Midterm Exam 3 - Siena Sciencemmccolgan/GP130F12/Exams/midterm3_all.pdf · Midterm Exam 3 November...

Physics 130 General Physics Fall 2012

Midterm Exam 3

November 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. One or more free-body or force-interaction diagrams, as appropriate,

with labeled 1D or 2D coordinate axes.5. Algebraic expression for the net force along each dimension, as appro-

priate.6. Algebraic expression for the conservation of energy or momentum equa-

tions, as appropriate.7. An algebraic solution of the unknown variables in terms of the

known variables.8. A final numerical solution, including units, with a box around it.9. 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 m/s when a 60 kg skydiver drops out by releasing his grip on theglider. Using the appropriate conservation law, determine the glider’s velocity just afterthe 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 skydiver’s motion in the vertical direction has no influence on the glider’s hori-zontal motion.

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:

This is an inelastic two-dimensional conservation of momentum problem. Applyingconservation of momentum in the x-direction, the final momentum in the x-directionis

pfx “ pix “ m1pvixq1 ` m2pvixq2 (1)

“ p0.020 kgqp2.0 m{sq ´ p0.030 kgqp1.0 m{sq cos 30˝ (2)

“ 0.0140 kgm{s (3)

Applying conservation of momentum in the y-direction, the final momentum in they-direction is

pfy “ piy “ m1pv ´ iyq1 ` m2pviyq2 (4)

“ p0.020 kgqp0 m{sq ´ p0.030 kgqp1.0 m{sq sin 30˝ (5)

“ ´0.0150 kg m{s (6)

Next, we find the magnitude of the final momentum by combining the x- and y-components using the Pythagorean theorem:

pf “a

pfx ` pfy (7)

“a

p0.0140 kg m{sq2 ` p´0.0150 kg m{sq2 (8)

“ 0.0205 kg m{s (9)

3


Using the final momentum, we can solve for the final speed:

pf “ pm1 ` m2qvf (10)

ñ vf “ pf

m1 ` m2

(11)

“ 0.0205 kg m{s0.020 kg ` 0.030 kg

(12)

“ 0.41 m{s (13)

(14)

The direction is given by

θ “ tan´1

ˆ

pfy

pfx

˙

(15)

“ tan´1

ˆ´0.0150 kg m{s0.0140 kg m{s

˙

(16)

“ ´47˝ (17)

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 ˆ 1013 kg hits the earth with an impact speed of 4.0 ˆ 104 m{s.(a) What is the earth’s recoil speed after such a collision? (Hint: Use a reference frame

in which the earth is initially at rest.)

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

Solution:

(a) This is an inelastic collision, and therefore a conservation of momentum problem.At the start, the asteroid has a speed of 4.0 ˆ 104 m{s while the earth is notmoving.

pix “ pfx (1)

Mearthpviqe ` mapviqa “ pMearth ` maqpvf q (2)

(3)

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

vf “ Mearthpviqe ` mapviqapMearth ` maq (4)

5


“ 0 kg m{s ` p1.0 ˆ 1013 kgqp4.0 ˆ 104 m{sqp5.98 ˆ 1024 kg ` 1.0 ˆ 1013 kgq (5)

“ 6.7 ˆ 10´8 m{s (6)

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

vE “ 2πrearth´sun

T“ 2πp1.50 ˆ 1011 mq

3.15 ˆ 107 s“ 3.0 ˆ 104 m{s (7)

(8)

Therefore,

vf{vE “ p6.7 ˆ 10´8 m{sq{p3.0 ˆ 104 m{sq (9)

“ 2.2 ˆ 10´12 (10)

“ 2.2 ˆ 10´10 %. (11)

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. Because momentum is conserved, we have

pix “ pfx (1)

mBpviqB ´ mF pviqF “ pmB ` mF qpvf q (2)

Solving for the final velocity of Brutus and Fred,

vf “ mBpviqB ´ mF pviqFpmB ` mF q (3)

“ p120 kgqp4.0 m{sq ´ p60 kgqp6.0 m{sqp120 kg ` 60 kgq (4)

“ 0.667 m{s (5)

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, and thereforethe force of friction must act in the negative x-direction. The net force is equal tothe total mass (the mass of both players) times the acceleration.

