problem 7 - university of icelandisa13/efni/einn/f.hl/edlisfraedi 1 v/skiladaemi 6.pdf · answer:...

skiladæmi 6Due: 11:59pm on Wednesday, October 14, 2015

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Problem 7.54

A 64.0kg skier starts from rest at the top of a ski slope of height 61.0 .

Part A

If frictional forces do −1.03×104 of work on her as she descends, how fast is she going at the bottom of the slope?

Take free fall acceleration to be = 9.80 .

ANSWER:

Correct

Part B

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.24. If the patch isof width 60.0 and the average force of air resistance on the skier is 180 , how fast is she going after crossing thepatch?

ANSWER:

Correct

Part C

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.0 into it before comingto a stop. What is the average force exerted on her by the snowdrift as it stops her?

ANSWER:

Correct

A Ball Hits a Wall Elastically

A ball of mass moving with velocity strikes a vertical wall. The angle between the ball's initial velocity vector and thewall is as shown on the diagram, which depicts the situation as seen from above. The duration of the collision betweenthe ball and the wall is , and this collision is completely elastic. Friction is negligible, so the ball does not start spinning.In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.

m

J

g m/s2

= 29.6 v m/s

m N

= 15.9 v m/s

m

= 4060 F N

m v iθi

Δt

Typesetting math: 65%

Part A

What is the final angle that the ball's velocity vector makes with the negative y axis?

Express your answer in terms of quantities given in the problem introduction.

Hint 1. How to approach the problem

Relate the vector components of the ball's initial and final velocities. This will allow you to determine in termsof .

Hint 2. Find the y component of the ball's final velocity

What is , the component of the final velocity of the ball?

Express your answer in terms of quantities given in the problem introduction and/or and , the and components of the ball's initial velocity.

Hint 1. How to approach this part

There is no force on the ball in the y direction. From the impulsemomentum theorem, this means thatthe change in the y component of the ball's momentum must be zero.

ANSWER:

Hint 3. Find the component of the ball's final velocity

What is , the component of the ball' final velocity?

Express your answer in terms of quantities given in the problem introduction and/or and , the and components of the ball's initial velocity.

Hint 1. How to approach this problem

Since energy is conserved in this collision, the final speed of the ball must be equal to its initial speed.

ANSWER:

θf

θfθi

vfy y

vix viy x

y

= vfy − cos( )vi θi

x

vfx x

vix viy x

y


Hint 4. Putting it together

Once you find the vector components of the final velocity in terms of the initial velocity, use the geometry ofsimilar triangles to determine in terms of .

ANSWER:

Correct

Part B

What is the magnitude of the average force exerted on the ball by the wall?

Express your answer in terms of variables given in the problem introduction and/or .

Hint 1. What physical principle to use

Use the impulsemomentum theorem, , along with the definition of impulse, . In this

case, only one force is acting, so . Putting everything together, .

Hint 2. Change in momentum of the ball

The fact that implies that the component of the ball's momentum does not change during thecollision. What is , the magnitude of the change in the ball's momentum?

Express your answer in terms of quantities given in the problem introduction and/or .

ANSWER:

ANSWER:

Correct

Pucks on Ice

Two hockey players, Aaron and Brunnhilde, are pushing two pucks on a frictionless ice rink. The pucks are initially at reston the starting line. Brunnhilde is pushing puck B, which has a mass three times as great as that of puck A, which Aaron ispushing. The players exert equal constant forces of magnitude on their pucks, directed horizontally, towards the finishline. They start pushing at the same time, and each player pushes his or her puck until it crosses the finish line, a distance away.

= vfx − sin( )vi θi

θf θi

= θf θi

F

vix

= −J p f p i = ∑ ΔtJ F

= FΔt∣∣J ∣∣ F = −p f p iΔt

=θf θi yΔpx x

vix

= Δpx 2m sin( )vi θi

= F2mvix

Δt

F

dTypesetting math: 65%

Part A

Which puck reaches the finish line first?

Hint 1. Compute the relative acceleration of the pucks

If and are the magnitudes of the accelerations of pucks A and B, respectively, what is the value of theratio ?

