Chapter 08 Homework

4/24/2014 Chapter 8 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2818796 1/40

Chapter 8 HomeworkDue: 10:00pm on Monday, March 31, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Momentum and Kinetic Energy

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moveswith speed and has mass . Object 2 moves with speed and has mass .

Part A

Which object has the larger magnitude of its momentum?

Hint 1. Momentum

Recall that an object's momentum is given by.

Hint 2. Compare the momenta

Perhaps the simplest way to compare the momenta of Object 1 and Object 2 is to carry out the steps thatfollow:

1. Write down the momentum equation for Object 1: .

2. For each known variable in the equation, substitute the value provided in the problemintroduction.

3. Simplify the equation.4. Repeat these steps for Object 2.5. Compare your equation for to that for .

Complete the first three steps to find an expression for .

Write in terms of and .

ANSWER:

ANSWER:

Correct

= vv1 = 2mm1 = vv2 2√ = mm2

= mp v

= ⋯p1

p1 p2

p1

p1 m v

= p1 2mv

Object 1 has the greater magnitude of its momentum.

Object 2 has the greater magnitude of its momentum.

Both objects have the same magnitude of their momenta.



Part B

Which object has the larger kinetic energy?

Hint 1. Kinetic energy and momentum

Kinetic energy is given by

.

Hint 2. Compare the kinetic energies

Perhaps the simplest way to compare the kinetic energies of Object 1 and Object 2 is to carry out the stepsthat follow:

1. Write down the kinetic energy equation for Object 1: .

2. For each known variable in the equation, substitute the value provided in the problemintroduction.

3. Simplify the equation.4. Repeat these steps for Object 2.5. Compare your equation for to that for .

Complete the first three steps to find an expression for .

Write in terms of and .

ANSWER:

ANSWER:

Correct

Many students confuse the quantities momentum and kinetic energy and think that they are the same thing.However, as this problem has demonstrated, they are two very different physical quantities and even if twoobjects have different momenta, they can still have the same amount of kinetic energy.

Exercise 8.9

A 0.160- hockey puck is moving on an icy, frictionless, horizontal surface. At the puck is moving to the right at3.10 .

K = m =12

v2 p2

2m

= ⋯K1

K1 K2

K1

K1 m v

= K1 mv2

Object 1 has the greater kinetic energy.

Object 2 has the greater kinetic energy.

The objects have the same kinetic energy.

kg t = 0m/s



Part A

Calculate the magnitude of the velocity of the puck after a force of 25.4 directed to the right has been applied for

4.5×10−2 .

Express your answer using two significant figures.

ANSWER:

Correct

Part B

What is the direction of the velocity of the puck after a force of 25.4 directed to the right has been applied for

4.5×10−2 .

ANSWER:

Correct

Part C

If instead, a force of 11.4 directed to the left is applied from to 4.5×10−2 , what is the magnitude of thefinal velocity of the puck?


ANSWER:

Correct

Part D

What is the direction of the final velocity of the puck in this case?

ANSWER:

Ns

= 10 v m/s

Ns

to the right

to the left

N t = 0 t = s

= 0.11 v m/s



Correct

Exercise 8.13

A 2.00 stone is sliding to the right on a frictionless horizontal surface at 4.00 when it is suddenly struck by anobject that exerts a large horizontal force on it for a short period of time. The graph in the figure shows the magnitude ofthis force as a function of time.

Part A

What impulse does this force exert on the stone?

ANSWER:

Correct

Part B

Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the right.

ANSWER:

to the right

to the left

kg m/s

= 2.50 p kg ⋅ m/s

= 5.25 v m/s



Correct

Part C

Just after the force stops acting, find the direction of the stone's velocity if the force acts to the right.

ANSWER:

Correct

Part D

Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the left.

ANSWER:

Correct

Part E

Just after the force stops acting, find the direction of the stone's velocity if the force acts to the left.

ANSWER:

Correct

Momentum and Internal Forces

Learning Goal:

To understand the concept of total momentum for a system of objects and the effect of the internal forces on the totalmomentum.

