midterm exam 3 - siena collegemmccolgan/gp130f12/exams/midterm3_all_mm.pdfmidterm exam 3 november...

Physics 130 General Physics Fall 2012

Midterm Exam 3November 29, 2012

Name:

Instructions

1. This examination is closed book and closed notes. All your belongingsexcept a pen or pencil and a calculator should be put away and yourbookbag should be placed on the floor.

2. You will find one page of useful formulae on the last page of the exam.

3. Please show all your work in the space provided on each page. If youneed more space, feel free to use the back side of each page.

4. Academic dishonesty (i.e., copying or cheating in any way) willresult in a zero for the exam, and may cause you to fail theclass.

In order to receive maximum credit,each solution should have:

1. A labeled picture or diagram, if appropriate.2. A list of given variables.3. A list of the unknown quantities (i.e., what you are being asked to find).4. A force-interaction diagram, if appropriate.5. One or more free-body diagrams, as appropriate, with labeled 1D or 2D

coordinate axes.6. Algebraic expression for the net force along each dimension, as appro-

priate.7. Algebraic expression for the conservation of energy or momentum equa-

tions, as appropriate.8. An algebraic solution of the unknown variables in terms of the

known variables.9. A final numerical solution, including units, with a box around it.

10. An answer to additional questions posed in the problem, if any.

1


1. A 10 m long glider with a mass of 680 kg (including the passengers) is gliding horizontallythrough the air at 30 ms when a 60 kg skydiver drops out by releasing his grip on theglider. What is the glider’s velocity just after the skydiver lets go?

Solution:

This is a conservation of momentum problem. At the start, both the glider and theskydiver are traveling at the same speed.

pfx � pix (1)

mgpvf qg �mspvf qs � pmg �msqpviq (2)

Immediately after release, the skydiver’s horizontal velocity is still 30 ms. Solvingfor the final velocity of the glider,

pvf qg � pmg �msqpviq �mspvf qsmg

(3)

pvf qg � p680kgqp30m{sq � 60kgp30m{sqs620kg

(4)

pvf qg � 30m{s (5)

The skydivers motion in the vertical direction has no influence on the gliders hor-izontal motion. Notice that we did not need to invoke conservation of momentumto solve this problem. Because there are no external horizontal forces acting on ei-ther the skydiver or the glider, neither will change their horizontal speed when theskydiver lets go!

2


2. A 20 g ball of clay traveling east at 2.0 m{s collides with a 30 g ball of clay traveling30� south of west at 1.0 m{s. What are the speed and the direction of the resulting 50 gball of clay?

Solution:

Applying conservation of momentum in the x-direction, find the final momentum inthe x-direction

pfx � pix � m1pv � ixq1 �m2pvixq2 (1)

pfx � p0.02kgqp2.0m{sq � p0.03kgqp1.0m{sq cosp30�q � 0.0140kgms (2)

(3)

Applying conservation of momentum in the y-direction, find the final momentum inthe y-direction

pfy � piy � m1pv � iyq1 �m2pviyq2 (4)

pfy � p0.02kgqp0m{sq � p0.03kgqp1.0m{sq sinp30�q � �0.0150kgm{s (5)

(6)

Find the magnitude of the final momentum using the pythagorean theorem to com-bine the x- and y-components

pf � apfx � pfy (7)

pf �ap0.0140kgm{sq2 � p�0.0150kgm{sq2 � 0.0205kgm{s (8)

(9)

We know that

pf � pm1 �m2qvf � 0.0205kgm{s (10)

(11)

3


Solving for the final speed,

vf � pfm1 �m2

(12)

vf � 0.0205kgm{s0.02kg � 0.03kg

(13)

vf � �.41m{s (14)

(15)

The direction is given by

θ � tan�1 pfypfx

� 0.014kgm{s0.0150kg

m{s (16)

θ � 47� (17)

(18)

4


3. Most geologists believe that the dinosaurs became extinct 65 million years ago when alarge comet or asteroid struck the earth, throwing up so much dust that the sun wasblocked out for a period of many months. Suppose an asteroid with a diameter of 2.0 kmand a mass of 1.0� 1013 kg hits the earth with an impact speed of 4.0� 104 m{s.(a) What is the earth’s recoil speed after such a collision? (Use a reference frame in

which the earth was initially at rest.)

(b) What percentage is this of the earth’s speed around the sun? (Use the astronomicaldata provided.)

