value of the x velocity positive, negative, or zero? Look at the x component of the first arrow. ANSWER: positive negative zero
Hint C.2
Specify the shape of the
graph
Does the x velocity remain constant or does it change? You can determine this by comparing the x components of the arrows. ANSWER: It remains constant. It changes.
ANSWER:
View
Part D Construct a possible graph for the y velocity versus time, Hint D.1 Determine the initial value of .
Is the initial value of the y velocity positive, negative, or zero? Look at the y component of the first arrow.
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ANSWER:
positive negative zero
Hint D.2
Specify the shape of the
graph
Does the y velocity remain constant or does it change? You can determine this by comparing the y components of the arrows. ANSWER: It remains constant. It changes.
Hint D.3 Specify the rate of change of Does the y velocity change at a constant rate or a changing rate? It may be helpful to write down the y components of successive arrows. See if the differences between successive arrows' y components stay the same. ANSWER: constant changing
Since the y velocity changes at a constant rate (implying a constant y acceleration), it must be represented by a graph with a constant slope. ANSWER:
View
An Object Accelerating on a Ramp
Description: Find the direction of the acceleration vector at various points of the trajectory of a ball sliding down a curved ramp. Learning Goal: Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the x-y plane), the definition of acceleration is in the limit .
In picturing this vector derivative you can think of the derivative of a vector as an instantaneous quantity by thinking of the velocity of the tip of the arrow as the vector changes in time. Alternatively, you can (for small ) approximate the acceleration as .
Obviously the difference between same direction, will be parallel to
and
is another vector that can lie in any direction. If it is longer but in the has the same magnitude as can differ from but is in a slightly
. On the other hand, if . In general, .
different direction, then direction, hence
will be perpendicular to
in both magnitude and
can have any direction relative to
This problem contains several examples of this.Consider an object sliding on a frictionless ramp as depicted here. The object is
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already moving along the ramp toward position 2 when it is at position 1. The following questions concern the direction of the object's acceleration vector, . In this problem, you should find the direction of the acceleration vector by drawing the velocity vector at two points near to the position you are asked about. Note that since the object moves along the track, its velocity vector at a point will be tangent to the track at that point. The acceleration vector will point in the same direction as the vector difference of the two velocities. (This is a result of the equation given above.)
Part A Which direction best approximates the direction of Hint A.1 Consider the change in velocity when the object is at position 1?
At this point, the object's velocity vector is not changing direction; rather, it is increasing in magnitude. Therefore, the object's acceleration is nearly parallel to its velocity. ANSWER: straight up downward to the left downward to the right straight down
Part B Which direction best approximates the direction of Hint B.1 Consider the change in velocity is not changing. Therefore, no component of the is changing there is an acceleration. when the object is at position 2?
At this point, the speed has a local maximum; thus the magnitude of
acceleration vector is parallel to the velocity vector. However, since the direction of ANSWER:
straight up upward to the right straight down downward to the left
Even though the acceleration is directed straight up, this does not mean that the object is moving straight up. Part C Which direction best approximates the direction of Hint C.1 Consider the change in velocity is not changing. Therefore, no component of the is changing there is an acceleration. when the object is at position 3?
At this point, the speed has a local minimum; thus the magnitude of
acceleration vector is parallel to the velocity vector. However, since the direction of ANSWER:
upward to the right to the right straight down downward to the right
Advice for the Quarterback
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Description: Find the initial velocity needed to throw a football a given distance and have it arrive at a given time. A quarterback is set up to throw the football to a receiver who is running with a constant velocity quarterback and is now a distance angle directly away from the
away from the quarterback. The quarterback figures that the ball must be thrown at an after it is thrown to avoid
to the horizontal and he estimates that the receiver must catch the ball a time interval
having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem.
Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Equation of motion in y direction , the height of the ball as a function of time? .
Hint A.1
What is the expression for Answer in terms of , , and ANSWER: =
Hint A.2
Height at which the ball is caught, the ball was caught at the same height as it had been released. That is, and . .