We begin by introducing the following terms:

System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a

to the right

to the left

= 2.75 v m/s

to the right

to the left



certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial stepin solving the problem.

Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that bothinteracting objects must belong to the system.

External force: Any force interaction between objects at least one of which does not belong to the chosen system; inother words, at least one of the objects is external to the system.

Closed system: a system that is not subject to any external forces.

Total momentum: The vector sum of the individual momenta of all objects constituting the system.

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . Tosimplify the analysis, we will make several assumptions:

1. The blocks can move in only one dimension, namely, along the x axis.2. The masses of the blocks remain constant.3. The system is closed.

At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly,the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you willshow that the total momentum of the system is not changed by the presence of internal forces.

Part A

Find , the x component of the total momentum of the system at time .

Express your answer in terms of , , , and .

ANSWER:

Correct

Part B

Find the time derivative of the x component of the system's total momentum.

Express your answer in terms of , , , and .

Hint 1. Finding the derivative of momentum for one block

Consider the momentum of block 1: . Take the derivative of this expression with respect

to time, noting that velocity is a function of time, and mass is a constant:

.

Hint 2. The relationship between velocity and acceleration

m1 m2

t (t)v1 (t)a1(t)v2 (t)a2

p(t) t

m1 m2 (t)v1 (t)v2

= p(t) (t) + (t)m1v1 m2v2

dp(t)/dt

(t)a1 (t)a2 m1 m2

(t) = (t)p1 m1v1

= =d (t)p1

dt

d( (t))m1v1

dtm1

d (t)v1

dt

a(t) = dv(t)



Recall the definition of acceleration as .

ANSWER:

Correct

Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be usefulin reaching our desired conclusion, if only for this very special case.

Part C

The quantity (mass times acceleration) is dimensionally equivalent to which of the following?

ANSWER:

Correct

Part D

Acceleration is due to which of the following physical quantities?

ANSWER:

Correct

Part E

a(t) = dv(t)dt

= dp(t)/dt (t) + (t)m1a1 m2a2

ma

momentum

energy

force

acceleration

inertia

velocity

speed

energy

momentum

force



Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for theacceleration of block 1?

Hint 1. Force and acceleration

Since the system is closed, the only object that can affect block 1 is the other block in the system, block 2.

ANSWER:

Correct

Part F

What could be the reason for the acceleration of block 2?

ANSWER:

Correct

Part G

Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force

exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton's

second law?

ANSWER:

the large mass of block 1

air resistance

Earth's gravitational attraction

a force exerted by block 2 on block 1




F12

F21

and

and

and

and

=F12 m2a2 =F21 m1a1

=F12 m1a1 =F21 m2a2

=F12 m1a2 =F21 m2a1

=F12 m2a1 =F21 m1a2



Correct

Note that both and are internal forces.

Part H

Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2

on block 1 by . If we suppose that is greater than , which of the following statements about forces is

true?

Hint 1. Which of Newton's laws is useful here?

Newton's third law!

ANSWER:

Correct

Newton's third law states that forces and are equal in magnitude and opposite in direction. Therefore,

their x components are related by

Part I

Now recall the expression for the time derivative of the x component of the system's total momentum: . Considering the information that you now have, choose the best alternative for an equivalent

expression to .

Hint 1. What is ?

,

the total (internal) force on the system (as a whole). Use the information from the last part to simplify theright-hand side of the above equation.

ANSWER:

F12 F21

F12

F21 m1 m2

Both forces have equal magnitudes.

| | > | |F12 F21

| | > | |F21 F12

F 12 F

21

= −F12 F21

d (t)/dt =px Fx

d (t)/dtpx

Fx

= +Fx F12 F21



Correct

The derivative of the total momentum is zero; hence the total momentum is a constant function of time. Wehave just shown that for the special case of a closed two-block system, the internal forces do not change thetotal momentum of the system. It can be shown that in any system, the internal forces do not change the totalmomentum: It is conserved. In other words, total momentum is always conserved in a closed system ofobjects.

Collisions in One Dimension

On a frictionless horizontal air table, puck A (with mass 0.253 ) is moving toward puck B (with mass 0.372 ), whichis initially at rest. After the collision, puck A has velocity 0.118 to the left, and puck B has velocity 0.650 to theright.