Solution:

(a) This is a conservation of momentum problem. At the start, the asteroid has aspeed of 4.0� 104 m{swhile the earth is not moving.

pix � pfx (1)

mepviqe �mapviqa � pme �maqpvf q (2)

(3)

Solving for the final velocity of the earth and asteroid system,

vf � mepviqe �mapviqapme �maq (4)

5


vf � 0 kgm{s� p1.0� 1013kgqp4.0� 104m{sqp1.0� 1013kg � 5.98� 1024kgq (5)

vf � 6.8� 10�8m{s (6)

(b) The speed of the earth going around the sun is

vE � 2πr

T� 2πp1.50� 1011mq

3.15� 107s� 3.0� 104m{s (7)

(8)

6


4. Fred (mass 60 kg) is running with the football at a speed of 6.0 m{s when he is methead-on by Brutus (mass 120 kg), who is moving at 4.0 m{s. Brutus grabs Fred in atight grip, and they fall to the ground. Which way do they slide, and how far? Thecoefficient of kinetic friction between football uniforms and Astroturf is 0.30.

Solution:

This is an inelastic collision in which Fred and Brutus are moving toward each other.Assume that Brutus is moving in the positive x-direction and Fred is moving in thenegative x-direction. Momentum is conserved so that

pix � pfx (1)

mBpviqB �mF pviqF � pmB �mF qpvf q (2)

(3)

Solving for the final velocity of Brutus and Fred,

vf � mBpviqB �mF pviqFpmB �mF q (4)

vf � p120kgqp4.0m{sq � p60kgqp6.0m{sqp180kgq (5)

vf � 0.667m{s (6)

(7)

Now that we know that Brutus and Fred moved to the right after the collision, wecan find the distance that they traveled. Friction slows them down. The force offriction is in the negative x-direction. The sum of the forces is equal to the masstimes the acceleration.¸

Fx � mTa (8)

fk � �µFn � �µpmB �mF qg � pmB �mF qa (9)

a � �µg � �p0.30qp9.8m{s2q (10)

a � 2.94m{s2 (11)

7


Using the kinematic equation,

pvf q2 � pviq2 � 2ad � �2ad (12)

d � pvf q22a

� p0.667m{s2q2p2qp2.94m{s2q � 0.076m (13)

d � 7.6cm (14)

(15)

8


5. A neutron is an electrically neutral subatomic particle with a mass just slightly greaterthan that of a proton. A free neutron is radioactive and decays after a few minutesinto other subatomic particles. In one experiment, a neutron at rest was observed todecay into a proton (mass 1.67� 10�27 kg) and an electron (mass 9.11� 10�31 kg). Theproton and electron were shot out back-to-back. The proton speed was measured to be1.0� 105 m{s and the electron speed was 3.0� 107 m{s. No other decay products weredetected.

(a) Was momentum conserved in the decay of this neutron?

(b) If a neutrino was emitted in the above neutron decay, in which direction did ittravel? Explain your reasoning.

(c) How much momentum did this neutrino ”carry away” with it?

Solution:

(a) The initial momentum is pi � 0 since the neutron is not moving. The momentumafar the decay is

pf � meve �mpvp (1)

pf � p9.11� 10�31kgqp3.0� 107m{sq � p1.67� 10�27kgqp1.0� 105m{sq (2)

pf � �1.4� 10�22kg m{s (3)

(4)

Momentum is not conserved.

(b) Because the initial momentum is zero,

pe � pp � pn � 0 (5)

pn � 1.4� 10�22kg m{s (6)

(7)

(c) The neutrino must ”carry away” 1.4� 10�22kg m{s.

9


6. A two-stage rocket is traveling at 1200 m{s with respect to the earth when the firststage runs out of fuel. Explosive bolts release the first stage and push it backward witha speed of 35 m{s relative to the second stage. The first stage is three times as massiveas the second stage. What is the speed of the second stage after the separation?

Solution:

The rocket is traveling at an initial speed before the explosion. The two parts separateafter the explosion. To find the speed of the second stage, we’ll use the conservationof momentum equation.

pf � pi (1)

pm1 �m2qvi � m1v1f �m2v2f (2)

p3m2 �m2qvi � 3m2v1f �m2v2f � m2p3v1f � v2f q (3)

4m2vi � m2p3v1f � v2f q (4)

4vi � 3v1f � v2f (5)

(6)

The fact that the first stage is pushed backward at 35 m{s relative to the second canbe written

v1f � �35m{s� v2f (7)

4vi � 3p�35m{s� v2f q � v2f (8)

4vi � 105m{s� 4v2f (9)

(10)

10


Solving for the velocity of the second stage v2f

v2f � 4vi � 105m{s4

(11)

v2f � 4 � 1200m{s� 105m{s4

(12)

v2f � 1.2 km{s (13)

(14)

11


7. A 500 g rubber ball is dropped from a height of 10 m and undergoes a perfectly elasticcollision with the earth.

(a) For an elastic collision, what quantities are conserved?