Remember that after time Express ANSWER: = in terms of
Part B Find , the initial horizontal component of velocity of the ball. Receiver's position
Hint B.1 Find
, the receiver's position before he catches the ball. , , and .
Answer in terms of ANSWER:
=
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Hint B.2 Find
Football's position
, the horizontal distance that the ball travels before reaching the receiver. and .
Answer in terms of ANSWER:
=
Express your answer for ANSWER:
in terms of
,
, and
.
=
Part C Find the speed Hint C.1 with which the quarterback must throw the ball.
How to approach the problem
Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. Answer in terms of ANSWER: = , , , and .
Part D Assuming that the quarterback throws the ball with speed it. Hint D.1 Find angle from and and . Find the angle that this vector would make with , find the angle above the horizontal at which he should throw
Think of velocity as a vector with Cartesian coordinates the x axis using the results of Parts A and B.
Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . ANSWER:
=
Direction of Acceleration of Pendulum
Description: Questions about the direction and relative magnitude of acceleration at various points in the pendulum trajectory. Conceptual.
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Learning Goal: To understand that the direction of acceleration is in the direction of the change of the velocity, which is unrelated to the direction of the velocity. The pendulum shown makes a full swing from to .
Ignore friction and assume that the string is massless. The eight labeled arrows represent directions to be referred to when answering the following questions.
Part A Which of the following is a true statement about the acceleration of the pendulum bob, . ANSWER: is equal to the acceleration due to gravity. is equal to the instantaneous rate of change in velocity. is perpendicular to the bob's trajectory. is tangent to the bob's trajectory.
Part B What is the direction of Hint B.1 when the pendulum is at position 1?
Velocity at position 1
What is the velocity of the bob when it is exactly at position 1? ANSWER: =
Hint B.2
Velocity of bob after it has descended
What is the velocity of the bob just after it has descended from position 1? ANSWER: very small and having a direction best approximated by arrow D very small and having a direction best approximated by arrow A very small and having a direction best approximated by arrow H The velocity cannot be determined without more information.
Enter the letter of the arrow parallel to ANSWER: H
.
Part C What is the direction of Hint C.1 at the moment the pendulum passes position 2?
Instantaneous motion
At position 2, the instantaneous motion of the pendulum can be approximated as uniform circular motion. What is the direction of acceleration for an object executing uniform circular motion?
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Enter the letter of the arrow that best approximates the direction of
.
We know that for the object to be traveling in a circle, some component of its acceleration must be pointing radially inward. Part D What is the direction of Hint D.1 when the pendulum reaches position 3?
Velocity just before position 3
What is the velocity of the bob just before it reaches position 3? ANSWER: very small and having a direction best approximated by arrow B very small and having a direction best approximated by arrow C very small and having a direction best approximated by arrow H The velocity cannot be determined without more information.
Hint D.2
Velocity of bob at position 3
What is the velocity of the bob when it reaches position 3? ANSWER: =
Give the letter of the arrow that best approximates the direction of
.
Part E As the pendulum approaches or recedes from which position(s) is the acceleration vector vector . position 2 only positions 1 and 2 positions 2 and 3 positions 1 and 3 almost parallel to the velocity
ANSWER:
Shooting over a Hill
Description: Finding the necessary parameters so that a shooting is succesfull. A projectile is fired with speed . at an angle from the horizontal as shown in the figure
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Part A Find the highest point in the trajectory, Hint A.1 Velocity at the top , . .
At the highest point of the trajectory Hint A.2 Which equation to use
The three kinematic equations that govern the motion in the y direction are , , and . The third equation contains the height variable , and all the other quantities are known (at find . This would be the simplest method. ), so you could use it to
Alternately, if you prefer, you could first find the time required to reach the maximum height from the first equation, and then use this time in the second equation to solve for . Express the highest point in terms of the magnitude of the acceleration due to gravity , the initial velocity angle . ANSWER: = , and the
Part B What is the range of the projectile, Hint B.1 ?
