Download HW #8: Chapter 9--Momentum

3/9/2016
HW #8: Chapter 9--Momentum
HW #8: Chapter 9­­Momentum
Due: 11:59pm on Tuesday, March 15, 2016
To understand how points are awarded, read the Grading Policy for this assignment.
Impulse on a Baseball
Learning Goal:
To understand the relationship between force, impulse, and momentum.
The effect of a net force acting on an object is related both to the force and to the total time the force acts on the object. The physical quantity impulse
is a measure of both these effects. For a constant net force, the impulse is given by
.
The impulse is a vector pointing in the same direction as the force vector. The units of are or .
Recall that when a net force acts on an object, the object will accelerate, causing a change in its velocity. Hence the object's momentum (
change. The impulse­momentum theorem describes the effect that an impulse has on an object's motion:
) will also
.
So the change in momentum of an object equals the net impulse, that is, the net force multiplied by the time over which the force acts. A given change in
momentum can result from a large force over a short time or a smaller force over a longer time.
In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 to the
ball for a time of .
Part A
Assuming that this force is constant, what is the magnitude of the impulse on the ball?
Enter your answer numerically in newton seconds using two significant figures.
ANSWER:
= 8.4 Correct
We often visualize the impulse by drawing a graph showing the force versus time. For a constant net force such as that used in the previous part, the graph
showing the magnitude of the force versus time will look like the one shown in .
Part B
The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have physical meaning.
ANSWER:
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HW #8: Chapter 9--Momentum
length
height
For this graph, the area
of the rectangle corresponds to the impulse.
slope
Correct
The assumption of a constant net force is idealized to make the problem easier to solve. A real force, especially in a case like the one presented in
Parts A and B, where a large force is applied for a short time, is not likely to be constant.
A more realistic graph showing the magnitude of the force that the swinging bat applies to the baseball will show the force building up to a maximum value
as the bat comes into full contact with the ball. Then as the ball loses contact with the bat, the graph will show the magnitude of the force decaying to zero. It
will look like the graph in .
Part C
If both the graph representing the constant net force and the graph representing the variable net force represent the same impulse acting on the
baseball, which geometric properties must the two graphs have in common?
ANSWER:
maximum force
area
slope
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HW #8: Chapter 9--Momentum
Correct
When the net force varies over time, as in the case of the real net force acting on the baseball, you can simplify the problem by finding the average
net force acting on the baseball during time . This average net force is treated as a constant force that acts on the ball for time impulse on the ball can then be found as . The
.
Graphically, this method states that the impulse of the baseball can be represented by either the area under the net force versus time curve or the
area under the average net force versus time curve. These areas are represented in the figure as the areas shaded in red and blue respectively.
The impulse of an object is also related to its change in momentum. Once the impulse is known, it can be used to find the change in momentum, or if either
the initial or final momentum is known, the other momentum can be found. Keep in mind that . Because both impulse and
momentum are vectors, it is essential to account for the direction of each vector, even in a one­dimensional problem.
Part D
Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to
the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction.
ANSWER:
The impulse on the ball caused by the bat will be in the positive
negative
x direction.
Correct
Part E
Now assume that the pitcher in Part D throws a 0.145­ baseball parallel to the ground with a speed of 32 in the +x direction. The batter then hits
the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an
impulse of to the baseball?
Enter your answer numerically in meters per second using two significant figures.
ANSWER:
= ­26 Correct
The negative sign in the answer indicates that after the bat hits the ball, the ball travels in the opposite direction to that defined to be positive.
Colliding Balls
Balls A and B roll across a table, then collide and bounce off each other. The paths of the two balls are pictured (viewed from above) in the diagram.
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HW #8: Chapter 9--Momentum
Part A
Which set of arrows best represents the change in momentum for balls A and B?
Hint 1. Consider the components of the momentum vector
Recall that momentum is a vector quantity. It is helpful to consider the horizontal and vertical components of the momenta separately. How do
the vertical components of the momenta change during the collision? How do the horizontal components change? (Do the horizontal
components of the momenta change?)
ANSWER:
A
B
C
D
E
Correct
Part B
Which of the following arrows indicates the direction of the impulse applied to ball A by ball B?
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HW #8: Chapter 9--Momentum
Hint 1. Definition of impulse
Impulse is defined as the change in momentum. How does the momentum of ball A change during the collision? (If you get stuck, try looking
back at part A. There, you found the change in momenta for each of the balls.)
