Physics 303K Test 3 Solutions

Version 023 – Test 3 ver2 – florin – (57850) 1

This print-out should have 20 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

A particular flywheel that rotates about itscenter of mass has a moment of inertia I =3

4MR2, where R = the radius.

What is the moment of inertia if the fly-wheel is rotated about a point on its rim?

1.3

4MR2

2.7

4MR2 correct

3.3

2MR2

4.5

4MR2

5. 3MR2

Explanation:

I = MR2 +3

4MR2 =

7

4MR2

002 10.0 points

A sphere of uniform density with mass 24 kg,and radius 0.7 m is spinning, making onecomplete revolution every 0.3 s. The center ofthe mass of the sphere has a speed of 5 m/s.What is the total kinetic energy of the sphere?1. 195.7172. 598.3113. 1331.74. 494.8455. 471.2686. 148.2177. 430.7848. 989.5689. 1476.8610. 1959.35

Correct answer: 1331.7.

Explanation:

The total kinetic energy is the sum of therotational and translational kinetic energy.

The rotational kinetic energy is given by:

Krot =1

2Iω2

In order to find ω we utilize the period of therotation, when was given as T = 0.3 s:

ω =2π

T

=2π

0.3 s= 20.944 rad/s

Next we need to find the moment of inertia Iof the sphere:

I =2

5mR2

=2

5(24 kg)(0.7 m)2

= 4.704 kg ·m2

Nowwe solve for the rotational kinetic energy:

Krot =1

2Iω2

=1

2(4.704 kg ·m2)(20.944 rad/s)2

= 1031.7 J

Our translational kinetic energy is given by:

Ktrans =1

2mv2

=1

2(24 kg)(5 m/s)2

= 300 J

Thus, the total kinetic energy is:

Ktot = Krot +Ktrans

= (1031.7 J) + (300 J)

= 1331.7 J

003 10.0 points

A familiar toy consists of an aligned row ofidentical steel balls that are suspended bymonofilament so they just touch. When theballs collide the collisions are very close toperfectly elastic. When two balls are lifted


from one end together and released, two ballspop out from the other end.If instead one ball popped out with twice

the speed of the initial set of two balls, whatlaw of physics would be violated?

1. Newton’s Third Law (Reciprocity)

2. The Momentum Principle (conservationof momentum)

3. Newton’s First Law

4. Both the Momentum and Energy Princi-ples

5. The Energy Principle correct

6. No physical law’s would be violated, it isa possible outcome.

7. Conservation of Mass

8. Newton’s First and Third Laws

Explanation:

The original KE of the system is

1

2(2m) v2 = mv2

(two balls). If one ball pops out the other endwith speed 2 v, its kinetic energy is

1

2m (2 v)2 = 2mv2 .

But the overall potential energy of the systemdoes not change and nothing external to thesystem is doing any work on it, so there is noway for the total KE of the system to double,or change in any way other than to decreaseslightly (since real collisions are not preciselyelastic). This (impossible) process violatesthe Energy Principle.

004 10.0 points

A particle oscillates up and down in simpleharmonic motion. Its height y as a functionof time t is shown in the diagram.

1 2 3 4 5

5

5

y(cm) t (s)

At what time t in the period shown doesthe particle achieve its maximum negativeacceleration?

1. t = 0 s

2. None of these; the acceleration is con-stant.

3. t = 5 s

4. t = 1 s

5. t = 2 s

6. t = 3 s correct

7. t = 4 s

Explanation:

This oscillation is described by

y(t) = − sin

(

π t

2

)

,

v(t) =d y

dt= −

π

2cos

(

π t

2

)

a(t) =d2 y

dt2

=(π

2

)2

sin

(

π t

2

)

.

The maximum negative acceleration will oc-

cur when sin

(

π t

2

)

= −1, or at t = 3 s .

From a non-calculus perspective, the veloc-ity is positive just before t = 3 s but decreas-ing since the particle is slowing down. Att = 3 s, the particle is momentarily at restand v = 0. Just after t = 3 s , the velocityis negative since the slope of the position ver-sus time plot has a negative slope. Remember

that a =∆v

∆t, acceleration is a negative max-


imum because the velocity is changing from apositive to a negative value.

005 10.0 points

The following graph represents a hypotheticalpotential energy curve for a particle of massm.

r

U(r)

O

3U0

2U0

U0

r0 2 r0

If the particle is released from rest at posi-tion r0, its speed ‖~v‖ at position 2 r0 is mostnearly

1. ‖~v‖ =

√

2U0

m.