ÿ

Fx “ ´fk “ pmB ` mF qa (6)

´µn “ pmB ` mF qa (7)

´µpmB ` mF qg “ pmB ` mF qa (8)

ñ a “ ´µg (9)

“ ´p0.30q ˆ p9.8 m{s2q (10)

“ ´2.94 m{s2 (11)

7


Finally, we can use the kinematic equation to find the distance traveled with vf “ 0:

v2f “ v2i ` 2ad (12)

ñ d “ ´ v2i2a

(13)

“ ´p0.667 m{s2q22 ˆ p´2.94 m{s2q (14)

“ 0.076 m (15)

“ 7.6 cm (16)

8


5. 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)

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

v1f “ ´35 m{s ` v2f (6)

4vi “ 3p´35 m{s ` v2f q ` v2f (7)

4vi “ ´105 m{s ` 4v2f (8)

Solving for the velocity of the second stage v2f

v2f “ 4vi ` 105 m{s4

(9)

9


“ 4 ˆ 1200 m{s ` 105 m{s4

(10)

“ 1.2 km{s (11)

10


6. 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 rubber ball 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

11


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 ´ mevef

mb

(10)

vef “ 2mbvbi

me ` 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 ˆ 1024kg

“ 2.3 ˆ 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 “ xf

vi“ 0.001m

2.3 ˆ 10´24m{s “ 4.3 ˆ 1020 sec “ 1.36 ˆ 1013 years (14)

12


7. 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)

13


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)

(18)

(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 .

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 (19)

(20)

14


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 (21)

1

2mv2 ` 0 “ 0 ` mgh (22)

h “ v2

2g“ p´2.56m{sq2

2p9.8m{s2q “ 0.33 m (23)

15


8. 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)

where the minus sign means that the centripetal acceleration points inward,toward the center of the circular track.

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

ÿ

F “ ma “ ´mv2

R(2)

FN ´ Fg “ ´mv2

R(3)

When the normal force is greater than zero, the car is still touching the track.Therefore, by setting the normal force equal to zero we can find the maximumspeed that the car can have and still stay on the track.

Fg “ mv2max

R(4)

16


mg “ mv2max

R(5)

ñ vmax “a

gR (6)

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

Kf ` Uf ` Ki ` Ui (7)

1

2mv2max ` mgyf “ 1

2mv2i ` mgyi (8)

1

2mv2max ` mgyf “ 0 ` mgyi (9)

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

1

2v2max ` gR “ 0 ` gyi (10)

yi “ v2max

2g` R (11)

We can now plug in our results for vmax from above

yi “ p?gRq22g

` R (12)

“ R

2` R (13)

“ 3R

2(14)

(b) Plugging into the above equation, hmax for a roller coaster that has R “ 10 mwe get

yi “ 3R

2“ 3 ˆ 10 m

2“ 15 m (15)

17


9. 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

2mpvf q2 ` mgyf “ 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

ÿ

F “ ma “ ´mv2

R(6)

FN ´ Fg cosφ “ ´mv2

R(7)

18


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 “a

gR 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´1p23

q Ñ φ “ 48˝ (14)

19


10. 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)

20


11. 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 combi-nation of a slight upward slope and a large coefficient of rolling resistance as the trucktires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopesupward at 6.0˝ and the coefficient of rolling friction is 0.40. Use work and energy to findthe 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)

The thermal energy created by friction is

∆Eth “ fkp∆xq (3)

“ pµkFNqp∆xq (4)

“ pµkmg cos θqp∆xq (5)

“ pµkmg cos θqp∆xq (6)

Using geometry, the final gravitationl potential energy of the truck is

Uf “ mgyf (7)

“ mgp∆xq sin θ (8)

Finally, the initial kinetic energy is simply

Ki “ 1

2mv2i (9)

21


Plugging everything into the conservation of energy equation, we get

mgp∆xq sin θ ` µkmg cos θp∆xq “ 1

2mv2i (10)

g∆xpsin θ ` µk cos θq “ 1

2v2i (11)

ñ ∆x “ v2i2gpsin θ ` µk cos θq (12)

“ p35 m{sq22 ˆ p9.8 m{s2q ˆ psin 6˝ ` 0.4 ˆ cos 6˝q (13)

“ 124 m (14)

“ 0.12 km. (15)

22


12. 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 theincline 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 cosp18˝q (1)