ANSWER:
A
B
C
D
E
Correct
Video Tutor: Off­Center Collision
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window
and answer the question on the right. You can watch the video again at any point.
Consider the video you just watched. The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of
linear momentum. However, since the two pucks mutual center of mass does not coincide with either of the pucks velocity vectors, they have angular
momentum. This becomes evident after the collision when due to conservation of angular momentum the two pucks spin around their mutual center of
mass.
Part A
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the
same. Compared to the pucks in the video, this pair of pucks will rotate
Hint 1. How to approach the problem
This question asks you to relate the mass of a rotating object to the object’s angular speed (rate of rotation).
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HW #8: Chapter 9--Momentum
You can do that using angular momentum: for linear momentum.
. In this equation, the moment of inertia is the “mass” term, equivalent to in the equation
So, how does the pucks moment of inertia relate to their mass ? Is the relationship linear ( )? Or is it quadratic ( square ( ) or inverse
)? To see the answer, try writing the general equation that defines .
Now, how will doubling the mass of the pucks affect , , and ?
ANSWER:
twice as fast.
at the same rate.
four times as fast.
one­half as fast.
one­fourth as fast.
Correct
By doubling the mass but keeping the velocities unchanged, we doubled the angular momentum of the two­puck system. However, we also
doubled the moment of inertia. Since , the rotation rate of the two­puck system must remain unchanged.
Impulse and Momentum Ranking Task
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all
automobiles are brought to rest.
Part A
Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below.
ANSWER:
Correct
Part B
Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.
Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below.
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HW #8: Chapter 9--Momentum
Hint 1. Relating impulse to momentum
The impulse applied to an object is equal to the object’s change in momentum. Therefore, the impulse needed to stop them should be equal to
the difference between the initial momentum and the final momentum. (The final momentum is zero since the car is brought to a stop.)
Hint 2. Apply the impulse­momentum theorem
All of the cars are brought to rest. What is the final momentum of each automobile?
Enter your answer in kilogram meters per second to three significant figures.
Hint 1. How to find momentum
Recall that an object's momentum is equal to the product of its mass and velocity.
ANSWER:
= 0 ANSWER:
Correct
Part C
Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.
Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below.
Hint 1. How to approach the problem
You know the impulse needed to stop the cars. However, this impulse could be a very large force exerted over a fraction of a second, a very
small force exerted over several minutes, or any situation in between, so long as the force multiplied by the time gives the proper impulse. You
must know the time intervals over which the stopping forces are exerted to determine the magnitudes of the stopping forces from the impulses.
ANSWER:
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HW #8: Chapter 9--Momentum
Correct
The more momentum an object has, the more impulse is needed to stop it. However, this impulse can be provided via a large force acting over a
short time interval or a relatively small force acting over a relatively long time interval. If you are driving down the highway at 55 , you can stop
your car by either lightly pressing on the brakes and traveling a long time before stopping, or pressing more firmly on the brakes and stopping more
quickly. In both cases, your braking system has applied the same amount of impulse to your car.
± Catching a Ball on Ice
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A
friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.8 . Olaf's mass is 66.4 .
Part A
If Olaf catches the ball, with what speed do Olaf and the ball move afterward?
Express your answer numerically in meters per second.
Hint 1. How to approach the problem
Using conservation of momentum and the fact that Olaf's initial momentum is zero, set the initial momentum of the ball equal to the final
momentum of Olaf and the ball, then solve for the final velocity.
Hint 2. Find the ball's initial momentum
What is , the initial momentum of the ball?
Express your answer numerically in kilogram meters per second.
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ANSWER:
= 4.72 ANSWER:
= 7.07×10−2 Correct
Part B
If the ball hits Olaf and bounces off his chest horizontally at 7.80 in the opposite direction, what is his speed after the collision?
Express your answer numerically in meters per second.
Hint 1. How to approach the problem
The initial momentum of the ball is the same as in Part A. Apply conservation of momentum, keeping in mind that both Olaf and the ball have a
nonzero final momentum.
Hint 2. Find the ball's final momentum
Taking the direction in which the ball was initially traveling to be positive, what is , the ball's final momentum?
Express your answer numerically in kilogram meters per second.
ANSWER:
= ­3.12 ANSWER:
= 0.118 Correct
Momentum in an Explosion
A giant "egg" explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with piece B
moving in the positive x direction. The masses of both pieces are indicated in , shown traveling in
opposite directions.
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HW #8: Chapter 9--Momentum
Part A
What is the magnitude of the momentum |
| of piece A before the explosion?