2. ‖~v‖ =

√

U0

m.

3. ‖~v‖ =

√

U0

8m.

4. ‖~v‖ =

√

U0

4m.

5. ‖~v‖ =

√

8U0

m.

6. ‖~v‖ =

√

6U0

m.

7. ‖~v‖ =

√

U0

2m.

8. ‖~v‖ =

√

4U0

m. correct

9. ‖~v‖ =

√

U0

6m.

Explanation:

The total energy of the particle is con-served. So the change of the potential en-ergy is converted into the kinetic energy ofthe particle, which gives

1

2mv2 = 3U0 − U0

Therefore

v =

√

4U0

m.

006 10.0 points

The energy levels in hydrogen are given by

En =

(

−13.6 eV

n2

)

.

If the hydrogen atom goes from its third ex-

cited state to the ground state, what would bethe energy of the emitted photon?

1. 10.2 eV

2. 12.09 eV

3. Since energy is conserved, there would beno emitted photon.

4. −12.75 eV

5. None of the above.

6. 12.75 eV correct

7. −12.09 eV

8. Not enough information.

9. −10.2 eV

Explanation:

If the hydrogen atom goes from its thirdexcited state (n = 4) to its ground state (n =1), the energy difference (positive value) givesthe emitted photon energy:

Ephoton(4 → 1) =E4 −E1

=− 13.6

(

1

42−

1

1

)

eV

= 12.75 eV

007 10.0 points

A block of mass 0.5 kg is initially at a heightof 1.3 m above the ground and has a velocityv1 in the downward direction at time t = 0.


Resting on the ground directly beneath thisblock is a spring with natural length 1 m andspring constant 147 N/m. The block landson the spring and reaches a lowest height of0.8 m. Find the initial velocity v1. Theacceleration due to gravity is 9.8 m/s2.1. 1.42. 1.9493. 1.2494. 1.5365. 2.1076. 1.3427. 1.2818. 1.8659. 1.810. 2.341

Correct answer: 1.4 m/s.

Explanation:

Let : m = 0.5 kg ,

yi = 1.3m,

yf = 0.8 m ,

l = 1 m , and

ks = 147 N/m .

By the conservation of energy ∆E = 0.

∆K +∆U = 0

Kf −Ki + Uf − Ui = 0

Kf −Ki +Ufgrav − Uigrav + Ufsp − Uisp = 0

Ki =1

2mv2i

The final kinetic energy is zero. The changein the gravitational potential energy is

Ufgrav − Uigrav = mg(yf − yi)

The initial spring potential energy is zero sothe change is

Ufsp =1

2ks(l − yf )

2

vi =

√

2g(yf − yi) +k

m(l − yf )

2

008 10.0 points

A mass of 4m is moving with a velocityv1 and collides with a mass of 2m, whichis suspended by a string of length L. Thetwo masses stick together as the result of thecollision, and the compound system swings tothe right, passes point B as shown, and stopsat the horizontal level.

v14m B

6m

L

A

2m

θ

Find the initial speed v1 of the 4m mass.

1. v1 = 7/2√

2 g L

2. v1 = 9/4√

2 g L

3. v1 = 9/2√

2 g L

4. v1 = 7/5√

2 g L

5. v1 = 9/5√

2 g L

6. v1 = 5/2√

2 g L

7. v1 = 7/4√

2 g L

8. v1 = 3/2√

2 g L correct

9. v1 = 4/3√

2 g L

10. v1 = 5/3√

2 g L

Explanation:

First we need to find the kinetic energy Kof the compound system immediately afterthe collision. The acceleration of gravity is g .


Let : m1 = 4m and

m2 = 2m.

Momentum is conserved in the collision, so

pf = pi

(m1 +m2) v = m1 v1

(6m) v = (4m) v1

v =2

3v1 .

Immediately after the collision, the kineticenergy of the compound system is

K =1

2(m1 +m2) v

2

=1

2(6m)

(

2

3v1

)2

=4

3mv21 .

If we choose a system of the combinedmasses and the earth, there are no exter-nal forces or work done on the system so theEnergy Principle gives

Ei = Ef

Ui +Ki = Uf +Kf

0 +1

2(m1 +m2) v

2 = (m1 +m2) g L+ 0

1

2

(

2

3v1

)2

= g L

v1 =3

2

√

2 g L .

