midterm exam 3 - siena collegemmccolgan/gp130f12/exams/... · physics 130 general physics fall 2012...

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. 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.

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1. Fred (mass 70 kg) is running with the football at a speed of 7.0 m{s when he is methead-on by Brutus (mass 130 kg), who is moving at 5.0 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 0.25.

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 pviqF

pmB �mF q(3)

�p130 kgqp5.0 m{sq � p70 kgqp7.0 m{sq

p130 kg � 70 kgq(4)

� 0.80 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.

¸Fx � �fk � pmB �mF qa (6)

�µn � pmB �mF qa (7)

�µpmB �mF qg � pmB �mF qa (8)

ñ a � �µg (9)

� �p0.25q � p9.8 m{s2q (10)

� �2.45 m{s2 (11)

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

v2f � v2i � 2ad (12)

ñ d � �v2i2a

(13)

��p0.80 m{s2q2

2 � p�2.45 m{s2q(14)

� 0.13 m (15)

� 13 cm (16)

3


2. A two-stage rocket is traveling at 1100 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 25 m{s relative to the second stage. The first stage is four 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)

p4m2 �m2qvi � 4m2v1f �m2v2f � m2p4v1f � v2f q (3)

5m2vi � m2p4v1f � v2f q (4)

5vi � 4v1f � v2f (5)

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

v1f � �25 m{s � v2f (6)

5vi � 4p�25 m{s � v2f q � v2f (7)

5vi � �100 m{s � 5v2f (8)

Solving for the velocity of the second stage v2f

v2f �5vi � 100 m{s

5(9)

4


�5 � 1100 m{s � 100 m{s

5(10)

� 1.1 km{s (11)

5


3. 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)

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Cancelling the m’s

v1i � 3vf (7)

Solving for vf , we find

vf �v1i3�

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

2m1pv1iq

2 �1

2m2pv2iq

2 �1

2m1pv1f q

2 �1

2m2pv2f q

2 (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)

7


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)

8


4. Bob can throw a 700 g rock with a speed of 32 m/s. He moves his hand forward 0.8 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 q

2 �1

2mpviq

2 (2)

W �1

2mpvf q

2 � 0 (3)

W �1

20.7kgp32m{sq2 (4)

W � 358 J (5)

(b) Calculate force

W � F � ∆r � F∆r (6)

F �W

∆r�

358 J

.8m(7)

F � 448 N (8)

(c) Calculate power

P � F � v � Fv (9)

P � 448 Np32m{sq (10)

P � 14, 336 W (11)

9


5. 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 8.0� and the coefficient of rolling friction is 0.42. Use work and energy to findthe length of a ramp that will stop a 16, 000 kg truck that enters the ramp at 30 m/s(� 70 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)

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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 �v2i

2gpsin θ � µk cos θq(12)

�p30 m{sq2

2 � p9.8 m{s2q � psin 8� � 0.42 � cos 8�q(13)

� 82.7 m (14)

� 0.083 km. (15)

11


6. In a hydroelectric dam, water falls 28 m and then spins a turbine to generate electricity.

(a) What is ∆U of 1.0 kg of water?

(b) Suppose the dam is 78% efficient at converting the water’s potential energy toelectrical energy. How many kilograms of water must pass through the turbineseach second to generate 60 MW of electricity?

Solution:

(a) The change in the potential energy of 1.0 kg of water falling 28 m is

∆Ug � �mgh (1)

� �p1.0 kgqp9.8 m{s2qp28 mq (2)

� �274 J. (3)

(b) The dam is required to produce 60 MW � 50 � 106 W of power every second,or in other words 60 � 106 J of energy per second. If the dam is 78% efficient atconverting the water’s potential energy into electrical energy, then every kilogram ofwater that falls produces 0.78 � 274 J � 214 J of electrical energy. Therefore, thetotal mass of water needed every second is

60 � 106 J �1.0 kg

214 J� 2.5 � 105 kg. (4)

This is a typical value for a small hydroelectric dam.

12


7. Water flowing out of a 15 mm diameter faucet fills a 1.0 L bottle in 8 s. At what distancebelow the faucet has the water stream narrowed to 12 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 1.0 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 �p1.0 Lq � p10�3 m3{Lq

8 s(1)

� 1.25 � 10�4 m3{s. (2)

Next, we use this result to find the velocity, v1, with which the water leaves thefaucet through the D1 � 15 � 10�3 m diameter aperture (at point 1):

Q � v1A1 (3)

ñ v1 �Q

A1

(4)

�Q

πpD1{2q2(5)

�1.25 � 10�4 m3{s

π � p7.5 � 10�3 mq2(6)

� 0.707 m{s. (7)

13


Now, from the continuity equation we have

v2A2 � v1A1 (8)

ñ v2 � v1A1

A2

(9)

� v1πpD1{2q

2

πpD2{2q2(10)

� v1

�D1

D2

2

(11)

� 0.707 m{s

�15 � 10�3 m

12 � 10�3 m

2

(12)

� 1.11 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 � v21q

2g(17)

�p1.11 m{sq2 � p0.707 m{sq2

2 � 9.8 m{s2(18)

� 0.037 m (19)

� 3.7 cm. (20)

14


8. A 120 g ice cube at �12� C is placed in an aluminum cup whose initial temperature is65� C. The system comes to an equilibrium temperature of 15� 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 �12 �C to �0 �C;(2) the ice becomes water at 0 �C; (3) the water temperature increases from 0 �C to15 �C; and (4) the cup temperature decreases from 65 �C to 15 �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)

� p0.120 kgq � r2090 J{pkg Kqs � p12 Kq

� 3010 J.

Q2 � Mice Lf (3)

� p0.120 kgq � p3.33 � 105 J{kgq

� 40, 000 J.

Q3 � Mice cwater ∆T (4)

� p0.120 kgq � r4190 J{pkg Kqs � p15 Kq

� 7540 J.

Q4 � MAl cAl ∆T (5)

� MAl � r900 J{pkg Kqs � p�50 Kq

� �p45, 000 J{kgqMAl.

Inserting everything into equation (1), we get

50, 550 J � p45, 000 J{kgqMAl � 0

ñMAl � 1.1 kg. (6)

15


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 � rθc � 2πr

vt � ωr

ar �v2

r� ω2r

v �2πr

T

Forces

~Fnet � Σ~F � ma

~Fnet � Σ~Fr � ma � mv2

r

Fg � 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 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 � ~d

P � ~F � ~v

P � ∆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 � 1024 kgrearth�sun � 1.50 � 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 collegemmccolgan/gp130f12/exams/... · physics 130 general physics fall 2012...

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