ANSWER:

ANSWER:

Correct

Part B

Let be the magnitude of the kinetic energy of puck A at the instant it reaches the finish line. Similarly, is themagnitude of the kinetic energy of puck B at the (possibly different) instant it reaches the finish line. Which of thefollowing statements is true?

Hint 1. Determine the simplest way to answer this question

There are several possible approaches to this problem. Which is the simplest?

Choose the best option.

ANSWER:

aA aB

/aA aB

= 3/aA aB

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


Hint 2. Work done on puck A

Find , the work done on puck A over the distance .

ANSWER:

Hint 3. Work done on puck B

Find , the work done on puck B over the distance .

ANSWER:

ANSWER:

Correct

Part C

Let be the magnitude of the momentum of puck A at the instant it reaches the finish line. Similarly, is themagnitude of the momentum of puck B at the (possibly different) instant it reaches the finish line. Which of thefollowing statements is true?

Choose the best option.

Hint 1. Method 1: Compute the ratio of the pucks' velocities

The momentum of an object is the product of its mass and velocity. From the problem introduction, you knowthat . Find , the ratio of the velocity of puck A at the instant it reaches the finish line to the

velocity of puck B at the (possibly different) instant it reaches the finish line.

Hint 1. How to find the final velocities

You can easily compute the ratio using the (already determined) fact that the final kinetic energy of bothpucks is the same. Write the kinetic energy of each puck in terms of its velocity (for example,

). Set these expressions equal, and use the known ratio of the masses.

Use (force equals mass times acceleration) to find the acceleration of each puck.

Use (relating distance traveled to acceleration and time) to find the time to the finish line.

Use the workenergy theorem.

Apply conservation of momentum and energy.

F = ma

d = a12 t2

WA d

= WA Fd

WB d

= WB Fd

You need more information to decide.

=KA KB

<KA KB

>KA KB

pA pB

=mA

mB

13

/vA vB

=KA12

mAvA2


ANSWER:

Hint 2. Method 2: Use the impulsemomentum theorem

The impulsemomentum theorem states that.

You are given that both forces are the same, and you have compared the times in an earlier part.

ANSWER:

Correct

± The ImpulseMomentum Theorem

Learning Goal:

To learn about the impulsemomentum 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 . The left side is called the impulse of

the net force and is denoted by . Then equation (2) can be rewritten as

.

This equation is known as the impulsemomentum theorem. It states that the change in an object's momentum is equal tothe impulse of the net force acting on the object. In the case of a constant net force acting along the direction ofmotion, the impulsemomentum theorem can be written as

. (3)

Here , , and are the components of the corresponding vector quantities along the chosen coordinate axis. If themotion in question is twodimensional, it is often useful to apply equation (3) to the x and y components of motionseparately.

= 1.73/vA vB

Δp = FΔt

You need more information to decide.

=pA pB

<pA pB

>pA pB

Σ =F dp dt

ΣF F net

dp dt

t1 t2

Σ dt = dt∫ t2t1

F ∫ t2t1

dp dt

−p2→

p1→

J

= −J p2→

p1→

F net

F( − ) = m − mt2 t1 v2 v1

F v1 v2


The following questions will help you learn to apply the impulsemomentum theorem to the cases of constant and varyingforce 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:

Correct

Part B

The particle starts from rest at . 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:

Correct

Part C

The particle has momentum of magnitude at a certain instant. What is , the magnitude of its momentum\texttip\Delta tDeltat seconds later?

Express your answer in terms of any or all of \texttipp_\rm 1p_1, \texttipmm, \texttipFF, and\texttip\Delta tDeltat.

ANSWER:

Correct

Part D

The particle has momentum of magnitude \texttipp_\rm 1p_1 at a certain instant. What is \texttipv_\rm 2v_2,the magnitude of its velocity \texttip\Delta tDeltat seconds later?

Express your answer in terms of any or all of \texttipp_\rm 1p_1, \texttipmm, \texttipFF, and\texttip\Delta tDeltat.

ANSWER:

mF F

t = 0 p tt > 0

m F t

= p F(t)

t = 0 v tt > 0

m F t

= v F(t)m

p1 p2

\texttipp_\rm 2p_2 = F\left(\Deltat\right)+p_1


Correct

Let us now consider several twodimensional situations.

A particle of mass \texttipmm is moving in the positive x direction at speed \texttipvv. After a certain constant force isapplied to the particle, it moves in the positive y direction at speed 2v.