“ p120 Nq ˆ p5.0 mq ˆ pcos 18˝q (2)

“ 570J (3)

Wg “ Fg ¨ ∆r “ mg∆x cos 120˝ (4)

“ p6 kgq ˆ p9.8 m{s2q ˆ p5.0 mq ˆ pcos 120˝q (5)

“ ´196 J (6)

WN “ 0 J (7)

(b) Find the increase in thermal energy

∆Eth “ fk ¨ ∆r “ Fx∆x “ µkFn∆x (8)

(9)

To find the normal force,

ÿ

Fy “ 0 (10)

Fn ´ Fg cos 30˝ ` T sin 18˝ “ 0 (11)

ñ Fn “ Fg cosp30˝q ´ T sin 18˝ (12)

“ p8.0 kgq ˆ p9.8 m{s2q ˆ pcos 30˝q ´ p120 Nq ˆ psin 18˝q (13)

“ 30.814 N (14)

23


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

∆Eth “ µkFn∆x (15)

“ p0.25q ˆ p30.814 Nq ˆ p5.0 mq (16)

“ 38.5 J (17)

(18)

24


13. 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 outputduring the pushoff period?

Solution:

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

∆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)

25


14. 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 ˆ 106 W of power every second,or in other words 50 ˆ 106 J of energy per second. If the dam is 80% efficient atconverting the water’s potential energy into electrical energy, then every kilogramof water that falls produces 0.8 ˆ 250 J “ 200 J of electrical energy. Therefore, thetotal mass of water needed every second is

50 ˆ 106 J ˆ 1.0 kg

200 J“ 2.5 ˆ 105 kg. (4)

This is a typical value for a small hydroelectric dam.

26


15. 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 itis 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

exerted by the water. Since the statue is in static equilibrium, we know that thenet force is zero. The rest of the solution follows through a series of substitutions,recalling that the 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 ˆ p28.4 Nqp28.4 N ´ 17.0 Nq (10)

“ 2490 kg m´3. (11)

27


16. 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 waterlevel?

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 beL, and the cross-sectional area of the can be A. The can is in static equilibrium, sofrom the free-body diagram sketched above we have

ÿ

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

ˆ

Vcan

2

˙

(5)

“ p1000 kg m´3q ˆˆ

355 ˆ 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 ˆ 10´6 m3

πp0.031 mq2 (9)

“ 0.1176 m. (10)

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

d “ 0.1176 m ´ p0.020 kg ` 0.1775 kgqπ ˆ p0.031 mq2 ˆ p1000 kg m´3q (11)

“ 0.0522 m (12)

“ 5.2 cm. (13)

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17. 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 pressure atpoint 1 is p1 and the pressure at point 1 is p2. Both p1 and p2 are atmosphericpressure. The velocity and the area at point 1 are v1 and A1, and at point 2 theyare 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, therate at which water is flowing out of the faucet, remembering that one liter equals10´3 m3:

Q “ p2.0 Lq ˆ p10´3 m3{Lq10 s

(1)

“ 2.0 ˆ 10´4 m3{s. (2)

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

Q “ v1A1 (3)

ñ v1 “ Q

A1

(4)

“ Q

πpD1{2q2 (5)

“ 2.0 ˆ 10´4 m3{sπ ˆ p8 ˆ 10´3 mq2 (6)

“ 1.0 m{s. (7)

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Now, from the continuity equation we have

v2A2 “ v1A1 (8)

ñ v2 “ v1A1

A2

(9)

“ v1πpD1{2q2πpD2{2q2 (10)

“ v1

ˆ

D1

D2

˙2

(11)

“ 1.0 m{sˆ

16 ˆ 10´3 m

10 ˆ 10´3 m

˙2

(12)

“ 2.56 m{s. (13)

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

p1 ` 1

2ρv21 ` ρgy1 “ p2 ` 1

2ρv22 ` ρgy2 (14)

ρgpy1 ´ y2q “ 1

2ρpv22 ´ v21q (15)

gd “ 1

2pv22 ´ v21q (16)

d “ pv22 ´ v21q2g

(17)

“ p2.56 m{sq2 ´ p1.0 m{sq22 ˆ 9.8 m{s2 (18)

“ 0.283 m (19)

« 28 cm. (20)

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18. A hurricane wind blows across a 6.0 m ˆ15.0 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

2ˆ p1.28 kg{m3q ˆ

ˆ

130 km

hrˆ 1000 m

kmˆ 1 hr

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 ˆ p6.0 m ˆ 15.0 mq (6)

“ 7.5 ˆ 104 N. (7)

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

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19. A gas is compressed from 600 cm´3 to 200 cm´3 at a constant pressure of 400 kPa. Atthe same time, 100 J of heat energy is transferred out of the gas. What is the change inthermal energy of the gas during this process?