(b) Write out the conservation of momentum equation for this problem.

(c) Write out the conservation of energy equation for this problem.

(d) Which quantities are not known in these equations?

(e) Explain how to obtain the formulas for these unknown quantities.

(f) What is the earth’s velocity after the collision? Assume the earth was at rest justbefore the collision.

(g) How many years would it take the earth to move 1.0 mm at this speed?

Solution:

(a) In an elastic collision, both energy and momentum are conserved.

(b) The conservation of momentum equation is

pi � pf (1)

mbvbi �mevei � mbvbf �mevef (2)

mbvbi � 0 � mbvbf �mevef (3)

(c) The conservation of energy equation is

Ki � Ui � Kf � Uf (4)

1

2mbpvbiq2 � 1

2mepveiq2 � 1

2mbpvbf q2 � 1

2mepvef q2 (5)

1

2mbpvbiq2 � 0 � 1

2mbpvbf q2 � 1

2mepvef q2 (6)

mbpvbiq2 � 0 � mbpvbf q2 �mepvef q2 (7)

(d) The quantities that are not known in these equations are vbf and vef .

(e) We’re given the mass of the asteroid and it’s initial velocity. We also knowthe mass of the earth. Therefore, we have two unknowns and two equations.The unknowns are vbf and vef . To find the final velocity of the earth, we can

12


solve for vbf in the momentum equation and plug this into the energy equation toeliminate vbf . This will give us the equation for vef . Once we have an expressionfor vef , we can plug this into the momentum equation to get an expression forvbf .

(f) We’ll assume that the earth is at rest right before the collision. To find thespeed of the rubber ball right before the collision, we can use kinematics.

pvfBq2 � ppviBq2 � 2a∆h � 2p9.8m{s2qp10mq (8)

vfB �b

2p9.8m{s2qp10mq � 14.0m{s (9)

By continuing with the derivations above, we find that the equations for vbfareand vef

vbf � mbvbi �mevefmb

(10)

vef � 2mbvbime �mb

(11)

We found that the velocity of the ball right before it collided with the earth was14.0m{s. This is our initial velocity of the ball in our conservation of momentumand energy equations. So, vbi � 14.0m{s. Plugging this into the above equation,we solve for vef .

vef � 2p0.500kgqp14.0m{sq5.98� 1024kg

� 2.3� 10�24m{s (12)

(g) The time it would take the earth to move 1.0mm at this speed is given by thekinematics equation.

xf � xi � vi∆t� 1

2a∆t2 � 0� vi∆t� 0 (13)

∆t � xfvi� 0.001m

2.3� 10�24m{s � 4.3� 1020 sec � 1.36� 1013 years (14)

13


8. A package of mass m is release from rest at a warehouse loading dock and slides downa 3.0 m high frictionless chute to a waiting truck. Unfortunately, the truck driver wenton a break without having removed the previous package, of mass 2m, from the bottomof the chute.

(a) Suppose the packages stick together. What is their common speed after the colli-sion?

(b) For an elastic collision, what quantities are conserved?

(c) Write out the conservation of momentum equation for this problem.

(d) Write out the conservation of energy equation for this problem.

(e) Which quantities are not known in these equations?

(f) Explain how to obtain the formulas for these unknown quantities.

(g) If the collision between the packages is perfectly elastic, to what height does thepackage of mass m rebound?

Solution:

(a) From conservation of energy, we can find the speed of the package at the bottomof the chute.

Ki � Ui � Kf � Uf (1)

0�mgh � 1

2mv2 (2)

v �a

2gh �b

2p9.8m{s2qp3.0mq � 7.668m{s (3)

If the packages stick together, the collision is inelastic and conservation of mo-mentum is conserved.

pi � pf (4)

m1v1i �m2v2i � pm1 �m2qvf (5)

mv1i � 0 � 3mvf (6)

14


Cancelling the m’s

v1i � 3vf (7)

Solving for vf , we find

vf � v1i

3� 7.668m{s

3� 2.56m{s (8)

(9)

The packages move together with a speed of 2.56m{s.(b) In an elastic collision, both energy and momentum are conserved.