Part E

Find the magnitude of the impulse \texttipJJ delivered to the particle.

Express your answer in terms of \texttipmm and \texttipvv. Use three significant figures in the numericalcoefficient.


This is a twodimensional situation. It is helpful to find the components \texttipJ_\mit xJ_x and\texttipJ_\mit yJ_y separately and then use the Pythagorean theorem to find \texttipJJ.

Hint 2. Find the change in momentum

Find \Delta p_x, the magnitude of the change in the x component of the momentum of the particle.

Express your answer in terms of \texttipmm and \texttipvv.

ANSWER:

ANSWER:

Correct

Part F

Which of the vectors below best represents the direction of the impulse vector \vecJ?

ANSWER:

\texttipv_\rm 2v_2 = \large\fracF\left(\Deltat\right)+p_1m

\Delta p_x = m v

\texttipJJ = \sqrt5mv


Correct

Part G

What is the angle \texttip\theta theta between the positive y axis and the vector \vecJ as shown in the figure?

ANSWER:

Correct

Part H

If the magnitude of the net force acting on the particle is \texttipFF, how long does it take the particle to acquire itsfinal velocity, 2v in the positive y direction?

Express your answer in terms of \texttipmm, \texttipFF, and \texttipvv. If you use a numericalcoefficient, use three significant figures.

ANSWER:

Correct

1

2

3

4

5

6

7

8

26.6 degrees

30 degrees

60 degrees

63.4 degrees

\texttiptt = \large\fracmvF\sqrt5


So far, we have considered only the situation in which the magnitude of the net force acting on the particle was eitherirrelevant to the solution or was considered constant. Let us now consider an example of a varying force acting on aparticle.

Part I

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. A varying force F(t)=6.00t^24.00t+3.00 is acting onthe particle between t=0.00 seconds and t=5.00 seconds. Find the speed \texttipvv of the particle at t=5.00seconds.

Express your answer in meters per second to three significant figures.

Hint 1. Use the impulsemomentum theorem

In this case, v_1=0 and v_2=v. Therefore,

\large\int_0.00^5.00F\,dt=\Delta mv.

Hint 2. What is the correct antiderivative?

Which of the following is an antiderivative \large\int(6.00t^24.00t+3.00)dt?

ANSWER:

ANSWER:

Correct

Ballistic Pendulum

In a ballistic pendulum an object of mass \texttipmm is fired with an initial speed \texttipv_\rm 0v_0 at a pendulumbob. The bob has a mass \texttipMM, which is suspended by a rod of length \texttipLL and negligible mass. After thecollision, the pendulum and object stick together and swing to a maximum angular displacement \texttip\theta theta asshown .

6.00t^34.00t^2+3.00t

6.00t4.00

2.00t^32.00t^2+3.00t

12.00t4.00

\texttipvv = 43.0 m/s


Part A

Find an expression for \texttipv_\rm 0v_0, the initial speed of the fired object.

Express your answer in terms of some or all of the variables \texttipmm, \texttipMM, \texttipLL, and\texttip\theta theta and the acceleration due to gravity, \texttipgg.


There are two distinct physical processes at work in the ballistic pendulum. You must treat the collision and thefollowing swing as two separate events. Identify which physical law or principle applies to each event, write anexpression to describe the collision, write an expression to describe the swing, and then relate the twoexpressions to find \texttipv_\rm 0v_0.

Hint 2. Determine which physical laws and principles apply

Which of the following physical laws or principles can best be used to analyze the collision between the objectand the pendulum bob? Which can best be used to analyze the resulting swing?

A. Newton's first lawB. Newton's second lawC. Newton's third lawD. Conservation of mechanical energyE. Conservation of momentum

Enter the letters corresponding to the correct answer, with a letter first for the collision and then asecond letter for the swing separated by a comma.

ANSWER:

Hint 3. Describe the collision

Compose an expression that describes the collision between the object and the pendulum bob. Put thisexpression in the form v_0 = \cdots.

E,D


Express your answer in terms of some or all of the variables \texttipmm, \texttipMM, \texttipv_\rm0v_0, \texttipvv, \texttipLL, and \texttip\theta theta and the acceleration due to gravity,\texttipgg.