Part A

What was the speed of puck A before the collision?

Hint 1. How to approach the problem

Apply the conservation of momentum equation. Keep in mind that momentum is a vector quantity, with bothmagnitude and direction. In a one-dimesional problem like this one, the direction of the momentum vectorcan be indicated by its sign. For the subparts that follow, take the positive direction to be to the right, theinitial direction of puck A.

Hint 2. The initial momentum

The initial momentum is given by

0.253 .

Hint 3. Find the final momentum of puck A

Find the final momentum of puck A, including the sign. Assume that the positive x direction is to the

right.

ANSWER:

Hint 4. Find the final momentum of puck B

0

nonzero constant

kt

kt2

kg kgm/s m/s

vAi

pAi

= (pAi kg ) ⋅ vAi

pAf

= −2.99×10−2 pAf kg ⋅ m/s



Find the final momentum of puck B, including the sign. Assume that the positive x direction is to the

right.

ANSWER:

ANSWER:

Correct

If you are required to use the answer obtained for a subsequent hint or part, use your unrounded/full precisionanswer.

Part B

Calculate , the change in the total kinetic energy of the system that occurs during the collision.


Use the velocity and mass of each puck before and after the collision to find their respective kinetic energies.The change in kinetic energy is then the final kinetic energy of the system minus the initial kinetic energy.

Hint 2. Find the initial kinetic energy of puck A

Find , the initial kinetic energy of puck A.

Express your answer in joules.

ANSWER:

Hint 3. Find the final kinetic energy of puck A

Find , the final kinetic energy of puck A.

ANSWER:

Hint 4. Find the final kinetic energy of puck B

Find , the final kinetic energy of puck B.

ANSWER:

pBf

= 0.242 pBf kg ⋅ m/s

= 0.838 vAi m/s

ΔK

KAi

= 8.88×10−2 KAi J

KAf

= 1.76×10−3 KAf J

KBf



ANSWER:

Correct

Exercise 8.22

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causingless damage. In one such accident, a 1800 car traveling to the right at 1.60 collides with a 1450 car going tothe left at 1.10 . Measurements show that the heavier car's speed just after the collision was 0.260 in itsoriginal direction. You can ignore any road friction during the collision.

Part A

What was the speed of the lighter car just after the collision?

ANSWER:

Correct

Part B

Calculate the change in the combined kinetic energy of the two-car system during this collision.

ANSWER:

Correct

Exercise 8.23

Two identical 1.40 masses are pressed against opposite ends of a spring of force constant 1.55 , compressingthe spring by 25.0 from its normal length.

Part A

= 7.86×10−2 KBf J

= −8.43×10−3 ΔK J

kg m/s kgm/s m/s

= 0.563 v m/s

= -2890 ΔK J

kg N/cmcm



Find the maximum speed of each mass when it has moved free of the spring on a smooth, horizontal lab table.

ANSWER:

Correct

Exercise 8.31: Asteroid Collision

Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing blow. Asteroid ,which was initially traveling at = 40.0 with respect to an inertial frame in which asteroid was at rest, isdeflected 30.0 from its original direction, while asteroid travels at 45.0 to the original direction of , as shown in .

Part A

Find the speed of asteroid after the collision.

ANSWER:

Correct

Part B

Find the speed of asteroid after the collision.

ANSWER:

= 1.86 vmax m/s

AvA1 m/s B

∘ B ∘ A

A

= 29.3 vA2 m/s

B



Correct

Part C

What fraction of the original kinetic energy of asteroid dissipates during this collision?

ANSWER:

Correct

Exercise 8.38

Two cars collide at an intersection. Car , with a mass of 2000 , is going from west to east, while car , of mass1400 , is going from north to south at 17.0 . As a result of this collision, the two cars become enmeshed andmove as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision,the enmeshed cars moved at an angle of 65.0 south of east from the point of impact.

Part A

How fast were the enmeshed cars moving just after the collision?

ANSWER:

Correct

Part B

How fast was car going just before the collision?