(c) In this conservation of momentum equation, we’ll use the velocity calculatedabove for the initial velocity before the collision. Also, the package at thebottom is at rest before the collision. The conservation of momentum equationgives

pi � pf (10)

m1v1i �m2v2i � m1v1f �m2v2f (11)

m1v1i � 0 � m1v1f �m2v2f (12)

mv1i � mv1f � p2mqv2f (13)

v1i � v1f � 2v2f (14)

v1f � v1i � 2v2f (15)

(d) The conservation of energy equation is

Ki � Ui � Kf � Uf (16)

1

2m1pv1iq2 � 1

2m2pv2iq2 � 1

2m1pv1f q2 � 1

2m2pv2f q2 (17)

1

2m1pv1iq2 � 0 � 1

2m1pv1f q2 � 1

2me2v2f q2 (18)

mpv1iq2 � mpv1f q2 � 2mpv2f q2 (19)

pv1iq2 � pv1f q2 � 2pv2f q2 (20)

(e) The quantities that are not known in these equations are v1f and v2f .

(f) We have two unknowns and two equations. The unknowns are v1f and v2f .To find the final velocity of the second package, we can solve for v1f in themomentum equation and plug this into the energy equation to eliminate v1f .This will give us the equation for v2f . Once we have an expression for v2f , wecan plug this into the momentum equation to get an expression for v1f .

15


From the derivations above, the equation relating the final speed of the firstmass to it’s initial speed is

v1f � m� 2m

m� 2mv1i � �1

3p7.668m{sq � �2.56m{s (21)

(22)

The package rebounds and goes back up the ramp. To determine how high thepackage goes, we can use the conservation of energy.

Ki � Ui � Kf � Uf (23)

1

2mv2 � 0 � 0�mgh (24)

h � v2

2g� p�2.56m{sq2

2p9.8m{s2q � 0.33 m (25)

16


9. A roller coaster car on the frictionless track shown below starts from rest at height h.The track’s valley and hill consist of circular-shaped segments of radius R.

(a) What is the maximum height hmax from which the car can start so as not to fly offthe track when going over the hill? Give your answer as a multiple of R. Hint Firstfind the maximum speed for going over the hill.

(b) Evaluate hmax for a roller coaster that has R � 10 m.

Solution:

(a) When the car is in the valley and on the hill it can be considered to be in circularmotion. We start at the top of the hill and determine the maximum speed sothat the car stays on the track. We can use the equations for circular motion.The acceleration is given by,

a � v2

r(1)

The free-body diagram for the car at the top of the hill, illustrates the two forcesacting on the car. Using Newton’s law, we know,

¸F � ma � �mv2

R(2)

FN � Fg � �mv2

R(3)

To stay on the track, the normal force has to be greater than zero. If we set thenormal force equal to zero, this is the maximum speed of the car that allows itto stay on the track.

Fg � mv2

R(4)

mg � mv2

R(5)

v2 � gRÑ v �agR (6)

17


Now that we know the maximum height at the top of the hill, we can use theconservation of energy to find the starting height.

Kf � Uf � Ki � Ui (7)

1

2mpvf q2 �mgyf � 1

2mpviq2 �mgyi (8)

1

2mpvf q2 �mgyf � 0�mgyi (9)

We’ll use yf � R for the height at the top of the hill and factor the mass term,

1

2pvf q2 � gR � 0� gyi (10)

yi � pvf q22g

�R (11)

We can now plug in our results for vmax from above

yi � p?gRq22g

�R (12)

yi � R

2�R � 3R

2(13)

(14)

(b) Plugging into the above equation, hmax for a roller coaster that has R � 10 m

yi � 3R

2� 3 � 10m

2� 15m (15)

18


10. A pendulum is formed from a small ball of mass m on a string of length L. As thefigure shows, a peg is height h � L{3 above the pendulum’s lowest point. From whatminimum angle θ must the pendulum be released in order for the ball to go over the

top of the peg without the string going slack?

Solution: This is a two part problem. First, we need to find the velocity for the ballto go over the peg without the string going slack. Then we need to find the potentialenergy to match that velocity.

(a) Find velocity of ball so that string doesn’t go slack.

We’ll set our origin on the peg. This means that once the string touches thepeg, the ball will be moving in circular motion with a radius of L{3.

When the ball is let go, it moves below the peg and then moves in circularmotion. When it reaches the top of the circle and is directly above the peg, wecan use Newton’s second law and the fact that the ball is moving in circularmotion. ¸

F � ma � �mv2

R(1)

�T � Fg � �mv2

R(2)

(3)

19


The critical speed is just as the tension equals zero. Factoring the negatives andthe masses, and setting Fg � mg and T � 0,

0�mg � mpvcq2R

(4)

pvcq �agR (5)

Setting R � L{3,

pvcq �cgL

3(6)

(b) Find the angle to release the ball. Now that we know the velocity for the stringto remain taut, we’ll use the conservation of energy to find the height to releasethe ball.