009 10.0 points

A body oscillates with simple harmonic mo-tion along the x-axis. Its displacement varieswith time according to the equation

x(t) = A sin(ω t+ φ) .

If A = 3 m, ω = 2.056 rad/s, and φ =1.0472 rad, what is the acceleration of thebody at t = 3 s? Note: The argument of thesine function is in radians rather than degrees.1. 26.64592. 2.401893. 15.0372

4. 20.65895. -10.18096. 7.531177. -65.68088. -41.29139. 4.1252510. 33.5888

Correct answer: −10.1809 m/s2.

Explanation:

Let : A = 3 m ,

ω = 2.056 rad/s ,

φ = 1.0472 rad , and

t = 3 s .

x = A sin(ω t+ φ)

v =d x

dt= ω A cos(ω t+ φ)

a =d v

dt= −ω2A sin(ω t+ φ)

= −ω2A sin(ω t+ φ)

= −(2.056 rad/s)2(3 m)

× sin[(2.056 rad/s)(3 s) + 1.0472 rad]

= −10.1809 m/s2 .

010 10.0 points

A(n) 3.2 kg object moving with a speed of6.5 m/s collides with a(n) 0.5 kg object mov-ing with a velocity of 8.3 m/s in a direction38.0653◦ from the initial direction of motionof the 3.2 kg object.

6.5 m/s

8.3m/s

0.5 kg

3.2 kg

38.0653◦

What is the speed of the two objects afterthe collision if they remain stuck together?1. 7.697482. 7.992793. 6.54134


4. 6.63325. 6.889466. 6.985697. 6.785768. 7.365749. 9.8863910. 6.3792


Explanation:

Let : m1 = 3.2 kg ,

m2 = 0.5 kg ,

mf = m1 +m2 = 3.7 kg ,

v1 = 6.5 m/s ,

v2 = 8.3 m/s ,

p1 = m1 v1 = 20.8 kgm/s ,

p2 = m2 v2 = 4.15 kgm/s ,

2 p1 p2 = 2m1 v1m2 v2

= 172.64 kg2m2/s2 ,

px = p1 + p2 cos θ

= 24.0673 kgm/s ,

py = p2 sin θ

= 2.55872 kgm/s ,

θ = 38.0653◦ , and

π − θ = 141.935◦ .

v1

v 2

m2

m1

θ

vfφ

mf

The final momentum is

pf = (m1 +m2) vf . (1)

Momentum is conserved

~p1 +~p2 = ~pf . (2)

Using the law of cosines, we have

p21 + p22 − 2 p1 p2 cos(π − θ) = p2f .

Solving for vf , we have

vf =

√

p21 + p22 − 2 p1 p2 cos(π − θ)

m1 +m2

=

[

1

(3.2 kg) + (0.5 kg)

]

×[

(432.64 kg2m2/s2)2

+ (17.2225 kg2m2/s2)2

− (172.64 kg2m2/s2) cos(141.935◦)]1/2

= 6.54134 m/s .

Alternate Solution: Since

p2f = p2x + p2y ,

we have

vf =

√

p2x + p2y

m1 +m2

=

[

1

(3.2 kg) + (0.5 kg)

]

×[

(579.236 kg2m2/s2)

+ (6.54705 kg2m2/s2)]1/2

= 6.54134 m/s .

011 10.0 points

The escape speed from a very small asteroidis only 23 m/s. If you throw a rock away fromthe asteroid at a speed of 49 m/s, what will

be its final speed? G = 6.7× 10−11 N ·m2

kg2.

1. 35.77712. 40.97563. 34.6414. 39.23015. 18.06. 36.20777. 43.26668. 28.26669. 45.254810. 26.8328



Explanation:

vesc =

√

2GM

R

v2esc =2GM

R

Use the Energy Principle.

Ei = Ef

Ui +Ki = Uf +Kf

−GMm

ri+

1

2mv2i = 0 +

1

2mv2f

Where Uf is zero.

v2f = v2i −2GM

R

v2f = v2i − v2esc

vf =√

v2i − v2esc

=√

(49 m/s)2 − (23 m/s)2

vf = 43.2666 m/s

012 10.0 points

A test car of mass 745 kg is moving at aspeed of 7.5 m/s when it crashes into a wallto test its bumper. If the car comes to rest in0.39 s, how much average power is expendedin the process?1. 49000.02. 55249.23. 51840.04. 64034.05. 57600.26. 53726.07. 58719.08. 38729.29. 45161.510. 50050.0

Correct answer: 53726 W.

