Version 100/ABCBA – Exam 3 – haley – (56465) This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Planet A and planet B of equal mass m orbit the same star of mass M in circular trajectories of radii rA = R and rB = 2R respectively. Calculate the ratio of the kinetic energy of A to the kinetic energy of B.
1
where K is the total kinetic energy of the atoms (in the rest frame of the molecule) and U is the potential energy stored in their interaction (U = 0 at large distance). Assume there is no rotation. Which is true regarding the rest mass of diatomic oxygen? 1. mO2 c2 = 2mO c2 + Eint correct 2. mO2 c2 = 2mO c2 − Eint 3. mO2 c2 = 2mO c2
1. 8 2. 2 correct
Explanation: A bound system has negative total energy by definition, so Eint < 0. Since the internal energy of O2 is negative, the molecule’s rest mass will be less than the sum of its constituent oxygen atoms’ rest masses. mO2 c2 = 2mO c2 + Eint satisfies this requirement.
3. 1/8 4. 16 5. 1/16 6. 1/2 Explanation: Since A has a circular orbit, 2 m vA GmM = , R R2
which implies
003 (part 2 of 2) 5.0 points Which is true regarding the binding energy of diatomic oxygen at any time after the chemical bonding has taken place? 1. |Ebind | = |Eint | correct
1 GmM 2 KEA = m vA = . 2 2R Similarly, for B, 2 m vB GM m = , 2R 4 R2
which implies 1 GM m 2 KEB = m vB = . 2 4R Therefore, KEA = 2. KEB 002 (part 1 of 2) 5.0 points Two oxygen atoms pass near each other and form a bound state of diatomic oxygen, O2 . Its internal energy is given by Eint = K + U,
2. |Ebind | = K 3. |Ebind | = |U | Explanation: The amount by which it is lower is called the binding energy. Thus, |Ebind | = |Eint | is correct. 004 (part 1 of 2) 5.0 points Consider the following situations and choices of system: A. A hair dryer’s heating element is kept at a constant high temperature while it receives an input of electric energy. (System: hair dryer)
Version 100/ABCBA – Exam 3 – haley – (56465) B. A hot Aluminum ingot is placed into a well-insulated Styrofoam container full of cold water. The ingot and water temperatures equalize after a certain length of time. (System: ingot and water) C. A hot Aluminum ingot is placed into a well-insulated Styrofoam container full of cold water. The ingot and water temperatures equalize after a certain length of time. (System: ingot) Which of the following is the complete list of situations+systems for which Q is nonzero? 1. A, B 2. A, B, C 3. B, C 4. None 5. B 6. A, C correct 7. C 8. A Explanation: A: Constant temperature implies that ∆Ethermal = 0. Thus, the electrical energy input must have been converted to heat, so Q < 0. B: The container is insulated, so Q = 0. Since there are no energy inputs, ∆Ethermal = 0. C: The ingot cools by transmitting heat to the surrounding water, so Q < 0. As it cools, it loses thermal energy, so ∆Ethermal < 0. 005 (part 2 of 2) 5.0 points Which of the following is the complete list of situations+systems for which ∆Ethermal is nonzero?
2
3. A, B 4. C correct 5. B 6. A, B, C 7. None 8. B, C Explanation: See part 1 explanation. 006 10.0 points The escape speed from an asteroid whose radius is 10 km is only 8 m/s. If you throw a rock away from the asteroid at a speed of 16 m/s, what will be its final speed? Use G = 6.7 × 10−11 N · m2 /kg2 . 1. 17.3205 2. 22.5167 3. 19.0526 4. 25.9808 5. 8.66025 6. 13.8564 7. 12.1244 8. 20.7846 9. 10.3923 10. 24.2487 Correct answer: 13.8564 m/s. Explanation: First use the escape speed to get the mass of the asteroid: r
2GM R 2 1v R ⇒M = 2 G = 4.77612 × 1015 kg . vesc =
1. A 2. A, C
Now, if vi = 16 m/s, then vf is found from Ei = Ef :
Version 100/ABCBA – Exam 3 – haley – (56465) Ui + K i = Uf + K f GM m 1 1 ⇒− + m vi2 = 0 + m vf2 ri 2 2 r 2GM ⇒ vf = vi2 − R = 13.8564 m/s . 007 (part 1 of 2) 5.0 points A pendulum consists of a very light but stiff rod of length L hanging from a nearly frictionless axle, with a mass m at the end of the rod. Calculate the gravitational potential energy as a function of the angle, θ, measured from the vertical. Set U = 0 at the location of the mass when the pendulum is hanging straight down.