ANSWER:

Correct

Exercise 8.42

= 20.7 vB2 m/s

A

0.196

A kg Bkg m/s

∘

= 7.72 v m/s

A

= 5.55 v A m/s

kg



A bullet of mass 4.00 is fired horizontally into a wooden block of mass 1.28 resting on a horizontal surface. Thecoefficient of kinetic friction between block and surface is 0.210. The bullet remains embedded in the block, which isobserved to slide a distance 0.300 along the surface before stopping.

Part A

What was the initial speed of the bullet?

ANSWER:

Correct

Exercise 8.50

You are at the controls of a particle accelerator, sending a beam of 2.40×107 protons (mass ) at a gas target ofan unknown element. Your detector tells you that some protons bounce straight back after a collision with one of thenuclei of the unknown element. All such protons rebound with a speed of 2.10×107 . Assume that the initial speedof the target nucleus is negligible and the collision is elastic.

Part A

Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass .

ANSWER:

Correct

Part B

What is the speed of the unknown nucleus immediately after such a collision?

ANSWER:

Correct

Surprising Exploding Firework

A mortar fires a shell of mass at speed . The shell explodes at the top of its trajectory (shown by a star in the

g kg

m

= 357 v m/s

m/s m

m/s

m

15.0 m

3.00×106 m/s

m v0



figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smallerpiece of mass and a larger piece of mass . Both pieces land at exactly the same time. The smaller piece

lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance from themortar. If there had been no explosion, the shell would have landed a distance from the mortar. Assume that airresistance and the mass of the shell's explosive charge are negligible.

Part A

Find the distance from the mortar at which the larger piece of the shell lands.

Express in terms of .

Hint 1. Find the position of the center of mass in terms of

The two exploded pieces of the shell land at the same time. At the moment of landing, what is the distance from the mortar to the center of mass of the exploded pieces?

Express your answer in terms of .

Hint 1. Key idea

The explosion only exerts internal forces on the particles. The only external force acting on the two-piece system is gravity, so the center of mass will continue along the original trajectory of the shell.

ANSWER:

Hint 2. Find the position of the center of mass in terms of

The larger piece of the shell lands a distance from the mortar, and the smaller piece lands a distance zerofrom the mortar. What is , the final distance of the shell's center of mass from the mortar?

Express your answer in terms of .

m15

m45

dr

d

d r

r

xcm

r

= xcm r

d

d

xcm

d



Hint 1. A helpful figure

Here is a figure to help you visualize the situation.

ANSWER:

ANSWER:

Correct

The Center of Mass of the Earth-Moon-Sun System

A common, though incorrect, statement is, "The Moon orbits the Earth." That creates an image of the Moon’s orbit thatlooks like that shown in the figure.

The Earth's gravity pulls on the Moon, causing it to orbit. However, by Newton’s third law, it is known that the Moonexerts a force back on the Earth. Therefore, the Earth should move in response to the Moon. Thus a more accuratestatement is, "The Moon and the Earth both orbit the center of mass of the Earth-Moon system."

= xcm4d5

= d 1.25r



In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you willcalculate the center of mass of the Earth-Moon-Sun system. The mass of the Moon is 7.35×1022 , the mass of theEarth is 6.00×1024 , and the mass of the sun is 2.00×1030 . The distance between the Moon and the Earth is3.80×105 . The distance between the Earth and the Sun is 1.50×108 .

Part A

Calculate the location of the center of mass of the Earth-Moon system. Use a coordinate system in which the

center of the Earth is at and the Moon is located in the positive x direction.

Express your answer in kilometers to three significant figures.

Hint 1. Calculating the center of mass

The general equation for the center of mass for a system of two particles of masses and is

,

where and are the locations of the particles in the given coordinate system.

While the Earth and Moon are very large bodies, treating them as particles is reasonable in this problem,because the distance between them is much greater than their radii.

Hint 2. Find the coordinates of the Earth and Moon

Taking the center of the Earth as the origin of your coordinate system, what is the x coordinate of the Moon ?

Express your answer in kilometers to three significant figures.