Express your answer numerically in kilogram meters per second.
Hint 1. Initial momentum
The momentum of any object is determined by the product of the object's mass and velocity. The egg is initially at rest. Use this to find the initial
momentum.
ANSWER:
|
| = 0 Correct
Similarly, piece B has zero momentum before the collision. The total momentum of the "egg," the sum of the two individual momenta, is also zero.
Part B
During the explosion, is the magnitude of the force of piece A on piece B greater than, less than, or equal to the magnitude of the force of piece B on
piece A?
Hint 1. Forces in an explosion
The forces specified in this problem must obey Newton's third law, which states that every action has an equal­magnitude and oppositely­
directed reaction.
ANSWER:
greater than
less than
equal to
cannot be determined
Correct
Part C
The component of the momentum of piece B, of piece A after the explosion.
, is measured to be +500 after the explosion. Find the component of the momentum Enter your answer numerically in kilogram meters per second.
Hint 1. Conservation of momentum
The law of conservation of momentum states that the total momentum in an isolated system of objects must remain constant, regardless of the
interactions (or collisions) between the objects. Thus, the total momentum of the two pieces of the egg after the explosion must be equal to the
total momentum of the two pieces of the egg before the explosion.
ANSWER:
= ­500 Correct
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HW #8: Chapter 9--Momentum
Colliding Cars
In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel
off as a single unit. The collision is therefore completely inelastic.
The two cars shown in the figure, of masses and , collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of car 2 was traveling northward at a speed of . After the collision, the two cars stick together and
travel off in the direction shown.
, and
Part A
First, find the magnitude of , that is, the speed of the two­car unit after the collision.
Express in terms of , , and the cars' initial speeds and .
Hint 1. Conservation of momentum
Recall that conservation of linear momentum may be expressed as a vector equation,
.
Each vector component of linear momentum is conserved separately.
Hint 2. Determine the x and y components of momentum
The momentum of the two­car system immediately after the collision may be written as are the eastward and northward directions, respectively.
Find and , where the positive x and y directions
.
Express the two components, separated by a comma, in terms of , , , and .
ANSWER:
, = ,
Hint 3. A vector and its components
Recall that the square of the magnitude of a vector is given by the Pythagorean formula
.
Hint 4. Velocity and momentum
Find , the magnitude of the final velocity.
Express in terms of the magnitude of the final momentum and the masses and .
ANSWER:
= ANSWER:
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HW #8: Chapter 9--Momentum
Correct
Part B
Find the tangent of the angle .
Express your answer in terms of the magnitudes of the initial momenta of the two cars, introduction.
and , or the quantities given in the problem
Hint 1. Determine the x and y components of momentum
The momentum of the two­car system immediately after the collision may be written as are the eastward and northward directions, respectively.
Find , where the positive x and y directions
and Express the two components, separated by a comma, in terms of the magnitudes of the initial momenta of the two cars, the quantities given in the problem introduction.
, and , or
ANSWER:
, = ,
ANSWER:
= Correct
Part C
Suppose that after the collision, ; in other words, is degrees. Which quantities then must have been equal before the collision?
ANSWER:
The magnitudes of the momenta of the cars
The masses of the cars The speeds of the cars
Correct
Three­Block Inelastic Collision
A block of mass = 1.20 moving at = 1.50 undergoes a completely inelastic collision with a stationary block of mass = 0.600 . The
blocks then move, stuck together, at speed . After a short time, the two­block system collides inelastically with a third block, of mass = 3.00 ,
which is initially at rest. The three blocks then move, stuck together, with speed . Assume that the blocks slide without friction.
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HW #8: Chapter 9--Momentum
Part A
Find , the ratio of the velocity of the two­block system after the first collision to the velocity of the block of mass before the collision.
Express your answer numerically using three significant figures.
Hint 1. What physical principle to use
Apply the principle of conservation of linear momentum, noting that the mass of the two­block system is .
ANSWER:
= 0.667
Correct
Intuition and experience with the momentum equations lead to the following conclusions:
1. The blocks will slow down after collision (
, or ).
2. The greater the mass of block 1 for a fixed mass of block 2, the less the blocks will slow down after the collision (
increases as the
mass of block 1 increases with respect to the mass of block 2, but the ratio will still, of course, be less than 1).
The simplest equation that satisfies these criteria is . Try to use similar reasoning for the rest of this problem.