Kf � Uf � Ki � Ui (7)

1


2mpviq2 �mgyi (8)

1

2pvcq2 � gyf � 0� gyi (9)

Plugging in yf � L{3 and pvcq2 � gL{3,

gL

6� gL

3� gyi (10)

L

6� L

3� yi (11)

yi � L

2(12)

This result tells us that we need to release the ball at a height of L/2 above thepeg.

To find the angle, consider the third figure in the three-part figure above. Weneed to find the distance of the ball below the point where the string is attached.We can use the string length and result above.

h � L� L

2� L

3� L

6(13)

To find the angle, we can use cosine,

cos θ �L6

L� 1

6(14)

θ � 80.4� (15)

20


11. A sled starts from rest at the top of the frictionless, hemispherical, snow-covered hillshown below.

(a) Find an expression for the sled’s speed when it is at angle φ.

(b) Use Newton’s laws to find the maximum speed the sled can have at angle φ withoutleaving the surface.

(c) At what angle φmax does the sled “fly off” the hill?

Solution:

(a) Find sled’s speed When the sled is at a position that relates to angle φ, theheight is the R cosφ. Using conservation of energy to find the speed,

Kf � Uf � Ki � Ui (1)

1


2mpviq2 �mgyi (2)

1

2pvf q2 � gR cosφ � 0� gyi (3)

pvf q2 � 2gR � 2gR cosφ � 2gRp1� cosφq (4)

vf �a

2gRp1� cosφq (5)

(b) Find max speed ¸F � ma � �mv2

R(6)

FN � Fg cosφ � �mv2

R(7)

21


The critical speed is just as the normal force equals zero. Factoring the negativesand the masses, and setting Fg � mg and FN � 0,

g cosφ � pvcq2R

(8)

vc �agR cosφ (9)

(c) Find the angle when the sled gets air! We can set the speeds from the twoprevious parts equal to each other to find φ.a

gR cosφ �a

2gRp1� cosφq (10)

gR cosφ � 2gRp1� cosφq (11)

cosφ � 2� 2 cosφ (12)

3 cosφ � 2 (13)

φ � cos�1p23q Ñ φ � 48� (14)

22


12. Bob can throw a 500 g rock with a speed of 30 m/s. He moves his hand forward 1.0 mwhile doing so.

(a) How much work does Bob do on the rock?

(b) How much force, assumed to be constant, does Bob apply to the rock?

(c) What is Bob’s maximum power output as he throws the rock?

Solution:

(a) Calculate work

W � ∆K (1)

W � 1

2mpvf q2 � 1

2mpviq2 (2)

W � 1

2mpvf q2 � 0 (3)

W � 1

20.5kgp30m{sq2 (4)

W � 225 J (5)

(b) Calculate force

W � F �∆r � F∆r (6)

F � W

∆r� 225 J

1m(7)

F � 225 N (8)

(c) Calculate power

P � F � v � Fv (9)

P � 225 Np30m{sq (10)

P � 6750 W (11)

23


13. Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loosegravel are constructed to stop runaway trucks that have lost their brakes. The

combination of a slight upward slope and a large coefficient of rolling resistance as thetruck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp

slopes upward at 6.0� and the coefficient of rolling friction is 0.40. Usw work andenergy to find the length of a ramp that will stop a 15, 000 kg truck that enters the

ramp at 35 m/s (� 75 mph).

Solution:

We’ll identify the truck and the loose gravel as the system. We need the gravel insidethe system because friction increases the temperature of the truck and the gravel.We will also use the model of kinetic friction and the conservation of energy equation.

Kf � Uf �∆Eth � Ki � Ui �Wext (1)

0� Uf �∆Eth � Ki � 0� 0 (2)

(3)

The thermal energy created by friction is

∆Eth � fkp∆xq (4)

∆Eth � pµkFNqp∆xq (5)

∆Eth � pµkmgcosθqp∆xq (6)

∆Eth � pµkmg cos θqp∆xq (7)

Plugging this into the conservation of energy equation,

Uf � µkmg cos θp∆xq � Ki (8)

∆x � Ki � Uf

µkmg cos θ(9)

∆x �12mpviq2 �mgyf qµkmg cos θ

(10)

(11)

24


14. An 8.0 kg crate is pulled 5.0 m up a 30� incline by a rope angled 18� above the incline.The tension in the rope is 120 N, and the crate’s coefficient of kinetic friction on the

incline is 0.25.

(a) How much work is done by tension, by gravity, and by the normal force?

(b) What is the increase in thermal energy of the crate and incline?