ANSWER:

kgkg kg

km km

xcm

x = 0

xcm m1 m2

=xcm+m1x1 m2x2+m1 m2

x1 x2

xm

= 3.80×105 xm km



ANSWER:

Correct

Part B

Where is the center of mass of the Earth-Moon system?The radius of the Earth is 6378 and the radius of the Moon is 1737 . Select one of the answers below:

Choose the correct description of the location of the center of mass of the Earth-Moon system.

ANSWER:

Correct

As you can see, the center of mass for the Earth-Moon system actually lies within the radius of the Earth. Forthis reason, saying that the Moon orbits the Earth is often a good approximation, though in fact, both the Earthand the Moon orbit that point with a period of 28 days. The Moon makes large orbits around the center of massof the Earth-Moon system, whereas the center of the Earth makes small orbits.

Part C

Calculate the location of the center of mass of the Earth-Moon-Sun system during a full Moon. A full Moon occurswhen the Earth, Moon, and Sun are lined up as shown in the figure. Use a coordinate system in which the center ofthe sun is at and the Earth and Moon both lie along the positive x direction.

4600

km km

The center of mass is exactly in the center between the Earth and the Moon.

The center of mass is nearer to the Moon than the Earth, but outside the radius of the Moon.

The center of mass is nearer to the Earth than the Moon, but outside the radius of the Earth.

The center of mass is inside the Earth.

The center of mass is inside the Moon.

x = 0



Express your answer in kilometers to threesignificant figures.

Hint 1. Calculating the center of mass

The general equation for the center of mass for a system of thee particles of masses , , and

is

,

where , , and refer to the distances to each of the three particles from the origin.

ANSWER:

Correct

The equatorial radius of the Sun is 695,000 . As you can see, the center of mass for the Sun-Earth-Moonsystem is well within the Sun. However, if you were to find the center of mass of the Jupiter-Sun system, youwould find that it is slightly above the surface of the Sun at 780,000 from the center of the Sun. A distantalien civilization would not be able to see Jupiter directly, because it is far too faint, but they would be able tosee the Sun move back and forth as it orbited the center of mass with Jupiter. Because the sun is "wobbling,"alien scientists would be able to infer that there was a planet around the Sun. This is one of the methods thathuman scientists are using to identify planets around other stars.

Exercise 8.54

A 1200- station wagon is moving along a straight highway at 12.0 . Another car, with mass 1800 and speed20.0 , has its center of mass 40.0 ahead of the center of mass of the station wagon .

xcm m1 m2 m3

=xcm+ +m1x1 m2x2 m3x3+ +m1 m2 m3

x1 x2 x3

= 456 xcm km

km

km

kg m/s kgm/s m



Part A

Find the position of the center of mass of the system consisting of the two automobiles.

ANSWER:

Correct

Part B

Find the magnitude of the total momentum of the system from the above data.

ANSWER:

Correct

Part C

Find the speed of the center of mass of the system.

ANSWER:

Correct

= 16.0 behind the leading car xcm m

= 5.04×104 Px1 kg ⋅ m/s

= 16.8 vcm,x m/s



Part D

Find the total momentum of the system, using the speed of the center of mass.

ANSWER:

Correct

Part E

Compare your result of part D with that of part B.

ANSWER:

Correct

Conservation of Momentum in Inelastic Collisions

Learning Goal:

To understand the vector nature of momentum in the case in which two objects collide and stick together.

In this problem we will consider a collision of two moving objects such that after the collision, the objects stick togetherand travel off as a single unit. The collision is therefore completely inelastic.

You have probably learned that "momentum is conserved" in an inelastic collision. But how does this fact help you tosolve collision problems? The following questions should help you to clarify the meaning and implications of thestatement "momentum is conserved."

Part A

What physical quantities are conserved in this collision?

ANSWER:

Correct

= 5.04×104 Px2 kg ⋅ m/s

>

<

=

Px1 Px2

Px1 Px2

Px1 Px2

the magnitude of the momentum only

the net momentum (considered as a vector) only

the momentum of each object considered individually



Part B

Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speedsare and . What is the speed of the two-car system after the collision?


Think about how you would calculate the final speed of the two cars with the information provided and usingthe idea of conservation of momentum. Better yet, try the calculation out. What do you get?