3
008 (part 2 of 2) 5.0 points Suppose that you hit the stationary hanging mass so it has an initial speed vi . What is the minimum initial speed needed for the pendulum to go over the top (θ = 180◦ )? p 1. vi = g L p 2. vi = 2 g L correct p 3. vi = 4 g L s gL 4. vi = 2 p 5. vi = 8 g L 1p gL 2 p 7. vi = 2 g L 6. vi =
1. U = −m g L sin(θ) 2. U = −m g L cos(θ) 3. U = m g L θ 4. U = m g L(1 − sin(θ)) 5. U = m g L(1 − cos(θ)) correct 6. U = m g L sin(θ) 7. U = m g L cos(θ) Explanation: We must find the vertical distance the mass is from the bottom so we can use the potential energy formula for gravity on Earth. We can see that the height y can be found by subtracting L cos(θ) from L. The first value is the vertical distance of the mass from the pivot, while hanging at an angle of θ, and the second is the distance of the U = 0 location from the pivot. Subtracting these two values gives us the height of the mass from the U = 0 location. Therefore, the potential energy is given by: U =m g y =m g (L − L cos(θ)) U =m g L(1 − cos(θ)) .
Explanation: At the initial state, right after the instant that the mass is given the initial speed, we have zero potential energy and nonzero kinetic energy. At the final state, if we had given the pendulum the absolute minimum speed for the pendulum to go over the top, then it would have zero kinetic energy at the top and nonzero potential energy. Thus we conserve energy to get: Ei =Ef KEi + Ui =KEf + Uf 1 m vi2 + 0 =0 + m g L(1 − cos(180◦ )) 2 vi2 =2 g L(2) p vi =2 g L 009 10.0 points Many heavy nuclei undergo spontaneous “alpha decay,” in which the original nucleus emits an alpha particle (a helium nucleus containing two protons and two neutrons), leaving behind a “daughter” nucleus that has two fewer protons and two fewer neutrons than
Version 100/ABCBA – Exam 3 – haley – (56465) the original nucleus. Consider a radium-220 nucleus that is at rest before it decays to radon-216 by alpha-decay. The mass of the radium-220 nucleus is 219.962 u (unified atomic mass units) where 1 u = 1.6603 × 10−27 kg (approximately the mass of one nucleon). The mass of a radon-216 nucleus is 215.953 u, and the mass of an alpha particle is 4.00151 u. Radium has 88 protons, radon 86, and an alpha particle 2. Calculate the final kinetic energy of the alpha particle. Assume that its speed is small compared to the speed of light. Use c = 3 × 108 m/s. 1. 1.09885 × 10−12 J correct 2. 9.88962 × 10−13 J
ticle was at rest (pRa = 0). By conservation of momentum, the sum of the two final momenta must be zero, meaning their squares will be equal. Let p2α 1 1 β≡ . + 2 mRn mα Now we solve for pα : β = Erest,Ra − Erest,Rn − Erest,α = mRa c2 − mRn c2 − mα c2 = 1.11921 × 10−12 J 2(1.11921 × 10−12 J) ⇒ = 1 1 + mRn mα ⇒ pα = 1.20834 × 10−19 kg · m/s . p2α
Don’t forget to make the appropriate conversions. Now that we know the momentum, the kinetic energy is just
3. 9.3402 × 10−13 J 4. 1.26367 × 10−12 J 5. 1.15379 × 10−12 J
KE =
6. 1.20873 × 10−12 J
p2α 2 mα
= 1.09885 × 10−12 J .
7. 1.37356 × 10−12 J
010 10.0 points The force due to a deflected cantilever beam can be modeled using a linear and cubic term, i.e.