Solution:

(a) Find work done by tension, gravity and the normal

WT � T �∆r � T∆x cos 18� � p120Nqp5.0mqpcos 18� � 570J (1)

Wg � Fg �∆r � mg∆x cos 120� � p6kgqp9.8m{s2qp5.0mqpcos 120� � �196J(2)

WN � 0J (3)

(b) Find the increase in thermal energy

∆Eth � fk �∆r � Fx∆x � µkFn∆x (4)

(5)

To find the normal force,¸Fy � 0 (6)

Fn � Fg cos 30� � T sin 18� � 0 (7)

Fn � Fg cos 30� � T sin 18� � p8.0kgqp9.8m{s2q cos 30� � p120Nqpsin 18�q(8)

Fn � 30.814N (9)

Plugging this result into the equation for the thermal energy, we find

∆Eth � µkFn∆x � p0.25qp30.814Nqp5.0mq � 38.5J (10)

(11)

25


15. The spring shown in the figure below is compressed 50 cm and used to launch a 100 kgphysics student. The track is frictionless until it starts up the incline. The student’s

coefficient of kinetic friction on the 30� incline is 0.15.

(a) What is the student’s speed just after losing contact with the spring?

(b) How far up the incline does the student go?

Solution:

(a) This is a conservation of energy problem. We want to know the initial velocityimmediately after the spring is no longer in contact with the student.

Kf � Ugf � Usf �∆Eth � Ki � Ugi � Usi �Wext (1)

1


2kp∆xf q2 � 0J � 1

2mpviq2 �mgyi � 1

2kp∆xiq2 � 0J(2)

26


1

2mpvf q2 � 1

2kp∆sq2 (3)

v �ck∆x

m� 14.14m{s (4)

(b) Using the conservation of energy equation again, we can how far the studenttravels up the incline. .Let’s define the y-height when the student reaches thehighest point as yf � p∆sq sin 30�.

Kf � Ugf � Usf �∆Eth � Ki � Ugi � Usi �Wext (5)

1


2kp∆xf q2 � 0J � 1

2mpviq2 �mgyi � 1

2kp∆xiq2 � 0J(6)

mgp∆sq sin 30� � µkFn∆s � mgyi � 1

2p∆xq2 (7)

(8)

From the sum of the forces in the y-direction on the incline, the normal force isFn � mg cos 30�. Plugging this in,

mgp∆sq sin 30� � µkmgcos30�∆s � mgyi � 1

2p∆xq2 (9)

∆sqmgpsin 30� � µk cos 30�q � mgyi � 1

2p∆xq2 (10)

∆s � mgyi � 12kp∆xq2

mgpsin 30� � µk cos 30�q (11)

∆s � 32.1m (12)

27


16. A 5.0 kg cat leaps from the floor to the top of a 95 cm high table. If the cat pushesagainst the floor for 0.20 s to accomplish this feat, what was her average power output

during the pushoff period?

Solution:

The average power output during the push-off period is equal to the work done by thecat divided by the time the cat applied the force. Since the force on the floor by thecat is equal in magnitude to the force on the cat by the floor, work done by the cat canbe found using the workkinetic-energy theorem during the push-off period: Wnet �Wfloor � ∆K. We do not need to explicitly calculate Wcat, since we know that thecat’s kinetic energy is transformed into its potential energy during the leap. In otherwords,

∆Ug � mgpy2 � y1q (1)

� p5.0 kgqp9.8 m{s2qp0.95 mq (2)

� 46.55 J. (3)

Thus, the average power output during the push-off period is

P � Wnet

t(4)

� 46.55 J

0.20 s(5)

� 0.23 kW. (6)

28


17. In a hydroelectric dam, water falls 25 m and then spins a turbine to generate electricity.

(a) What is ∆U of 1.0 kg of water?

(b) Suppose the dam is 80% efficient at converting the water’s potential energy toelectrical energy. How many kilograms of water must pass through the turbineseach second to generate 50 MW of electricity?

Solution:

(a) The change in the potential energy of 1.0 kg of water falling 25 m is

∆Ug � �mgh (1)

� �p1.0 kgqp9.8 m{s2qp25 mq (2)

� �250 J. (3)

(b) The dam is required to produce 50 MW � 50�106 W of power every second, or inother words 50�106 J of energy per second. If the dam is 80% efficient at convertingthe water’s potential energy into electrical energy, then every kilogram of water thatfalls produces 0.8�250 J � 200 J of electrical energy. Therefore, the total mass of wa-ter needed every second is

50� 106 J� 1.0 kg

200 J� 2.5� 105 kg. (4)

This is a typical value for a small hydroelectric dam.

29


18. (a) The 70 kg student in the figure below balances a 1200 kg elephant on a hydrauliclift. What is the diameter of the piston the student is standing on?

(b) When a second student joins the first, the piston sinks 35 cm. What is the secondstudent’s mass? Assume that the density of oil is 900 kg{m3.