Hint 1. Identify the type of collision

Is the collision between the object and the pendulum bob an elastic or inelastic collision?

ANSWER:

Hint 2. Find the momentum before the collision

Compose an expression for \texttipp_\rm beforep_before, the momentum of the object and pendulumbob before the collision when the object moves with speed \texttipv_\rm 0v_0.

Express your answer in terms of some or all of the variables \texttipmm, \texttipMM,\texttipv_\rm 0v_0, \texttipvv, \texttipLL, and \texttip\theta theta and the accelerationdue to gravity, \texttipgg.

Hint 1. Momentum

The momentum of an object of mass \texttipmm moving with speed \texttipvv is given bymv.

ANSWER:

Hint 3. Find the momentum after the collision

Compose an expression for \texttipp_\rm afterp_after, the momentum of the object and pendulumbob after the collision when they move with speed \texttipvv.


ANSWER:

elastic

inelastic

\texttipp_\rm beforep_before = m v_0


ANSWER:

Hint 4. Describe the swing

Compose an expression that describes the motion of the object and the pendulum bob after the collision. Putthis expression in the form v = \cdots.

Express your answer in terms of some or all of the variables \texttipmm, \texttipMM, \texttipv_\rm0v_0, \texttipvv, \texttipLL, and \texttip\theta theta and the acceleration due to gravity\texttipgg.

Hint 1. Identify the energy at the bottom of the swing

What is the mechanical energy \texttipE_\rm bottomE_bottom of the object and pendulum bob justafter the collision but while they are still located at the bottom of the swing? Assume that the height ofthe pendulum bob and object is zero at this location.


Hint 1. Mechanical energy

The mechanical energy of a system is the total kinetic and gravitational potential energy of thesystem. The kinetic energy \texttipKK of an object of mass \texttipmm moving with speed\texttipvv is

\largeK = \frac12mv^2.The gravitational potential energy \texttipUU of an object of mass \texttipmm a height\texttipyy above some reference point is

U = mgy.

Hint 2. Find the gravitational potential energy at the bottom of the swing

What is the gravitational potential energy \texttipU_\rm bottomU_bottom of the object andpendulum bob at the bottom of the swing? Keep in mind that the height of the pendulum bob andobject is zero at this location.

\texttipp_\rm afterp_after = \left(m+M\right) v

\texttipv_\rm 0v_0 = \large\fracm+Mm v


ANSWER:

Hint 3. Find the kinetic energy at the bottom of the swing

What is the kinetic energy \texttipK_\rm bottomK_bottom of the object and pendulum bob atthe bottom of the swing?

Express your answer in terms of some or all of the variables \texttipmm, \texttipMM,\texttipv_\rm 0v_0, \texttipvv, \texttipLL, and \texttip\theta theta and theacceleration due to gravity, \texttipgg.

ANSWER:

ANSWER:

Hint 2. Identify the energy at the top of the swing

What is the mechanical energy \texttipE_\rm topE_top of the object and pendulum bob at the top ofthe swing when it has reached its maximum angular displacement?


Hint 1. Mechanical energy

The mechanical energy of a system is the total kinetic and gravitational potential energy of thesystem. The kinetic energy \texttipKK of an object of mass \texttipmm moving with speed\texttipvv is

\largeK = \frac12mv^2.The gravitational potential energy \texttipUU of an object of mass \texttipmm a height\texttipyy above some reference point is

U_\rm bottom= mgL

U_\rm bottom= 0

U_\rm bottom= mgL

\texttipK_\rm bottomK_bottom = \large\frac12 \left(m+M\right) v^2

\texttipE_\rm bottomE_bottom = \large\frac12 \left(m+M\right) v^2


U = mgy.

Hint 2. Find the height at the top of the swing

What is the height \texttiphh of the object and pendulum bob at the top of the swing, when theyhave reached their maximum displacement? Keep in mind that the pendulum has a length\texttipLL and swings through an angle \texttip\theta theta.

Express your answer in terms of some or all of the variables \texttipLL and \texttip\thetatheta.

ANSWER:

Hint 3. Find the gravitational potential energy at the top of the swing

What is the gravitational potential energy \texttipU_\rm topU_top of the object and pendulumbob at the top of the swing?