Explanation:

The power expended is the energy lost perunit time. The energy that is lost in the

process of stopping the car is

−∆Ecar = −∆Kcar

=1

2mcar(vcar)

2 .

Thus, the power expended will be

P =−∆Ecar

t

=12mcar(vcar)

2

t

=12(745 kg)(7.5 m/s)2

0.39 s

= 53726 W .

013 10.0 points

A tennis ball of mass 0.1 kg is dropped froma very tall building and it falls straight down-ward. After a short time, it is found to befalling at constant velocity.If the ball falls 100 m at a constant velocity

of −̂ (50 m/s), how much work is done on theball by air friction over that distance? Theacceleration of gravity is 10 m/s2 .

1. −125 J.

2. −10 J.

3. −100 J. correct

4. 10 J.

5. 125 J.

6. 1000 J.

7. −1000 J.

8. 100 J.

9. −12.5 J.

10. 12.5 J.

Explanation:

Let : m = 0.1 kg ,

g = 10 m/s2 , and

h = 100 m .


The work done by air friction is

Wa = Ef −Ei .

Since K does not change, the work done byair friction will reduce to

Wa = Ef − Ei = Ugf − Ugi

= ∆Ug = −mg h

= −(0.1 kg) (10 m/s2) (100 m)

= −100 J .

014 10.0 points

This problem describes a method of deter-mining the moment of inertia of an irregu-larly shaped object such as the payload for asatellite. A mass m is suspended by a cordwound around the inner shaft (radius r) of aturntable supporting the object. When themass is released from rest, it descends uni-formly a distance h, acquiring a speed v.

m

Find moment of inertia of the equipment(including the turntable) in terms of magni-tudes of given variables.

1. I = mr2(

3 g h

2 v2− 2

)

2. I = mr2(

3 g h

v2− 1

)

3. I = mr2(

g h

v2− 1

)

4. I = mr2(

3 g h

2 v2− 1

)

5. I = mr2(

2 g h

v2− 1

)

correct

Explanation:

From conservation of energy,

∆Krot +∆Ktrans +∆U = 0

1

2I(v

r

)2

+1

2mv2 = mg h

Iv2

r2= 2mg h r2 −mr2 .

I = mr2(

2 g h

v2− 1

)

.

keywords:

015 10.0 points

In a certain time interval, natural gas with en-ergy content 10000 J was piped into a houseduring a winter day. In the same time in-terval sunshine coming through the windowsdelivered 3000 J of energy into the house.The temperature of the house didn’t change.What was ∆Ethermal of the house?

Ia. 10000 JIb. 0 JIc. 13000 JId. 3000 J

For the system of the house, what was Q,the energy transfer between the house and theair?

IIa. −3000 JIIb. 13000 JIIc. 3000 JIId. −13000 J

1. Id, IIb

2. Id, IIa

3. Ib, IId correct

4. Ib, IIb

5. Ia, IIa

6. Ia, IIc


7. Ic, IIc

8. Ic, IId

Explanation:

Since the temperature doesn’t change, theamount of thermal energy cannot change, so∆Ethermal = 0.Since we have no change in the thermal

energy, the energy that is being added by thenatural gas and the sunshine must transferout to the surrounding outside air. Hence,Q = −13000 J, where the sign is negativesince the energy flows from the system (thehouse) to the surroundings (the outside air).

016 10.0 points

A 85.8 kg man sits on the back end of a 4.6 mlong boat. The front of the boat touches thepier, but the boat isn’t tied. The man noticeshis mistake, stands up and walks to the boat’sfront, but by the time he reaches the front,it’s moved 2.38 m away from the pier.Assuming no water resistance to the boat’s

motion, calculate the boat’s mass (not count-ing the man).1. 64.93932. 29.38243. 129.2584. 80.03195. 45.24746. 21.12787. 81.0578. 35.59339. 327.35310. 57.3274

Correct answer: 80.0319 kg.

Explanation:

In the absence of external forces, the centerof mass of the man–boat system remains atrest. So if the man moves distance ∆Xman

and the boat moves distance ∆Xboat, then wemust have

∆XCM = ∆

(

MmanXman +MboatXboat

Mman +Mboat

)

=Mman∆Xman +Mboat∆Xboat

Mman +Mboat

= 0

and therefore

Mman∆Xman +Mboat∆Xboat = 0 .

Solving this equation for the boat’s mass, wefind

Mboat = Mman ×∆Xman

−∆Xboat

.