Solution:

To solve this problem we need the hydraulic lift equation,

F2 � A2

A1

F1 � ρghA2. (1)

(a) The hydraulic lift is in equilibrium and the pistons on the left and the right areat the same level. Therefore, h � 0 and the hydraulic lift equation simplifies to

Fleft

Aleft

� Fright

Aright

(2)

ñ FG,student

πr2student

� FG,elephant

πr2elephant

(3)

ñ rstudent �dFG,student

FG,elephant

relephant (4)

�cmstudent

melephant

relephant (5)

�d

70 kg

1200 kg� p1.0 mq (6)

� 0.242 m (7)

� 24 cm. (8)

30


(b) With the second student added, the pistons are no longer at the same height.Therefore, the hydraulic lift equation becomes

Fright

Aright

� Fleft

Aleft

� ρgh. (9)

Noting that the force on the piston on the left is Fleft � pms�70 kgqg, we can solve forthe second student’s mass, ms:

melephantg

Aright

� pms � 70 kgqgAleft

� ρgh (10)

ñ ms � 70 kg

Aleft

� melephant

Aright

� ρh (11)

ms � 70 kg ��melephant

Aright

� ρh

Aleft (12)

ms ��melephant

Aright

� ρh

Aleft � 70 kg (13)

Substituting everything, we get:

ma ��

1200 kg

πp1.0 mq2 � p900 kg{m3qp0.35 mq�� πp0.242 mq2 � 70 kg

� 58 kg (14)

31


19. You need to determine the density of a ceramic statue. If you suspend it from a springscale, the scale reads 28.4 N. If you then lower the statue into a tub of water, so that it

is completely submerged, the scale reads 17.0 N. What is the statue’s density?

Solution:

The buoyant force, FB, on the can is given by Archimedes’ principle.

The free-body diagram sketched above shows that the gravitational force of thestatue, FG,statue is balanced by the spring force, Fsp, and the buoyant force, FB exertedby the water. Since the statue is in static equilibrium, we know that the net forceis zero. The rest of the solution follows through a series of substitutions, recalling thatthe density of water is ρw � 1000 kg m�3:

FB � Fsp � FG,statue � 0 (1)

ñ FB � FG,statue � Fsp (2)

ρwVstatueg � FG,statue � Fsp (3)

Vstatue � FG,statue � Fsp

ρwg(4)

mstatue

ρstatue

� FG,statue � Fsp

ρwg(5)

(6)

ñ ρstatue

mstatue

� ρwg

FG,statue � Fsp

(7)

ρstatue � ρwmstatueg

FG,statue � Fsp

(8)

ρstatue � ρwFG,statue

FG,statue � Fsp

(9)

� p1000 kg m�3q � p28.4 Nqp28.4 N� 17.0 Nq (10)

� 2490 kg m�3. (11)

32


20. A 355 mL soda can is 6.2 cm in diameter and has a mass of 20 g. Such a soda can halffull of water is floating upright in water. What length of the can is above the water

level?

Solution:

The buoyant force, FB, on the can is given by Archimedes’ principle.

Let the length of the can above the water level be d, the total length of the can be L,and the cross-sectional area of the can be A. The can is in static equilibrium, so fromthe free-body diagram sketched above we have¸

Fy � FB � FG,can � FG,water � 0 (1)

ρwaterApL� dqg � pmcan �mwaterqg (2)

L� d � pmcan �mwaterqAρwater

(3)

ñ d � L� pmcan �mwaterqAρwater

(4)

Recalling that one liter equals 10�3 m3 and the density of water is ρwater � 1000 kg m�3,the mass of the water in the can is

mwater � ρwater

�Vcan

2

(5)

� p1000 kg m�3q ��

355� 10�6 m3

2

(6)

� 0.1775 kg. (7)

To find the length of the can, L, we have:

Vcan � AL (8)

ñ L � 355� 10�6 m3

πp0.031 mq2 (9)

� 0.1176 m. (10)

33


Finally, substituting everything into equation (4), we get:

d � 0.1176 m� p0.020 kg � 0.1775 kgqπ � p0.031 mq2 � p1000 kg m�3q (11)

� 0.0522 m (12)

� 5.2 cm. (13)

34


21. Water flowing out of a 16 mm diameter faucet fills a 2.0 L bottle in 10 s. At whatdistance below the faucet has the water stream narrowed to 10 mm in diameter?

Solution:

We treat the water as an ideal fluid obeying Bernoulli’s equation. The pressureat point 1 is p1 and the pressure at point 1 is p2. Both p1 and p2 are atmo-spheric pressure. The velocity and the area at point 1 are v1 and A1, and atpoint 2 they are v2 and A2. Let d be the distance of point 2 below point 1.