ANSWER:

Correct

Part C

Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of theirmomenta are and . After the collision, what is the magnitude of their combined momentum?


Think about how you would calculate the final momentum of the two cars using the information provided andthe idea of conservation of momentum. Better yet, try the calculation out. What do you get? Keep in mindthat momentum is a vector, but you are asked about the magnitude of the momentum, which is a scalar.

ANSWER:

v1 v2

The answer depends on the directions in which the cars were moving before the collision.

+v1 v2

−v1 v2

−v2 v1

v1v2− −−−√+v1 v2

2

+v12 v2

2− −−−−−−−√

p1 p2



Correct

Part D

Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and

. After the collision, their combined momentum is . Of what can one be certain?

Hint 1. Momentum is a vector

Momentum is a vector quantity, and conservation of momentum holds for two-dimensional and three-dimensional collisions as well as for one-dimensional collisions.

ANSWER:

Correct

You can decompose the vector equation that states the conservation of momentum into individual equations foreach of the orthogonal components of the vectors.

Part E

Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of theirmomenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be

certain?

Hint 1. How to approach the problem mathematically

Momentum is a vector quantity. It is impossible to make exact predictions about the direction of motion after

The answer depends on the directions in which the cars were moving before the collision.

+p1 p2

−p1 p2

−p2 p1

p1p2− −−−√+p1 p2

2

+p12 p2

2− −−−−−−−√

p 1p 2 p

= +p p1→

p2→

= −p p1→

p2→

= −p p2→

p1→

p1 p2 p



a collision if nothing is known about the direction of motion before the collision. However, one can put somebounds on the values of the final momentum. Start with the expression for from Part D:

.

Therefore,

,

where is the angle between and . (To derive the above, you would have to break each vector into

components.) So the value of is controlled by .

Hint 2. How to approach the problem empirically

Consider the directions for the initial momenta that will give the largest and smallest final momentum.

ANSWER:

Correct

When the two cars collide, the magnitude of the final momentum will always be at most (a value

attained if the cars were moving in the same direction before the collision) and at least (a value

attained if the cars were moving in opposite directions before the collision).

± A Superball Collides Inelastically with a Table

As shown in the figure , a superball with mass equal to 50 grams is dropped from a height of . It collideswith a table, then bounces up to a height of . The duration of the collision (the time during which thesuperball is in contact with the table) is . In this problem, take the positive y direction to be upward, anduse for the magnitude of the acceleration due to gravity. Neglect air resistance.

p

= +p p1→

p2→

|p| = + = =∣∣p1→

p2→∣

∣ | + | + 2 ⋅p1 |2 p2 |2 p1→

p2→− −−−−−−−−−−−−−−−−√ | + | + 2| || |cosθp1 |2 p2 |2 p1 p2

− −−−−−−−−−−−−−−−−−−−−√θ p 1 p 2

|p| θ

+ ≥ p ≥p1 p2 p1p2− −−−√

+ ≥ p ≥p1 p2+p1 p2

2+ ≥ p ≥ | − |p1 p2 p1 p2

+ ≥ p ≥p1 p2 +p12 p2

2− −−−−−−−√

+p1 p2

| − |p1 p2

m = 1.5 mhi= 1.0 mhf

= 15 mstcg = 9.8 m/s2



Part A

Find the y component of the momentum, , of the ball immediately before the collision.

Express your answer numerically, to two significant figures.

You did not open hints for this part.

ANSWER:

Part B

Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving thetable.



ANSWER:

pbefore,y

= pbefore,y kg ⋅ m/s

= pafter,y kg ⋅ m/s



Part C

Find , the y component of the impulse imparted to the ball during the collision.



ANSWER:

Part D

Find the y component of the time-averaged force , in newtons, that the table exerts on the ball.



ANSWER:

Part E

Find , the change in the kinetic energy of the ball during the collision, in joules.



ANSWER:

Jy

= Jy kg ⋅ m/s

Favg,y

= Favgy N

−Kafter Kbefore

= −Kafter Kbefore J



Exercise 8.15

To warm up for a match, a tennis player hits the 58.0 ball vertically with her racket.

