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6
PHYS2AB Winter 2006, UC San Diego
Homework-6
Assignment is due at 5:00pm on Tuesday, February 21, 2006
Credit for problems submitted late will decrease to 0% after the deadline has passed.
There is no penalty for wrong answers to free response questions. Multiple choice questions are penalized as described in the online help.
The unopened hint bonus is 5% per part.
You are allowed 7 attempts per answer.
Ups and Downs
Description: Several qualitative and conceptual questions involving objects launched upward in the gravitational field of
Earth in the absence of non-conservative forces.
Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the
earth.
In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is
conserved. This is one particular case of the law of conservation of energy.
In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy
and its gravitational potential energy
. The
transformations that take place involve the object's kinetic energy
law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not
change with time. This idea can be expressed by the equation
,
where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the
moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.
First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance.
Part A
As the projectile goes upward, what energy changes take place?
ANSWER:
Both kinetic and potential energy decrease.
Both kinetic and potential energy increase.
i Kinetic energy decreases; potential energy increases.
j
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j Kinetic energy increases; potential energy decreases.
k
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Part B
At the top point of the flight, what can be said about the projectile's kinetic and potential energy?
ANSWER:
Both kinetic and potential energy are at their maximum values.
Both kinetic and potential energy are at their minimum values.
j Kinetic energy is at a maximum; potential energy is at a minimum.
k
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n
i Kinetic energy is at a minimum; potential energy is at a maximum.
j
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Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system "Earth-ball." Although we will
often talk about "the gravitational potential energy of an elevated object," it is useful to keep in mind that the energy, in
fact, is associated with the interactions between the earth and the elevated object.
Part C
The potential energy of the object at the moment of launch __________.
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is negative
is positive
j is zero
k
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n
i depends on the choice of the "zero level" of potential energy
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j
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Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to
at the ground level--but this is not, by any means, the only choice!
assume that
Part D
Using conservation of energy, find the maximum height
to which the object will rise.
Express your answer in terms of and the magnitude of the acceleration of gravity .
ANSWER:
=
You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer.
Part E
At what height above the ground does the projectile have a speed of
?
Express your answer in terms of and the magnitude of the acceleration of gravity .
ANSWER:
=
Part F
What is the speed of the object at the height of
Hint F.1
?
How to approach the problem
You are being asked for the speed at half of the maximum height. You know that at the initial height (
), the speed is .
All of the energy is kinetic energy, and so, the total energy is
. At the maximum height, all of the energy is potential
energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been
converted to potential energy when the projectile is at
. Thus, the kinetic energy must be half of its original value (i.e.,
when
). You need to determine the speed, as a multiple of , that corresponds to such a
kinetic energy.
Express your answer in terms of and . Use three significant figures in the numeric coefficient.
ANSWER:
=
Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution
than can be produced by kinematics.
Part G
A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the
maximum height
of the ball's flight.
Part G.1
Find the final kinetic energy
Find the final kinetic energy of the ball. Here, the best choice of "final" moment is the point at which the ball reaches its
maximum height, since this is the point we are interested in.
Part G.1.a Find the speed at the maximum height
The speed of the ball at the maximum height is __________.
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ANSWER:
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0
n
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Express your answer in terms of , , and .
ANSWER:
=
Express your answer in terms of , , and .
ANSWER:
=
Part H
A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the
horizontal. Using conservation of energy, find the maximum vertical height
to which the ball will climb.
Express your answer in terms of , , and . You may or may not use all of these quantities.
ANSWER:
=
Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball
moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does.
Part I
A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides
up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical
to which the ball will climb.
height
Express your answer in terms of and .
ANSWER:
=
The profile of the hill does not matter; the equation
would have the same terms regardless of the steepness of the hill.
Potential Energy Graphs and Motion
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Description: Several questions that require the students to predict the characteristics of the particle's motion (direction of
force, acceleration, equilibrium positions, etc.) using a potential enrgy diagram U vs. x.