Express your answer in terms of some or all of the variables \texttipmm, \texttipMM,\texttipv_\rm 0v_0, \texttipvv, \texttipLL, and \texttip\theta theta and theacceleration due to gravity, \texttipgg.

ANSWER:

Hint 4. Find the kinetic energy at the top of the swing

What is the kinetic energy \texttipK_\rm topK_top of the object and pendulum bob at the topof the swing?

ANSWER:

ANSWER:

\texttiphh = L \left(1 \cos\left(\theta\right)\right)

\texttipU_\rm topU_top = \left(m+M\right) g L \left(1 \cos\left(\theta\right)\right)

\largeK_\rm top = \frac 12 m v_0^2

\largeK_\rm top = \frac 12 m v^2

\largeK_\rm top = \frac 12 (m + M) v^2

K_\rm top = 0


ANSWER:

Hint 5. Relating the two physical processes

By applying conservation of momentum to the collision, you found an expression for \texttipv_\rm 0v_0:

v_0 = \left ( m+M\over m \right ) v.By applying conservation of mechanical energy to the swing, you found an expression for \texttipvv:

v= \sqrt2gL[1 \cos(\theta)].Combine these two expression into just one expression for \texttipv_\rm 0v_0.

ANSWER:

Correct

The ballistic pendulum was invented during the Napoleonic Wars to aide the British Navy in making bettercannons. It has since been used by ballisticians to measure the velocity of a bullet as it leaves the barrel of a gun.In Part B you will use your expression for \texttipv_\rm 0v_0 to compare the initial speeds of bullets fired from9.0\rm mm and .44caliber handguns.

Part B

An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 \rm mm and a .44caliber. The guns are fired into a 10\rm kg pendulum bob of length \texttipLL. Assume that the 9.0\rm mm bullethas a mass of 6.0 \rm g and the .44caliber bullet has a mass of 12 \rm g . If the 9.0\rm mm bullet causes thependulum to swing to a maximum angular displacement of 4.3^\circ and the .44caliber bullet causes a displacement of10.1^\circ , find the ratio of the initial speed of the 9.0\rm mm bullet to the speed of the .44caliber bullet,(v_0)_9.0/(v_0)_44.

Express your answer numerically.


Use your expression from Part A to set up the ratio (v_0)_9.0/(v_0)_44. Try to cancel as many terms aspossible before plugging in your numbers to solve for a numeric answer.

ANSWER:

\texttipE_\rm topE_top = \left(m+M\right) g L \left(1 \cos\left(\theta\right)\right)

\texttipvv = \sqrt2 g L \left(1 \cos\left(\theta\right)\right)

\texttipv_\rm 0v_0 = \large\frac\left(m+M\right)\sqrt2gL\left(1\cos\left(\theta\right)\right)m

(v_0)_9.0/(v_0)_44 = 0.85


Correct

Police officers in the United States commonly carry 9.0\rm mm handguns because they are easier to handle,having a shorter barrel than typical .44caliber guns. Not only does the .44caliber bullet have more mass than the9.0\rm mm one, its passage through a longer gun barrel means that it also moves faster as it leaves the barrel,which makes the .44caliber Magnum a particularly powerful handgun. A .44caliber bullet can travel at speedsover 1000 \rm miles \;per\; hour (1600 \rm kilometers\;per\;hour).

Exercise 8.46

A 0.148 \rm kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.870 \rm m/s . It has aheadon collision with a 0.290 \rm kg glider that is moving to the left with a speed of 2.11 \rm m/s . Suppose the collisionis elastic.

Part A

Find the magnitude of the final velocity of the 0.148 \rm kg glider.

ANSWER:

Correct

Part B

Find the direction of the final velocity of the 0.148 \rm kg glider.

ANSWER:

Correct

Part C

Find the magnitude of the final velocity of the 0.290 \rm kg glider.

ANSWER:

Correct

Part D

Find the direction of the final velocity of the 0.290 \rm kg glider.

ANSWER:

3.08 m/s

to the right

to the left

9.61×10−2 m/s


Correct

Score Summary:Your score on this assignment is 102%.You received 6.13 out of a possible total of 6 points.

to the right

to the left


problem 7 - university of icelandisa13/efni/einn/f.hl/edlisfraedi 1 v/skiladaemi 6.pdf · answer:...

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