Now, let’s be careful about the displace-ments. Taking the back-to-front direction tobe positive, we have the boat moving back-ward, so

∆Xboat = −2.38 m < 0.

As to the man, his displacement relative to theboat is the boat’s full length (back to front),so

∆Xrel = +Lboat = +4.6 m,

but relative to the pier his displacement is only

∆Xman = ∆Xrel +∆Xboat

= +4.6 m− 2.38 m = +2.22 m .

Consequently,

Mboat = Mman ×∆Xman

−∆Xboat

= 85.8 kg×+2.22 m

+2.38 m= 80.0319 kg.

017 10.0 points

Starting from rest, a woman lifts a barbell ofmass mbb with a constant force F through adistance h, at which point she is still lifting,and the barbell has acquired a speed v. LetEwoman stand for the following energy termsassociated with the woman:

Ewoman = Echemical,woman

+Kwoman

+ Ugrav,woman+Earth

+ Ethermal,woman

The change in the kinetic energy of thebarbell is

1

2mbbv

2 − 0 =1

2mbbv

2.


The general statement of the energy principleis:

∆Esys = Wsurr

For which of the following systems will theleft hand side of this equation have ONLY the

terms +mbbgh and1

2mbbv

2?

1. barbell only

2. woman + Earth

3. woman + barbell

4. woman + barbell + Earth

5. barbell + Earth correct

6. Earth only

7. there is no such system

8. woman only

Explanation:

We may first note that +mbbgh on the lefthand side of the energy principle equationrepresents a potential energy change, whichonly multibody systems can exhibit. Thusthe single body answer choices cannot be cor-rect. Of the multibody choices, any contain-ing “woman”must have the term∆Ewoman onthe left side of the energy principle. The onlyremaining choice is barbell + Earth. This iscorrect because the change in potential energyof this system is +mbbgh due to the barbellrising, and the change in kinetic energy of this

system is1

2mbbv

2 due to the barbell gaining

speed v.

018 10.0 points

Calculate the moment of inertia for the sys-tem. The rods of length L are massless.

b b

bb

b

Axis

4m 3m

m 2m

L

L L

LL

1. 30mL2

2. 32mL2 correct

3. 52mL2

4. 16mL2

5. 41mL2

Explanation:

The moment of inertia is

I = mL2 + 2m (2L)2 + 3m[

(2L)2 + L2]

+4m (L2 + L2)

= mL2 + 8mL2 + 15mL2 + 8mL2

= (1 + 8 + 15 + 8)mL2

= 32mL2 .

019 10.0 points

Consider a diatomic molecule bound by inter-atomic forces. The figure below shows all ofthe quantized energies (bound states) for oneof these molecules. The energy for each stateis given on the graph, in electron volts (1 eV= 1.6× 10−19 J). What is the minimum en-ergy required to break a molecule apart, if itis initially in the ground state? (Note thatthe final state must be an unbound state; theunbound states are not quantized.)

r

Energy −0.16 eV

−0.4 eV

−0.88 eV

−1.84 eV


1. 1.842. 1.613. 2.074. 2.35. 3.686. 3.917. 3.228. 4.149. 2.7610. 2.53

Correct answer: 1.84 eV.

Explanation:

To break the molecule apart, we must putin enough energy to compensate for the depthof the ground state:

r

Energy −0.16 eV

−0.4 eV

−0.88 eV

−1.84 eV

So we must put in 1.84 eV.

020 10.0 points

Which of the following choices correspondsto a system of two electrons that start outfar apart, moving toward each other (that is,their initial velocities are nonzero and theyare heading straight at each other)? Notethat the horizontal and vertical axes in eachplot are the separation between the particlesand energy, respectively.

K + U

U

K(I)

r

K + U

U

K(II)

r

K + U

U

K(III)

r

K + UU

K

(IV)

r

K + U

U

K(V)

rK + U

U

K

(VI)

r

1. Figure II

2. Figure V

3. Figure III

4. Figure I correct

5. Figure VI

6. Figure IV

Explanation:

When the two electrons are very far awaytheir potential energy is 0, and since theyhave nonzero initial velocities, this means thatthey are unbounded and thus have an overallpositive energy at r = ∞, which is also equalto the kinetic energy at that location. Asthe electrons get closer, due to their Coulombrepulsion their kinetic energies drop to 0 whilethe potential energy rises. Thus the correctanswer is Figure (I).

Physics 303K Test 3 Solutions

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