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This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points Two pucks lie on ice and can slide with little friction. A string is attached to each puck and the string is pulled with a constant force F. The string is wound around the outer edge of puck 1 but attached to the center of puck 2. They both start from rest.
F
F
What happens in the next few seconds?
- 1 will move farther than 2
- 1 and 2 will move the same distance cor- rect
- More information is needed
- 2 will move farther than 1
Explanation: The momentum principle says that ∆p = Fnet∆t, so applying this to each block, we can see the right-hand sides will be equal. Therefore, the momentum change is the same in both cases, and the centers of mass move the same distance.
002 10.0 points The graph of potential energy versus inter- atomic distance for a particular molecule shows all of the quantized energies (bound states) for one of these molecules. The en- ergy for each state is given on the graph, in electron volts (1 eV = 1. 6 × 10 −^19 J).
r
Energy
− 0 .128 eV − 0 .32 eV − 0 .704 eV
− 1 .472 eV
What energy photon must be absorbed by the molecule to be promoted to the second excited state if it is initially in the ground state?
- − 1 .472 eV
- 1 .472 eV
- 1 .344 eV
- 0 .768 eV
- 1 .152 eV correct
- 0 .32 eV
Explanation: To promote the molecule to the second ex- cited state, we must put in enough energy to compensate for the difference between that state and the ground state:
r
Energy
− 0 .128 eV − 0 .32 eV − 0 .704 eV
− 1 .472 eV
So we must put in 1.152 eV.
003 10.0 points Consider a microscopic spring-mass system whose spring stiffness is 52 N/m, and the mass is 7 × 10 −^26 kg. What is the smallest amount of vibrational
energy that can be added to this system? Use the values
¯h = 6. 582 × 10 −^16 eV · s 1 eV = 1. 602 × 10 −^19 J c = 2. 998 × 108 m/s.
Correct answer: 0.0179395 eV.
Explanation:
ω 0 =
k m
=
52 N/m 7 × 10 −^26 kg = 2. 72554 × 1013 rad/s. For a quantum oscillator, the difference in successive energy levels is
∆E = ¯h ω 0 = (6. 582 × 10 −^16 eV · s) × (2. 72554 × 1013 rad/s) = 0 .0179395 eV. Note that this is in the IR range of the electromagnetic spectrum.
004 10.0 points A beam of high-energy π−^ (negative pions) is shot at a flask of liquid hydrogen, and some- times a pion interacts through the strong in- teraction with a proton in the hydrogen, in the reaction
π−^ + p+^ → π−^ + X+,
where X+^ is a positively charged particle of unknown mass.
Before After
π−^ p+
π−
X+
θ
The incoming pion momentum is 2920 MeV/c. The pion is scattered through 43 ◦, and its momentum is measured to be 1510 MeV/c. The X+^ particle is scattered through an unknown angle θ with an un- known momentum. A pion has a rest energy of 140 MeV, and a proton has a rest energy of 938 MeV. Find the scattering angle θ of the X+^ par- ticle.
Correct answer: 29.5614 Degrees.
Explanation: We conserve momentum in the x and y directions:
pπ,i =pπ,f cos 43◦^ + pX cos θ 0 =pπ,f sin 43◦^ − pX sin θ
We rearrange the second expression to solve for pX , and then plug it into the first expres- sion and solve for θ.
pπ,i = pπ,f cos 43◦^ + pX cos θ
pπ,i = pπ,f cos 43◦^ + pπ,f
sin 43◦ tan θ
than an object with a smaller β. The hollow sphere’s mass is distributed farther from its axis of rotation than a solid sphere’s, so the hollow sphere has a greater β. Therefore, the hollow sphere rolls down the wooden plank more slowly. The other items cannot mea- sure the distribution of mass, which is what is required.
007 10.0 points A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass m = 0.6 kg, at the ends of a massless rod of length d = 0.4 m. The barbell spins clockwise on the plane of the paper with an angular speed ω 0 = 18 rad/s (choice of axes: x to the right, y up, z out of the page; origin at A).
b
b
m
m
A
d
Calculate the z component of the total an- gular momentum for this system of two parti- cles.
- -1.
- -0.
- -3.
- -1.
- -0.
- -0.
- -1.
- -1.
- -0.