Learning Goal: To be able to interpret potential energy diagrams and predict the corresponding motion of a particle.
Potential energy diagrams for a particle are useful in predicting the motion of that particle. These diagrams allow one to
determine the direction of the force acting on the particle at any point, the points of stable and unstable equilibrium, the
particle's kinetic energy, etc.
Consider the potential energy diagram shown. The curve represents the value of potential energy
particle's coordinate . The horizontal line above the curve represents the constant value
of the total energy of the particle . The total energy is the sum of kinetic ( ) and
potential ( ) energies of the particle.
as a function of the
The key idea in interpreting the graph can be expressed in the equation
where
is the x component of the net force as function of the particle's coordinate .
Note the negative sign: It means that the x component of the net force is negative when
the derivative is positive and vice versa. For instance, if the particle is moving to the
right, and its potential energy is increasing, the net force would be pulling the particle to the left.
If you are still having trouble visualizing this, consider the following: If a massive particle is increasing its gravitational
potential energy (that is, moving upward), the force of gravity is pulling in the opposite direction (that is, downward).
If the x component of the net force is zero, the particle is said to be in equilibrium. There are two kinds of equilibrium:
Stable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle
back toward the equilibrium point (think of a ball rolling between two hills).
Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle
further away from the equilibrium point (think of a ball on top of a hill).
In answering the following questions, we will assume that there is a single varying force acting on the particle along the x
axis. Therefore, we will use the term force instead of the cumbersome x component of the net force.
Part A
The force acting on the particle at point A is __________.
Hint A.1
Sign of the derivative
If a function increases (as increases) in a certain region, then the derivative of the function in that region is positive.
Hint A.2
Sign of the component
If increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive x component points to the
right, and vice versa.
ANSWER:
directed to the right
directed to the left
j equal to zero
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i
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Consider the graph in the region of point A. If the particle is moving to the right, it would be "climbing the hill," and the
force would "pull it down," that is, pull the particle back to the left. Another, more abstract way of thinking about this is to
say that the slope of the graph at point A is positive; therefore, the direction of is negative.
Part B
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The force acting on the particle at point C is __________.
Hint B.1
Sign of the derivative
If a function increases (as increases) in a certain region, then the derivative of the function in that region is positive, and
vice versa.
Hint B.2
Sign of the component
If increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive x component points to the
right, and vice versa.
ANSWER:
directed to the right
directed to the left
j equal to zero
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i
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j
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Part C
The force acting on the particle at point B is __________.
Hint C.1
Derivative of a function at a local maximum
At a local maximum, the derivative of a function is equal to zero.
ANSWER:
directed to the right
directed to the left
i equal to zero
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The slope of the graph is zero; therefore, the derivative
, and
.
Part D
The acceleration of the particle at point B is __________.
Hint D.1
Relation between acceleration and force
The relation between acceleration and force is given by Newton's 2nd law,
.
ANSWER:
directed to the right
directed to the left
i equal to zero
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If the net force is zero, so is the acceleration. The particle is said to be in a state of equilibrium.
Part E
If the particle is located slightly to the left of point B, its acceleration is __________.
Hint E.1
The force on such a particle
To the left of B,
is an increasing function and so its derivative is positive. This implies that the x component of the force
on a particle at this location is negative, or that the force is directed to the left, just like at A. What can you say now about the
acceleration?
ANSWER:
directed to the right
directed to the left
j equal to zero
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j
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i
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Part F
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If the particle is located slightly to the right of point B, its acceleration is __________.
Hint F.1
The force on such a particle
To the right of B,
is a decreasing function and so its derivative is negative. This implies that the x component of the force
on a particle at this location is positive, or that the force is directed to the right, just like at C. What can you now say about
the acceleration?
ANSWER:
directed to the right
directed to the left
j equal to zero
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i
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As you can see, small deviations from equilibrium at point B cause a force that accelerates the particle further away; hence
the particle is in unstable equilibrium.