First, we use the time it takes to fill a 2.0 L bottle to compute the flow rate, Q, the rateat which water is flowing out of the faucet, remembering that one liter equals 10�3 m3:

Q � p2.0 Lq � p10�3 m3{Lq10 s

(1)

� 2.0� 10�4 m3{s. (2)

Next, we use this result to find the velocity, v1, with which the water leaves the faucetthrough the D1 � 16 � 10�3 m diameter aperture (at point 1):

Q � v1A1 (3)

ñ v1 � Q

A1

(4)

� Q

πpD1{2q2 (5)

� 2.0� 10�4 m3{sπ � p8� 10�3 mq2 (6)

� 1.0 m{s. (7)

35


Now, from the continuity equation we have

v2A2 � v1A1 (8)

ñ v2 � v1A1

A2

(9)

� v1πpD1{2q2πpD2{2q2 (10)

� v1

�D1

D2

2

(11)

� 1.0 m{s�

16� 10�3 m

10� 10�3 m

2

(12)

� 2.56 m{s. (13)

Finally, we turn to Bernoulli’s equation to find the height, d:

p1 � 1

2ρv2

1 � ρgy1 � p2 � 1

2ρv2

2 � ρgy2 (14)

ρgpy1 � y2q � 1

2ρpv2

2 � v21q (15)

gd � 1

2pv2

2 � v21q (16)

d � pv22 � v2

1q2g

(17)

� p2.56 m{sq2 � p1.0 m{sq22� 9.8 m{s2 (18)

� 0.283 m (19)

� 28 cm. (20)

36


22. A hurricane wind blows across a 6.0 m �15.0 m flat roof at a speed of 130 km/hr.

(a) Is the air pressure above the roof higher or lower than the pressure inside the house?Explain.

(b) What is the pressure difference?

(c) How much force is exerted on the roof? If the roof cannot withstand this muchforce, will it “blow in” or “blow out”?

Solution:

(a) The pressure above the roof is lower due to the higher velocity of the air.(b) Bernoulli’s equation with yinside � youtside is

pinside � poutside � 1

2ρairv

2 (1)

ñ ∆p � 1

2ρairv

2 (2)

� 1

2� p1.28 kg{m4q �

�130� 1000 m

3600 s

2

(3)

� 835 Pa. (4)

The pressure difference is 0.83 kPa.(c) The force on the roof is

Froof � p∆pqA (5)

� p835 Paq � p6.0 m� 15.0 mq (6)

� 7.5� 104 N. (7)

The roof will blow up, because the pressure inside the house is greater than the pres-sure on the top of the roof.

37


Kinematics and Mechanics

xf � xi � vxit� 1

2axt

2

vxf � vxi � axt

v2xf � v2

xi � 2axpxf � xiq

yf � yi � vyit� 1

2ayt

2

vyf � vyi � ayt

v2yf � v2

yi � 2aypyf � yiq

θf � θi � ωi∆t� 1

2α∆t2

ωf � ωi � α∆t

ωf2 � ωi

2 � 2α∆θ

s � rθc � 2πr

vt � ωr

ar � v2

r� ω2r

v � 2πr

T

Forces

~Fnet � Σ~F � ma

~Fnet � Σ~Fr � ma � mv2

rFg � mg

0 fs � µsFN

fk � µkFN

~FAonB � �~FBonA

Momentum

~pi � ~pf

~p � m~v

pvfx q1 � m1 �m2

m1 �m2

pvix q1

pvfx q2 � 2m1

m1 �m2

pvix q1Energy

38


Kf � Ugf � Ki � Ugi

∆K �∆U �∆Eth � ∆Emech �∆Eth � ∆Esys � Wext

K � 1

2mv2

U � mgy

∆Eth � fk∆s

W � ~F � ~dP � ~F � ~v∆E � P∆t

Fluids and Thermal Energy

P � F

A

ρ � m

VQ � vA

v1A1 � v2A2

p1 � 1

2ρv2

1 � ρgy1 � p2 � 1

2ρv2

2 � ρgy2

Qnet � Q1 �Q2 � ... � 0

Q � MLf for freezing

Q � Mc∆T

Constants

g � 9.8m

s2

ρoil � 900kg

m3

ρwater � 1 , 000kg

m3

ci � 2090J

kgK

cw � 4190J

kgK

cAl � 900J

kgK

Lf � 333 , 000J

kgice to water

39

midterm exam 3 - siena collegemmccolgan/gp130f12/exams/midterm3_all_mm.pdfmidterm exam 3 november...

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