Part A

If the ball is stationary just before it is hit and goes 5.30 high, what impulse did she impart to it?

ANSWER:

Pucks on Ice

Two hockey players, Aaron and Brunnhilde, are pushing two pucks on a frictionless ice rink. The pucks are initially atrest on the starting line. Brunnhilde is pushing puck B, whichhas a mass three times as great as that of puck A, whichAaron is pushing. The players exert equal constant forces ofmagnitude on their pucks, directed horizontally, towardsthe finish line. They start pushing at the same time, and eachplayer pushes his or her puck until it crosses the finish line, adistance away.

Part A

Which puck reaches the finish line first?


ANSWER:

g

m

= p kg ⋅ m/s

F

d



Part B

Let be the magnitude of the kinetic energy of puck A at the instant it reaches the finish line. Similarly, is

the magnitude of the kinetic energy of puck B at the (possibly different) instant it reaches the finish line. Which ofthe following statements is true?


ANSWER:

Part C

This question will be shown after you complete previous question(s).

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, itbreaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions.

Both pucks reach the finish line at the same time.

Puck A reaches the finish line first.

Puck B reaches the finish line first.

More information is needed to answer this question.

KA KB

You need more information to decide.

=KA KB

<KA KB

>KA KB



Part A

What is the momentum of piece A before the explosion?

Express your answer numerically in kilogram meters per second.


ANSWER:

Part B

During the explosion, is the magnitude of the force of piece A on piece B greater than, less than, or equal to themagnitude of the force of piece B on piece A?


ANSWER:

p A,i

= p A,i kg ⋅ m/s

greater than

less than

equal to

cannot be determined



Part C

The momentum of piece B, , is measured to be 500 after the explosion. Find the momentum of

piece A after the explosion.

Enter your answer numerically in kilogram meters per second.


ANSWER:

Exercise 8.20

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible frictionbetween your feet and the ice. A friend throws you a ball of mass 0.400 that is traveling horizontally at 10.0 .Your mass is 70.0 .

Part A

If you catch the ball, with what speed do you and the ball move afterwards?

ANSWER:

Part B

If the ball hits you and bounces off your chest, so afterwards it is moving horizontally at 7.50 in the opposite

direction, what is your speed after the collision?

ANSWER:

Exercise 8.27

p B,f kg ⋅ m/s p A,f

= p A,f kg ⋅ m/s

kg m/skg

= v cm/s

m/s

= v cm/s

kg kg



Two ice skaters, Daniel (mass 60.0 ) and Rebecca (mass 45.0 ), are practicing. Daniel stops to tie his shoelaceand, while at rest, is struck by Rebecca, who is moving at 13.0 before she collides with him. After the collision,Rebecca has a velocity of magnitude 6.00 at an angle of 55.1 from her initial direction. Both skaters move on thefrictionless, horizontal surface of the rink.

Part A

What is the magnitude of Daniel's velocity after the collision?

ANSWER:

Part B

What is the direction of Daniel's velocity after the collision?

ANSWER:

Part C

What is the change in total kinetic energy of the two skaters as a result of the collision?

ANSWER:

Exercise 8.28

You are standing on a large sheet of frictionless ice and are holding a large rock. In order to get off the ice, you throwthe rock so it has velocity relative to the earth of 11.5 at an angle of 36.0 above the horizontal.

Part A

If your mass is 70.0 and the rock's mass is 14.1 , what is your speed after you throw the rock?

ANSWER:

kg kgm/s

m/s ∘

= v m/s

= from the Rebecca's original direction θ ∘

= ΔK J

m/s ∘

kg kg



A Girl on a Trampoline

A girl of mass kilograms springs from a trampoline with an initial upward velocity of meters persecond. At height meters above the trampoline, the girl grabs a box of mass kilograms.

For this problem, use meters per second per secondfor the magnitude of the acceleration due to gravity.

Part A

What is the speed of the girl immediately before she grabs the box?

Express your answer numerically in meters per second.


ANSWER:

Part B

What is the speed of the girl immediately after she grabs the box?