Part G
Name all labeled points on the graph corresponding to unstable equilibrium.
Hint G.1
Definition of unstable equilibrium
Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle
further away from the equilibrium point (think of a ball on top of a hill).
List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your
answer as BDE.
ANSWER:
BF
Part H
Name all labeled points on the graph corresponding to stable equilibrium.
Hint H.1
Definition of stable equilibrium
Stable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle back
toward the equilibrium point. (Think of a ball rolling between two hills.)
List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your
answer as BDE.
ANSWER:
DH
Part I
Name all labeled points on the graph where the acceleration of the particle is zero.
Hint I.1
Relation between acceleration and force
The relation between acceleration and force is given by Newton's 2nd law,
.
List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your
answer as BDE.
ANSWER:
BDFH
Your answer, of course, includes the locations of both stable and unstable equilibrium.
Part J
Name all labeled points such that when a particle is released from rest there, it would accelerate to the left.
Part J.1
Determine the sign of the x component of force
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If the acceleration is to the left, so is the force. This means that the x component of the force is __________.
ANSWER:
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Part J.2
positive
negative
What is the behavior of
?
If the x component of the force at a point is negative, then the derivative of
is __________.
the region around the point
ANSWER:
i
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at that point is positive. This means that in
increasing
decreasing
List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your
answer as BDE.
ANSWER:
AE
Part K
Consider points A, E, and G. Of these three points, which one corresponds to the greatest magnitude of acceleration of the
particle?
Hint K.1
Acceleration and force
The greatest acceleration corresponds to the greatest magnitude of the net force, represented on the graph by the magnitude
of the slope.
ANSWER:
A
E
jG
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If the total energy
the particle since
of the particle is known, one can also use the graph of
to draw conclusions about the kinetic energy of
.
As a reminder, on this graph, the total energy
is shown by the horizontal line.
Part L
What point on the graph corresponds to the maximum kinetic energy of the moving particle?
Hint L.1
, , and
Since the total energy does not change, the maximum kinetic energy corresponds to the minimum potential energy.
ANSWER:
D
It makes sense that the kinetic energy of the particle is maximum at one of the (force) equilibrium points. For example,
think of a pendulum (which has only one force equilibrium point--at the very bottom).
Part M
At what point on the graph does the particle have the lowest speed?
ANSWER:
B
As you can see, many different conclusions can be made about the particle's motion merely by looking at the graph. It is
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helpful to understand the character of motion qualitatively before you attempt quantitative problems. This problem should
prove useful in improving such an understanding.
Problem 7.12
Tarzan and Jane. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an
angle of with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his
vine makes an angle of with the vertical.
Part A
Calculate Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.
ANSWER:
m/s
Problem 7.39
A man with mass sits on a platform suspended from a movable pulley, as shown in the figure , and raises himself at constant
speed by a rope passing over a fixed pulley. The platform and the pulleys have negligible mass. Assume
that there are no friction losses.
Part A
Find the force he must exert.
Take the free fall acceleration to be .
ANSWER:
Part B
Find the increase in the energy of the system when he raises himself a distance . (Answer by calculating the increase in
potential energy.)
Take the free fall acceleration to be .
ANSWER:
Part C
Find the increase in the energy of the system when he raises himself . (Answer by computing the product of the force on the
rope and the length of the rope passing through his hands.)
Take the free fall acceleration to be .
ANSWER:
Problem 7.42
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A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m.
When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope
.
Part A
What is the speed of the block as it slides along the horizontal surface after having left the spring?
ANSWER:
m/s
Part B
How far does the block travel up the incline before starting to slide back down?
ANSWER:
m
Problem 7.46
Riding a Loop-the-loop. A car in an amusement park ride rolls without friction around the track shown in the figure . It starts
from rest at point at a height above the bottom of the loop. Treat the car as a particle.
Part A
What is the minimum value of (in terms of ) such that the car moves around the loop without falling off at the top (point
)?