Part B
Find , the ratio of the velocity of the three­block system after the second collision to the velocity of the block of mass before the collisions.
Express your answer numerically using three significant figures.
Hint 1. How to approach the problem
This problem will be considerably easier if you take the initial state to be when the block alone is moving, rather than when both and are moving. This is a valid choice, since the forces on the blocks during the collision between and are internal forces (considering all
three blocks as a "system").
Hint 2. Total mass of the blocks
After the second collision, the mass of the system is .
ANSWER:
= 0.250
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HW #8: Chapter 9--Momentum
Correct
As long as friction is absent, the collision between the objects of masses and is irrelevant to the final velocity in this problem. This
yields another approach for calculating . For the purpose of calculating the final velocity, you could assume that moving with speed collides with a stationary mass equivalent to . This collision is illustrated in the figure.
This gives
.
You can check that this gives the same answer as you calculated in Part B.
Momentum in a Collision Graphing Question
Two asteroids, drifting at constant velocity, collide. The masses and velocities of the asteroids before the collision are indicated in the figure.
Part A
Sketch graphs of the momenta of asteroids A and B before the collision.
Hint 1. Initial momentum
The momentum of any object is determined by the product of the object's mass and velocity.
Hint 2. Shape of the momentum vs. time graph
The asteroids are drifting at constant velocity before the collision, and have constant mass.
ANSWER:
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HW #8: Chapter 9--Momentum
Correct
The total momentum of the system of two asteroids before the collision is equal to the sum of the individual momenta. After the collision, the two
asteroids join together to form a single megaasteroid. When two objects stick together after a collision, the collision is called perfectly inelastic.
Part B
During the collision, is the magnitude of the force of asteroid A on asteroid B greater than, less than, or equal to the magnitude of the force of asteroid B
on asteroid A?
Hint 1. Collision forces
The forces specified in this question must obey Newton's third law.
ANSWER:
greater
less
equal
cannot be determined
Correct
Part C
Sketch a graph of the total momentum in the system of the two asteroids after the collision.
Hint 1. How to approach the problem
The law of conservation of momentum states that the total momentum in an isolated system of objects must remain constant, regardless of the
interactions (or collisions) between the objects. Thus, the total momentum of the two asteroids after the collision must be equal to the total
momentum of the two asteroids before the collision.
ANSWER:
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HW #8: Chapter 9--Momentum
Correct
Part D
When the two asteroids collide, they stick together. Based on your graph in Part C, determine the velocity of the new megaasteroid.
Hint 1. Definition of momentum
Momentum is equal to (total) mass times velocity.
ANSWER:
16.7 Correct
Conservation of Momentum in Two Dimensions Ranking Task
Part A
The figures below show bird's­eye views of six automobile crashes an instant before they occur. The automobiles have different masses and incoming
velocities as shown. After impact, the automobiles remain joined together and skid to rest in the direction shown by . Rank these crashes
according to the angle , measured counterclockwise as shown, at which the wreckage initially skids.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Conservation of momentum in two dimensions
Since momentum is a vector quantity, the x component of momentum and the y component of momentum must be individually conserved in any
collision. Thus, the total x momentum before the collision must be equal to the total x momentum of the sliding wreckage after the collision. The
same is true for the total y momentum.
Hint 2. Determining the angle
Once the x and y momenta of the wreckage are determined, the exact angle through which the wreckage skids can be determined by
trigonometry. Determining the exact angle of this final momentum vector is accomplished the same way you would find the angle of any vector,
typically by finding the inverse tangent of the y component over the x component. (You can also determine the ranking without calculating the
exact angle at which the wreckage skids.)
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HW #8: Chapter 9--Momentum
ANSWER:
Correct
Change in Angular Velocity Ranking Task
A merry­go­round of radius , shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A
sandbag of mass is dropped onto the merry­go­round, at a position designated by . The sandbag does not slip or roll upon contact with the merry­go­
round.
Part A
Rank the following different combinations of go­round.
and on the basis of the angular speed of the merry­go­round after the sandbag "sticks" to the merry­
Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. How to approach the problem
Analyze the scenario using the conservation of angular momentum. The angular momentum of an object of moment of inertia and angular
speed is given by
.
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HW #8: Chapter 9--Momentum
Hint 2. Determining the change in moment of inertia
After the sandbag sticks to the merry­go­round, the moment of inertia of the merry­go­round system increases by an amount equal to the product
of the mass of the sandbag ( ) and the square of its distance from the rotation axis ( ).