- -2.
Correct answer: − 0 .864 kg · m^2 /s.
Explanation: First we must find the angular momentum of the individual masses. The magnitude of the angular momentum of mass 1 is given by
∣~L 1
∣ = |~r × ~p| = |~r⊥| |~p|.
Here, |~r⊥| =
d 2
and |~p| = m
d 2
ω 0. Putting these together, we have
∣ ∣ ∣~L 1
∣ =^ m
d 2
ω 0
= (0.6 kg)
0 .4 m 2
(18 rad/s)
= 0 .432 kg · m^2 /s. We do the exactly same thing for the second mass. Finally, we just add the two magnitudes we found in the first two parts.
~Ltot = ~L 1 + ~L 2
= 〈 0 , 0 , − 2 m
d 2
ω 0 〉
m d^2 ω 0 2
= 〈 0 , 0 , − 0 .864 kg · m^2 /s〉.
008 10.0 points Consider a particle at a location 〈 0 , 4 , 2 〉 m relative to a point A. Its momentum is 〈 0 , 0 , 9 〉 kg · m/s. What the x component of the particle’s angular momentum relative to A?
Correct answer: 36 kg · m^2 /s. Explanation: Angular momentum is given by ~LA = ~rA × ~p.
Performing this calculation using the component-wise definition of the cross prod- uct, we obtain
~LA = 〈ry pz, 0 , 0 〉 = 〈(4 m)(9 kg · m/s), 0 , 0 〉 = 〈 36 , 0 , 0 〉 kg · m^2 /s.
So the x component is 36 kg · m^2 /s.
009 10.0 points If all three collisions in the figure are totally inelastic, which cause(s) the most damage (deformation of objects, thermal energy in- crease, etc.)? Assume that the wall is station- ary and the car is completely stopped by it in the first diagram.
m
v
brick
wall
I
m^2 m v 0.5 v II
III^ m^ v^^2 v 0.5 m
- III correct
- II, III
- all three
- I
- II
- I, II
- I, III
Explanation: By momentum conservation the final veloc- ity in all three cases is zero, so in each case all the kinetic energy is lost, causing damage by crumpling the vehicles. The kinetic energy lost in case I is
KI =
mv^2
. The kinetic energy lost in case II is
KII =
mv^2 +
(2m)(0. 5 v)^2
mv^2.
The kinetic energy lost in case III is
KIII =
mv^2 +
m 2
(2v)^2
mv^2.
Since K 3 > K 2 > K 1 , case III causes the most damage.
010 10.0 points A skater pushes straight away from a wall. She pushes on the wall with a force whose magnitude is F , so the wall pushes on her with a force F (in the direction of her motion). As she moves away from the wall, her center of mass moves a distance d. What is the correct form of the energy principle for the point particle system for the skater?
- ∆Ktrans = −F d
- ∆Ktrans + ∆Einternal = 0
- ∆Ktrans = F d correct
- ∆Ktrans + ∆Einternal = F d
- ∆Ktrans + ∆Einternal = −F d Explanation: Since the center of mass moves a distance d, over which the force F is applied, the work done on the point particle (center-of-mass) system is WP P = F d. This work is the change in translational ki- netic energy, so
∆Ktrans = F d.
For the real system, no work is done since the contact point between the skater’s hand and
Two identical 0.4 kg blocks (labeled 1 and 2 in the diagram) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper portion of the figure.
95 N
95 N
x = 0 m 0.6 m
x = 0.2 m 2.0 m
Initial
Final
Then a constant force of 95 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower portion of the figure. At this final time, the system is moving to the right and also vibrating, and the spring is stretched. What is the final vibrational energy (Uspring + Kvib)of the two block and spring system?
Correct answer: 57 J.
Explanation: Let the distance the right hand block moved be called d and the distance the center of mass moved be called b. The translational kinetic energy of the system in the final state is
Ktrans,f = F b. The energy principle applied to the real
system yields
Ktrans,f + Evib,f = F d.
Combining the two equations,
Evib,f = F (d − b) = (95 N)(1.4 m − 0 .8 m) = 57 J.