Solution:

This is a first law of thermodynamics problem involving an isobaric (constant pres-sure) process. We know that W ą 0 because the gas is compressed, which transfersenergy into the system. Also, 100 J of heat is transferred out of the gas. The firstlaw of thermodynamics gives

∆Eth “ W ` Q (1)

“ ´p∆V ` Q (2)

“ ´p4.0 ˆ 105 Paq ˆ p200 ´ 600q ˆ 10´6 m´3 ´ 100 J (3)

“ 60 J. (4)

Therefore, the thermal energy increases by 60 J.

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20. A 100 g ice cube at ´10˝ C is placed in an aluminum cup whose initial temperature is70˝ C. The system comes to an equilibrium temperature of 20˝ C. What is the mass ofthe cup?

Solution:

There are two interacting systems: aluminum and ice. The system comes to thermalequilibrium in four steps: (1) the ice temperature increases from ´10 ˝C to ´0 ˝C;(2) the ice becomes water at 0 ˝C; (3) the water temperature increases from 0 ˝C to20 ˝C; and (4) the cup temperature decreases from 70 ˝C to 20 ˝C.

Since the aluminum and ice form a closed system, we have

Q “ Q1 ` Q2 ` Q3 ` Q4 “ 0. (1)

Each term is as follows:

Q1 “ Mice cice ∆T (2)

“ p0.100 kgq ˆ r2090 J{pkg Kqs ˆ p10 Kq“ 2090 J.

Q2 “ Mice Lf (3)

“ p0.100 kgq ˆ p3.33 ˆ 105 J{kgq“ 33, 300 J.

Q3 “ Mice cwater∆T (4)

“ p0.100 kgq ˆ r4190 J{pkg Kqs ˆ p20 Kq“ 8380 J.

Q4 “ MAl cAl∆T (5)

“ MAl ˆ r900 J{pkg Kqs ˆ p´50 Kq“ ´p45, 000 J{kgqMAl.

Inserting everything into equation (1), we get

43, 770 J ´ p45, 000 J{kgqMAl “ 0

ñ MAl “ 0.97 kg. (6)

34


Kinematics and Mechanics

xf “ xi ` vxit ` 1

2axt

2

vxf “ vxi ` axt

v2xf “ v2xi ` 2axpxf ´ xiq

yf “ yi ` vyit ` 1

2ayt

2

vyf “ vyi ` ayt

v2yf “ v2yi ` 2aypyf ´ yiq

θf “ θi ` ωi∆t ` 1

2α∆t2

ωf “ ωi ` α∆t

ωf2 “ ωi

2 ` 2α∆θ

s “ rθ

c “ 2πr

vt “ ωr

ar “ v2

r“ ω2r

v “ 2πr

T

Forces

~Fnet “ Σ~F “ ma

~Fnet “ Σ~Fr “ ma “ mv2

r

Fg “ mg

0 ă fs ă“ 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 q1

Energy

Kf ` Ugf “ Ki ` Ugi

∆K ` ∆U ` ∆Eth “ ∆Emech ` ∆Eth “ ∆Esys “ Wext

Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext

K “ 1

2mv2

U “ mgy

∆Eth “ fk∆s

W “ ~F ¨ ~dP “ ~F ¨ ~vP “ ∆E{∆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/melting

Q “ Mc∆T

Constants

g “ 9.8 m{s2

Mearth “ 5.98 ˆ 1024 kg

rearth´sun “ 1.50 ˆ 1011 m

ρwater “ 1000 kg{m3

ρair “ 1 .28 kg{m3

cice “ 2090J

kg K

cwater “ 4190J

kg K

cAl “ 900J

kg K

Lf “ 333 , 000J

kgice to water

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Midterm Exam 3 - Siena Sciencemmccolgan/GP130F12/Exams/midterm3_all.pdf · Midterm Exam 3 November...

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