ANSWER:
Correct
Spinning Situations
Suppose you are standing on the center of a merry­go­round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis
is pointing upward. The wheel is rotating counterclockwise when observed from above.
For this problem, neglect any air resistance or friction between the merry­go­round and its foundation.
Part A
Suppose you now grab the edge of the wheel with your hand, stopping it from spinning.
What happens to the merry­go­round?
Hint 1. Change in angular momentum
Consider yourself, the merry­go­round, and the bicycle wheel to be a single system. When you stop the wheel from spinning, what happens to
the angular momentum of the system about the vertical axis?
ANSWER:
It increases.
It decreases.
It remains unchanged.
ANSWER:
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HW #8: Chapter 9--Momentum
It remains at rest.
It begins to rotate counterclockwise (as observed from above).
It begins to rotate clockwise (as observed from above).
Correct
It may seem incredible, but as long as there are no external torques acting on the system (which includes yourself, the merry­go­round, and the
bicycle wheel) the angular momentum originally stored in the bicycle wheel is conserved.
What really happens?
To stop the wheel from spinning (counterclockwise), you must exert a clockwise torque on it. By extending Newton's third law, this means that the
wheel exerts a counterclockwise torque on you. If there is no friction between you and the merry­go­round, i.e., if the floor of the latter is completely
smooth, this torque from the wheel will make you spin counterclockwise. However, if there is friction between your shoe soles and the floor, to
prevent relative motion, this time, the floor exerts a clockwise torque on you, and you exert a counterclockwise torque on the floor of the merry­go­
round. If the axle is completely smooth, then this torque will now make the merry­go­round spin counterclockwise. Of course, the ride will spin
much slower than the wheel, because its moment of inertia is much larger.
Video Tutor: Spinning Person Drops Weights
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window
and answer the question at right. You can watch the video again at any point.
Part A
The experimenter from the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him and drop a long,
heavy bean bag straight down so that it lands across his lap, in his hands. What happens?
Hint 1. How to approach the problem
To make the experimenter spin faster or slower, the bean bag must either add to or subtract from his angular momentum.
Notice that the bag is not rotating before it contacts the experimenter, but is rotating once in his lap. Thus, the bag gains rotational momentum.
Rotational momentum will be conserved for the system of the bean bag and the experimenter. What does that imply?
Incidentally, since the bag drops close to the experimenter’s axis of rotation, any effect it has on his rate of rotation will be smaller than if it
dropped farther out.
ANSWER:
It's not possible to predict what will happen.
He continues spinning at the same speed.
He spins slower.
He spins faster.
Correct
The bean bag is not rotating before it contacts the experimenter, so (by conservation of rotational momentum) he must supply the rotational
momentum required to make it spin at the same rate as his lap. Therefore, he ends up spinning slower.
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HW #8: Chapter 9--Momentum
Twirling a Baton
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120 and length 80.0 .
Part A
Initially, the baton is spinning about a line through its center at angular velocity 3.00 What is its angular momentum?
.
Express your answer in kilogram meters squared per second.
Hint 1. Angular momentum for a rigid body rotating about an axis of symmetry
The angular momentum of a rigid body that rotates about an axis of symmetry at angular velocity is
,
where is the moment of inertia of the object about the rotational axis.
Hint 2. Moment of inertia
For a uniform rod of length with mass , the moment of intertia about an axis passing through its center perpendicular to the rod is
.
ANSWER:
1.92×10−2 Correct
Part B
With a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axis passing through its end at the same angular
velocity 3.00 as before. What is the new angular momentum of the rod?
Express your answer in kilogram meters squared per second.
Hint 1. How to approach the problem
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HW #8: Chapter 9--Momentum
If you know the moment of inertia of a rod about one of its ends, you can solve this part in the same way that you solved Part A.
Hint 2. Moment of inertia
For a rod of length with mass , the moment of intertia about an axis passing through either end is
.
ANSWER:
7.68×10−2 Correct
Here is another way to solve this problem. There is a theorem that relates the angular momentum of an object about an arbitrary axis to the
angular momentum of the object about the axis passing through its center of mass :
,
where is the mass of the object, is the length of the position vector of the center of mass with respect to the point chosen, and
is the
velocity of the center of mass with respect to the point chosen. Substituting for the values on the right­hand side would yield the same angular
momentum that you calculated.
Conceptual Question 9.14
To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object.
Part A
Should you choose to throw a rubber ball or a beanbag of equal size and weight?