015 (part 1 of 2) 5.0 points
A hoop of mass M and radius R rolls with- out slipping down a hill, as shown in the figure above. The lack of slipping means that when the center of mass of the hoop has speed v, the tangential speed of the hoop relative to the center of mass is also equal to vCM , since in that case the instantaneous speed is zero for the part of the hoop that is in contact with the ground (v−v = 0). Therefore, the angular speed of the rotating hoop is ω =
vCM R
The hoop is initially at rest, and the hill has a height h. What is the speed vf at the bottom of the hill?
gh correct
4 gh
2 gh
gR
M gh
2 gR
2 M gh Explanation:
By using the conservation of energy we ob- tain Ei = Ef
M gh =
M v^2 CM +
Iω^2
M gh =
M v^2 CM +
(M R^2 )
vCM R
M gh = M v^2 CM
vCM =
gh
016 (part 2 of 2) 5.0 points
Replace the hoop with a bicycle wheel whose rim has mass M and whose hub has mass m, as shown in the figure. The spokes have negligible mass and the hub has a neg- ligible moment of inertia. The speed of the bicycle wheel at the bottom of the hill is
- equal to the speed of the hoop
- greater than the speed of the hoop cor- rect
- less than the speed of the hoop
Explanation: For the bicycle wheel with hub of mass m, we can think about the limit of m very large. In this case, the rotational kinetic energy will be negligible compared to the translational ki- netic energy and vcm =
2 gh which is greater than that for the hoop. Alternatively, we can make use of the energy conservation equation for this case and solve it:
(M +m)gh =
(M +m)v CM^2 +
(M R^2 )
vCM R
vcm =
2 gh
M + m 2 M + m
017 (part 1 of 3) 5.0 points By calculating numerical quantities for a mul- tiparticle system, one can get a concrete sense of the meaning of the relationships ~psys = Mtot~vcm and Ktot = Ktrans + Krel. Consider an object consisting of two balls con- nected by a spring, whose stiffness is 400N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant the spring is stretched 0.3m, and the two balls at the ends of the spring have the following masses and veloci- ties:
m 1 = 2.2 kg m 2 = 4.6 kg
v ~ 1 = 〈 8 , − 7. 5 , 0 〉 m/s v ~ 2 = 〈− 11. 5 , 8 , 0 〉 m/s
Begin by finding the center of mass velocity for the two balls. The answer choices below are in the usual form
~vCM = 〈vx,CM, vy,CM, vz,CM〉 m/s.
- 〈 8 , − 7. 5 , 0 〉 m/s
- 〈− 7. 77941 , 5. 41176 , 0 〉 m/s
- 〈 1. 17647 , − 1. 10294 , 0 〉 m/s
- 〈− 3. 5 , 0. 5 , 0 〉 m/s
- 〈 2. 58824 , − 2. 42647 , 0 〉 m/s
- 〈− 5. 19118 , 2. 98529 , 0 〉 m/s correct
- 〈− 1. 69118 , 1. 17647 , 0 〉 m/s
- 〈− 35. 3 , 20. 3 , 0 〉 m/s
- 〈− 11. 5 , 8 , 0 〉 m/s
- 〈− 0. 514706 , 0. 0735294 , 0 〉 m/s Explanation:
4. ~L = −
m v^20 2 g
sin θ cos^2 θ kˆ
5. ~L = −
m v^30 2 g
sin^2 θ cos θ kˆ correct
- ~L is in the ˆı direction.
7. ~L = −
m v^30 2 g
sin θ cos^2 θ kˆ
8. ~L = −
m v^20 2 g
sin^2 θ cos θ kˆ
9. ~L = +
m v^20 2 g
sin θ cos^2 θ kˆ
- ~L is in the ˆ direction.
Explanation: Denote the position vector from the origin to the highest point of the trajectory as ~rh. At the highest point ~vh = ~vx = vo cos θ ˆı and
~L = ~rh × ~ph = ~rh × m~vh
so that
|~L| = m |~vh| |~rh| sin θh.
The maximum height is
h = |~rh| sin θh =
v^2 o sin^2 θ 2 g
and since |~vh| = |~v 0 x| = vx, the magnitude of the angular momentum is
|~L| = m h vx = m
v o^2 sin^2 θ 2 g
(vo cos θ)
m v o^3 2 g
sin^2 θ cos θ.
The direction is −kˆ by the right hand rule of the cross product.