Find the total time spent in air it takes to reach the highest point (found in
Find the total time the projectile spends in the air, by considering the time Part A) and then the time it takes to fall back to the ground. Express your answer in terms of ANSWER: = , , and .
Hint B.2
Find
What is the x coordinate of the projectile's position? Express your answer in terms of , ANSWER: , and .
=
Express the range in terms of
, , and .
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ANSWER: =
Consider your advice to an artillery officer who has the following problem. From his current postition, he must shoot over a hill of height at a target on the other side, which has the same elevation as his gun. He knows from his accurate map both the bearing and the distance to the target and also that the hill is halfway to the target. To shoot as accurately as possible, he
wants the projectile to just barely pass above the hill. Part C Find the angle Hint C.1 above the horizontal at which the projectile should be fired.
How to approach the problem and in terms of and . Solve these two equations to find in terms of
In the first half of this problem, you found and . Set up the ratio to .
Hint C.2 Find the ratio of
The only variable in your answer should be . ANSWER: =
Express your answer in terms of ANSWER: =
and
.
Recall the following trigonometry formulas: ,
,
and . In this case, since triangle with length , you can draw a right
as one of the angles, an "opposite" side of . You can , after you
, and an "adjacent" side of length and
then use this triangle to find
find the length of the hypotenuse using the Pythagorean Theorem.
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Part D What is the initial speed? Hint D.1 How to approach this part and . You will need to find an expression for and/or
Use one of the equations that you had derived for to find Hint D.2 . Find
Use the expression you derived for Leave your answer in terms of ANSWER: = and
and the Pythagorean theorem to find
.
Hint D.3
Find . and .
Now find the expression for Leave your answer in terms of ANSWER: =
Now use one of the equations that you had derived for previous hints to find Express in terms of , . , and .
and
and the expression for
and/or
from the
ANSWER: =
Part E Find , the flight time of the projectile. How to proceed in terms of and . You can use hints in Part B in the previous half.
Hint E.1 First, find ANSWER:
=
Now you can use the result that you derived for Express the flight time in terms of ANSWER: = and .
and
in terms of
and
to find
.
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Horizontal Cannon on a Cliff
Description: A cannonball is fired horizontally from the edge of a cliff. Three computational questions regarding its projectile motion. A cannonball is fired horizontally from the top of a cliff. The cannon is at height = 90.0 above ground level, and the ball is fired with initial horizontal speed gravity to be = 9.80 . . Assume acceleration due to
Part A Assume that the cannon is fired at time cannonball at the time Hint A.1 ? and that the cannonball hits the ground at time . What is the y position of the
How to approach the problem
In this problem, you are asked to find the height at a certain time. Nothing is asked or given about the distance coordinate . Therefore, you only need to consider the y equations of motion and variables. Write down the known and unknown y variables. Then find the appropriate equation(s) and substitute for the values. Hint A.2 Identify the knowns and unknowns The information given in the introduction can be used to determine the knowns and unknowns in the problem. For this part, you need to consider only the y variables. In terms of the given coordinate system, the initial height can be chosen to be 90.0 . Of course, the acceleration in the y direction is exclusively the acceleration due to gravity: . Which of the following quantities is/are also known? Check all that apply. ANSWER: at time at time
Thus the initial height is
90.0
, the final height is
, and the initial velocity in the y direction is
.
Hint A.3 Determine which equation to use to find the height at the requested time Three equations that describe motion in the y direction are given below. Which would you use to determine the height the cannonball at time ? ANSWER: of
When the known vairables are substituted into this equation, you get
, where
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denotes the height
at time
. In order to find
you need to find
and substitute for it in this equation .
Hint A.4 What is the value of ?
Find
Hint A.4.1 Identify which equation to use to find Which of the equations below could you use to find ANSWER: ?
Substitute the known variables into this equation to find Express your answer numerically in seconds. ANSWER: =
.
Now substitute for
into the equation for
to find
at time
.
Answer numerically in units of meters. ANSWER: =
The same answer can be obtained more easily (perhaps you did it this way) if you notice that the vertical displacement is given by and therefore is one-quarter of
. This means that ; then .