ANSWER:
=
Part B
If the car starts at height and the radius is , compute the speed of the passengers when the car is at point , which is at the
end of a horizontal diameter.
Take the free fall acceleration to be .
ANSWER:
=
Part C
If the car starts at height and the radius is , compute the radial acceleration of the passengers when the car is at point ,
which is at the end of a horizontal diameter.
Take the free fall acceleration to be .
ANSWER:
=
Part D
If the car starts at height and the radius is , compute the tangential acceleration of the passengers when the car is at point ,
which is at the end of a horizontal diameter.
Take the free fall acceleration to be .
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ANSWER:
Problem 7.62
A skier of mass
starts from rest at the top of a ski slope of height .
Part A
If frictional forces do
of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be .
ANSWER:
Part B
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is . If the patch is of width
and the average force of air resistance on the skier is , how fast is she going after crossing the patch?
ANSWER:
Part C
After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance
What is the average force exerted on her by the snowdrift as it stops her?
into it before coming to a stop.
ANSWER:
Problem 7.66
A truck with mass has a brake failure while going down an icy mountain road of constant downward slope angle
the truck is moving downhill at speed . After careening downhill a distance with
negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp
of constant upward slope angle . The truck ramp has a soft sand surface for which the
coefficient of rolling friction is .
. Initially
Part A
What is the distance that the truck moves up the ramp before coming to a halt? Use energy methods.
ANSWER:
Problem 7.68
A variable force is maintained tangent to a frictionless, semicircular surface . By slowly varying the force, a block with
weight is moved through the angla , and the spring to which it is attached is stretched
from position 1 to position 2. The spring has negligible mass and force constant . The
end of the spring moves in an arc of radius .
Part A
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Calculate the work done by the force .
ANSWER:
=
Problem 7.74
A 2.00-kg package is released on a
incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the
bottom of the incline . The coefficients of friction between the package and the incline are
and
. The mass of
the spring is negligible.
Part A
What is the speed of the package just before it reaches the spring?
ANSWER:
m/s
Part B
What is the maximum compression of the spring?
ANSWER:
m
Part C
The package rebounds back up the incline. How close does it get to its initial position?
ANSWER:
m
Problem 7.80
A hydroelectric dam holds back a lake of surface area that has vertical sides below the water level. The water level in the lake
is a height above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is
converted into electrical energy with efficiency.
Part A
If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the
water in the lake? The density of water is
.
Take free fall acceleration to be .
ANSWER:
Part B
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What volume of water must pass through the dam to produce an amount of electrical energy totalling
of electrical energy?
ANSWER:
Part C
What distance does the level of water in the lake fall when this much water passes through the dam?
ANSWER:
The Impulse-Momentum Theorem
Description: Several qualititative and computational questions involving derivation and applications of the
impulse-momentum theorem.
Learning Goal: To learn about the impulse-momentum theorem and its applications in some common cases.
Using the concept of momentum, Newton's second law can be rewritten as
, (1)
where
is the net force
acting on the object, and
is the rate at which the object's momentum is changing.
If the object is observed during an interval of time between times
and , then integration of both sides of equation (1) gives
. (2)
The right side of equation (2) is simply the change in the object's momentum
force and is denoted by . Then equation (2) can be rewritten as
. The left side is called the impulse of the net
.
This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the
acting along the direction of motion, the
impulse of the net force acting on the object. In the case of a constant net force
impulse-momentum theorem can be written as
. (3)
Here , , and are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in
question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.
The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying
force acting along the direction of motion. First, let us consider a particle of mass moving along the x axis. The net force
is acting on the particle along the x axis. is a constant force.
Part A
The particle starts from rest at
. What is the magnitude of the momentum of the particle at time ? Assume that
.
Express your answer in terms of any or all of , , and .
ANSWER:
=
Part B
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The particle starts from rest at
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. What is the magnitude of the velocity of the particle at time ? Assume that
.