Express your answer numerically in meters per second.

= v m/s

= 60m1 = 8.0vi

h = 2.0 = 15m2

g = 9.8

vbefore

= vbefore m/s

vafter




ANSWER:

Part C

This question will be shown after you complete previous question(s).

Part D

What is the maximum height that the girl (with box) reaches? Measure with respect to the top of the

trampoline.

Express your answer numerically in meters.


ANSWER:

Trading Momenta in a Collision

Two particles move perpendicular to each other until they collide. Particle 1 has mass and momentum of magnitude , and particle 2 has mass and momentum of magnitude . Note: Magnitudes are not drawn to scale in any of the

figures.

= vafter m/s

hmax hmax

= hmax m

m2p 2m p



Part A

Suppose that after the collision, the particles "trade" their momenta, as shown in the figure. That is, particle 1 nowhas magnitude of momentum , and particle 2 has magnitude of momentum ; furthermore, each particle is now

moving in the direction in which the other had been moving. How much kinetic energy, , is lost in the

collision?

Express your answer in terms of and .


ANSWER:

Part B

Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy is lost in this case?

Express your answer in terms of and .


ANSWER:

p 2p

Klost

m p

= Klost

Klost

m p



Sinking the 9-Ball

Jeanette is playing in a 9-ball pool tournament. She will win if she sinks the 9-ball from the final rack, so she needs toline up her shot precisely. Both the cue ball and the 9-ball have mass , and the cue ball is hit at an initial speed of .Jeanette carefully hits the cue ball into the 9-ball off center, so that when the balls collide, they move away from eachother at the same angle from the direction in which the cue ball was originally traveling (see figure). Furthermore, afterthe collision, the cue ball moves away at speed , while the 9-ball moves at speed .

For the purposes of this problem, assume that the collision isperfectly elastic, neglect friction, and ignore the spinning ofthe balls.

Part A

Find the angle that the 9-ball travels away from the horizontal, as shown in the figure.

Express your answer in degrees to three significant figures.


ANSWER:

Exercise 8.56

At one instant, the center of mass of a system of two particles is located on the -axis at = 2.0 and has a velocity

= Klost

m vi

θvf v9

θ

= θ ∘

x x m^



of (5.0 ) . One of the particles is at the origin. The other particle has a mass of 0.10 and is at rest on the -axis at = 8.0 .

Part A

What is the mass of the particle at the origin?


ANSWER:

Part B

Calculate the total momentum of this system.


ANSWER:

Part C

What is the velocity of the particle at the origin?


ANSWER:

Exercise 8.55

A machine part consists of a thin, uniform 4.00- bar that is 1.50 long, hinged perpendicular to a similar vertical barof mass 3.00 and length 1.80 . The longer bar has a small but dense 2.00- ball at one end

m/s i kg xx m

= m1 kg

= P kg ⋅ m/s i

= v1 m/s i

kg mkg m kg



Part A

By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivotedcounterclockwise through 90 to make the entire part horizontal? Find the magnitude of horizontal displacement.

ANSWER:

Part B

Find the direction of horizontal displacement.

ANSWER:

Part C

Find the magnitude of vertical displacement.

ANSWER:

∘

= |Δx| m

to le left

to the right

= |Δy| m



Part D

Find the direction of vertical displacement.

ANSWER:

Exercise 8.58

A system consists of two particles. At = 0 one particle is at the origin; the other, which has a mass of 0.50 , is onthe -axis at = 6.0 . At = 0 the center of mass of the system is on the -axis at = 2.4 . The velocity of thecenter of mass is given by ( 0.75 ) .

Part A

Find the total mass of the system.

ANSWER:

Part B

Find the acceleration of the center of mass at any time .


ANSWER:

Part C

Find the net external force acting on the system at = 3.0 .


ANSWER:

upward

downward

t kgy y m t y y m

m/s3t2 i

= +m1 m2 kg

t

= a cm ( )m/s3t i

t s



Score Summary:Your score on this assignment is 107%.You received 13.92 out of a possible total of 14 points, plus 1.02 points of extra credit.

= . ∑ F ext N i

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Chapter 08 Homework