ANSWER:
rubber ball
beanbag
Correct
Conceptual Question 9.17
If the earth warms significantly, the polar ice caps will melt. Water will move from the poles, near the earth's rotation axis, and will spread out around the
globe. In principle, this will change the length of the day.
Part A
Why? Will the length of the day increase or decrease?
ANSWER:
3740 Character(s) remaining
it would decrease because it would change the earth rotation
Submitted, grade pending
Multiple Choice Question 9.19
Curling is a sport played with 20 stones that slide across an ice surface. Suppose a curling stone sliding at 1.4 https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4141228
strikes another stone and comes to
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HW #8: Chapter 9--Momentum
Curling is a sport played with 20 rest in 2.6 .
stones that slide across an ice surface. Suppose a curling stone sliding at 1.4 strikes another stone and comes to
Part A
Approximately how much force is there on the stone during the impact?
ANSWER:
5.4×104 1.1×105 5400 1.1×104 Correct
Multiple Choice Question 9.22
Two friends are sitting in a stationary canoe. At the person at the front tosses a sack to the person in the rear, who catches the sack 0.2 later.
Part A
Which plot in the figure shows the velocity of the boat as a function of time? Positive velocity is forward, negative velocity is backward. Neglect any drag
force on the canoe from the water.
ANSWER:
A
B
C
D
Correct
Problem 9.8
A 60 tennis ball with an initial speed of 32 collision.
hits a wall and rebounds with the same speed. The figure shows the force of the wall on the ball during the
https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4141228
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HW #8: Chapter 9--Momentum
Part A
What is the value of , the maximum value of the contact force during the collision?
Express your answer using two significant figures.
ANSWER:
960 Correct
Problem 9.15
A 3.0 240 block of wood sits on a table. A 3.0 bullet, fired horizontally at a speed of 420 .
, goes completely through the block, emerging at a speed of
Part A
What is the speed of the block immediately after the bullet exits?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
= 0.18 Correct
Problem 9.24
The parking brake on a 2000 Cadillac has failed, and it is rolling slowly, at 4 , toward a group of small children. Seeing the situation, you realize you
have just enough time to drive your 2000 Volkswagen head­on into the Cadillac and save the children.
Part A
With what speed should you impact the Cadillac to bring it to a halt?
ANSWER:
4 Correct
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HW #8: Chapter 9--Momentum
Problem 9.44
Squids rely on jet propulsion, a versatile technique to move around in water. A 1.5 water backward to quickly get itself moving forward at 3.0 .
(the mass without water) squid at rest suddenly expels 0.10 of
Part A
If other forces (such as the drag force on the squid) are ignored, what is the speed with which the squid expels the water?
Express your answer using two significant figures.
ANSWER:
= 45 Correct
Problem 9.54
A 110 linebacker running at 2.0 onto the quarterback.
and an 82 quarterback running at 3.0 have a head­on collision in midair. The linebacker grabs and holds
Part A
Who ends up moving in his original direction of motion after they hit?
ANSWER:
Linebacker
Quarterback
ends up moving in his original direction of motion.
Correct
Problem 9.63
A 450 cannon fires a 11 cannonball with a speed of 200
relative to the muzzle. The cannon is on wheels that roll without friction.
Part A
When the cannon fires, what is the speed of the cannonball relative to the earth?
ANSWER:
195 Correct
Problem 9.76­79
Consider a golf club hitting a golf ball. To a good approximation, we can model this as a collision between the rapidly moving head of the golf club and the
stationary golf ball, ignoring the shaft of the club and the golfer. A golf ball has a mass of 46 . Suppose a 200 club head is moving at a speed of 40 just before striking the golf ball. After the collision, the golf ball’s speed is 60 .
Part A
What is the momentum of the club­ball system right before the collision?
ANSWER:
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HW #8: Chapter 9--Momentum
1.8 8.0 3220 8000 Correct
Part B
Immediately after the collision, the momentum of the club­ball system will be
ANSWER:
Less than before the collision.
The same as before the collision.
More than before the collision.
Correct
Part C
A manufacturer makes a golf ball that compresses more than a traditional golf ball when struck by a club. How will this affect the average force during
the collision?
ANSWER:
The force will decrease.
The force will not be affected.
The force will increase.
Correct
Part D
By approximately how much does the club head slow down as a result of hitting the ball?
ANSWER:
4 6 14 26 Correct
Score Summary:
Your score on this assignment is 91.5%.
You received 22.89 out of a possible total of 25 points.
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