Part B Given that the projectile lands at a distance projectile, Hint B.1 . How to approach the problem can be determined if you know either or and the angle of elevation of the cannon . Then you = 100 from the cliff, as shown in the figure, find the initial speed of the
The initial speed could use either
. or . In this case, , so there is no component of velocity in the vertical direction, and the second equation is not useful. You
need to determine from the given information. In this case, because the cannonball is launched with only an initial horizontal velocity. You are given the horizontal distance traveled by the cannonball and need to find its horizontal velocity (which is constant because the only force acting on the cannonball, gravity, acts exclusively in the vertical direction). You are not asked for or given any information about the y variables. Therefore, you need to consider only the x variables and equations. Hint B.2 Knowns and unknowns
Since you are asked to find , you need to determine the knowns and unknowns only for the x variables. In the coordinate system shown in the figure in the problem introduction, the known/given x variables are
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and 100 Hint B.3 The equation to use .
The equation that describes the motion in the x direction is . Substitute for the known quantities in this equation and solve for . Keep in mind that because the cannonball is launched with an initial horizontal velocity and no initial vertical velocity, and the horizontal component of the cannonball's velocity is constant. Express the initial speed numerically in meters per second. ANSWER: =
Part C What is the y position of the cannonball when it is at distance equation for this projectile, which gives in terms of directly: . from the hill? If you need to, you can use the trajectory
You should already know
from the previous part.
Express the position of the cannonball numerically in meters. ANSWER: =
Not surprisingly, the answer to this part is the same as that in Part A because a projectile travels equal horizontal distances in equal amounts of time.
Uniform Circular Motion
Description: Given an algebraic expression for the displacement vector of an object undergoing uniform circular motion (cartesian coordinates), differentiate to find velocity and acceleration. Learning Goal: To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration. Suppose that a particle's position is given by the following expression:
. Part A Choose the answer that best completes the following sentence: The particle's motion at can be described by ____________. ANSWER: an ellipse starting at time an ellipse starting at time on the positive x axis on the positive y axis
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a circle starting at time a circle starting at time The quantity
on the positive x axis on the positive y axis must have units of radians per second. If
is defined to be the angular velocity of the particle. Note that
is constant, the particle is said to undergo uniform circular motion. Part B When does the particle first cross the negative x axis? Express your answer in terms of some or all of the variables ANSWER: = , , and .
Now, consider the velocity and speed of the particle. Part C Find the particle's velocity as a function of time. Hint C.1 Derivative of . The velocity of the particle will be .
In the problem statement, you were given
Find the derivative of
with respect to time. , , , and .
Express your answer in terms of some or all of the variables ANSWER:
=
Express your answer using unit vectors (e.g., ANSWER: =
+
, where
and
are functions of
,
, , and
).
Part D Find the speed of the particle at time . Hint D.1 Definition of the magnitude of a vector is . Use this to find the speed, which is just the magnitude of the velocity.
The magnitude of
Hint D.2
Complete an mportant trig identity
Complete the following fundamental trigonometric identity: ANSWER: =
Express your answer in terms of some or all of the variables ANSWER:
,
, and
.
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ANSWER:
Note that the speed of the particle is constant:
.
Part E Now find the acceleration of the particle. Express your answer using unit vectors (e.g., ANSWER: = + , where and are functions of , , , and ).
Part F Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position . Express your answer in terms of some or all of the variables ANSWER: and .
=
Part G Now find the magnitude of the acceleration as a function of time. Express your answer in terms of some or all of the variables ANSWER: , , and .
=
Part H Finally, express the magnitude of the particle's acceleration in terms of speed of the particle. Express your answer in terms of one or both of the variables ANSWER: = and . and using the expression you obtained for the
There are three important things to remember about centripetal acceleration: 1. The centripetal acceleration is simply the acceleration of a particle going around in a circle. 2. It has magnitude of either or . 3. It is directed radially inward.
Score Summary:
Your score on this assignment is 0%. You received 0 out of a possible total of 120 points.
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