Express your answer in terms of any or all of , , and .
ANSWER:
=
Part C
The particle has momentum of magnitude
at a certain instant. What is , the magnitude of its momentum
Express your answer in terms of any or all of , , , and
ANSWER:
seconds later?
.
=
Part D
The particle has momentum of magnitude
at a certain instant. What is , the magnitude of its velocity
Express your answer in terms of any or all of , , , and
ANSWER:
seconds later?
.
=
Let us now consider several two-dimensional situations.
A particle of mass is moving in the positive x direction at speed . After a certain constant force is applied to the particle, it
moves in the positive y direction at speed .
Part E
Find the magnitude of the impulse
Hint E.1
delivered to the particle.
How to approach the problem
This is a two-dimensional situation. It is helpful to find the components
theorem to find .
Part E.2
Find
and
separately and then use the Pythagorean
Find the change in momentum
, the magnitude of the change in the x component of the momentum of the particle.
Express your answer in terms of
ANSWER:
=
Similarly, you can find
.
. Once you have obtained both
Express your answer in terms of
ANSWER:
and .
and
, you can find the magnitude of the delivered impulse
and . Use three significant figures in the numerical coefficient.
=
Part F
Which of the vectors below best represents the direction of the impulse vector ?
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2
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j5
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j6
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j8
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Part G
What is the angle between the positive y axis and the vector
ANSWER:
as shown in the figure?
26.6 degrees
30 degrees
j 60 degrees
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j 63.4 degrees
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Part H
If the magnitude of the net force acting on the particle is , how long does it take the particle to acquire its final velocity,
the positive y direction?
in
Express your answer in terms of , , and . If you use a numerical coefficient, use three significant figures.
ANSWER:
=
So far, we have considered only the situation in which the magnitude of the net force acting on the particle was either
irrelevant to the solution or was considered constant. Let us now consider an example of a varying force acting on a particle.
Part I
kilograms is at rest at
seconds. A varying force
A particle of mass
seconds and
seconds. Find the speed of the particle at
between
Hint I.1
is acting on the particle
seconds.
Use the impulse-momentum theorem
In this case,
and
. Therefore,
.
Part I.2
What is the correct antiderivative?
Which of the following is an antiderivative
ANSWER:
?
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j
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m
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Express your answer in meters per second to three significant figures.
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ANSWER:
Problem 8.8
Force of a Baseball Swing. A baseball has mass 0.145 kg.
Part A
If the velocity of a pitched ball has a magnitude of 45.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction,
find the magnitude of the change in momentum of the ball.
ANSWER:
Part B
Find the magnitude of the impulse applied to it by the bat.
ANSWER:
Part C
If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by the bat.
ANSWER:
N
Problem 8.10
An engine of the orbital maneuvering system (OMS) on a space shuttle exerts a force of
exhausting a negligible mass of fuel relative to the shuttle's mass of .
over a time interval of
,
Part A
What is the y component of the impulse of the force for this time interval of
?
ANSWER:
Part B
What is the y component of the shuttle's change in momentum from this impulse?
ANSWER:
Part C
What is the y component of the shuttle's change in velocity from this impulse?
ANSWER:
Problem 8.17
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Energy Change During a Hip Check. Ice-hockey star Wayne Gretzky is skating at 13.0 m/s toward a defender, who in turn is
skating at 5.0 m/s toward Gretzky . Gretzky's weight is 756 N; that of the defender is 900 N. Immediately after the collision,
Gretzky is moving at 1.5 m/s in his original direction. You can ignore external horizontal
forces applied by the ice to the skaters during the collision.
Part A
What is the velocity of the defender immediately after the collision?
ANSWER:
m/s
Part B
Calculate the change in the combined kinetic energy of the two players.
Take the free fall acceleration to be
ANSWER:
Summary
.
kJ
0 of 15 problems complete (0% avg. score)
0 of 150 points
2/28/2006 2:46 AM
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