8. 1.31862 × 10−12 J 9. 1.0439 × 10−12 J
Fbeam = −k1 δ − k3 δ 3
10. 8.79078 × 10−13 J Explanation: We begin by using conservation of energy to write Ei = Ef Erest,Ra = Erest,Rn + Erest,α + KRn + Kα = Erest,Rn + Erest,α +
4
p2Rn p2 + α 2mRn 2mα
= Erest,Rn + Erest,α 1 1 p2α + + 2 mRn mα
Note that p2Rn = p2α since the original par-
where δ is the deflection at the end of the beam. What work must be done on the beam to deflect it by 28 cm if k1 = 95 N/m and k3 = 30000 N/m3 ? 1. 2.25615 2. 44.64 3. 17.0116 4. 9.43189 5. 5.6355 6. 49.8232 7. 16.6635 8. 20.1351 9. 4.3578 10. 1.27655
Version 100/ABCBA – Exam 3 – haley – (56465) Correct answer: 49.8232 J. Explanation: The work done on the beam is given by Won beam = −Wby beam Z δmax Fbeam dδ =− 0 Z δmax k1 δ + k3 δ 3 dδ = 0 δmax 1 1 2 4 = k1 δ + k3 δ 2 4 0 = 49.8232 J
011 10.0 points Which of the following is a physical feature of a real spring which is NOT represented by the ideal spring potential energy graph, 1 Uideal = ks s2 − ES ? 2 A. A yield and breakage region B. The existence of bound states C. The existence of unbound states D. A region of compression where coils are touching E. An equilibrium point of minimum energy 1. A, B, D, E 2. A, C, D correct
5
Explanation: A real spring will break when stretched sufficiently, so A is true. The ideal spring also has bound states, so B is false. The ideal spring has no unbound states, so C is true. The ideal spring does not have a steep region of U to represent coils touching, so D is true. The ideal spring also has an equilibrium point, or local minimum, so E is false. 012 (part 1 of 2) 5.0 points Suppose a block of mass M is sliding down an inclined plane which makes an angle θ with the horizontal. The coefficient of kinetic friction between the plane and the block is µ (assume tan θ > µ). The block is initially at rest. What is the work done by friction as the block travels a distance d along the plane? 1. µ M g sin θd 2. −µ M g sin θd 3. −µ M g cos θd correct 4. µ M g tan θd 5. µ M g cos θd 6. −µ M g tan θd Explanation: From the free-body diagram of the block, the normal reaction force
3. A, C, E N = M g cos θ. 4. A, B, C, D, E The frictional force 5. A, C, D, E Ff = µ M g cos θ. 6. A, B, C, E 7. B, C, E
Since the block moves in a direction opposite to that of friction,
8. B, D, E
Wf = −Ff d = −µ M g cos θd.
9. B, C, D, E 10. A, B, C, D
013 (part 2 of 2) 5.0 points What is the speed of the block at this point?
Version 100/ABCBA – Exam 3 – haley – (56465) p 2 g d µ (sin θ − cos θ) p 2. µ 2 g d (sin θ − µ cos θ) p 3. 2 g d (tan θ − µ cos θ) p 4. 2 g d (sin θ + µ cos θ) p 5. 2 g d (sin θ − µ tan θ) p 6. 2 g d (sin θ − µ cos θ) correct 1.
K +U
(I)
(II)
K U
6
K
K +U r
r U
(III)
K
(IV)
K +U
U
K +U
Explanation:
r
r
Ei = M g d sin θ. U
The final energy 1 Ef = M vf2 . 2
K
K
(V)
K +U U
Therefore, energy conservation Ei + Wf = Ef implies,
(VI) r
1 M g d sin θ − µM g cos θ d = M vf2 , 2 or vf2 = 2 g d (sin θ − µ cos θ).
r
U 1. Figure V 2. Figure IV
Therefore, vf =
K +U K
p
3. Figure II correct 2 g d (sin θ − µ cos θ).
014 10.0 points Which of the following diagrams corresponds to a system of a proton and an electron that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)? Note that the horizontal and vertical axes in each plot are the separation between the particles and energy, respectively. To clarify, the kinetic energy line K is always the dotted line, the potential energy line U is always dashed, and the sum K + U is always a solid black line (sometimes lying on the x axis).
4. Figure VI 5. Figure III 6. Figure I Explanation: When the two particles are very far away, their potential energy is 0, and since they have nonzero initial velocities, this means that they are unbounded and thus have an overall positive energy at r = ∞, which is also equal to the kinetic energy at that location. As the electron and proton get closer, due to their Coulomb attraction their kinetic energies in-
Version 100/ABCBA – Exam 3 – haley – (56465) crease while the negative potential energy decreases even further. Thus the correct answer is Figure (II).
5. Statements II, III, IV
015 10.0 points The figure below is a graph of the energy of a system of a planet interacting with a star. The gravitational potential energy Ug is shown as the thick curve, and plotted along the vertical axis are various values of K + Ug .
7. Statements I, II, III, IV
7
6. Statements I, II, III
8. Statements II, IV, V, VI 9. Statements I, III, IV, V 10. Statements I, III, IV
K + Ug
r1 r2 r, from star to planet A B C
Suppose that K + Ug of the system is A. Which of the following statements are true? I. The potential energy of the system decreases as the planet moves from r1 to r2 . II. When the separation between the two bodies is r2 , the kinetic energy of the system is (A − B). III. The system is a bound system; the planet can never escape. IV. The planet will escape. V. When the separation between the two bodies is r2 , the kinetic energy of the system is B − C). VI. The kinetic energy of the system is greater when the distance between the star and planet is r1 than when the distance between the two bodies is r2 .
Explanation: Statement I is false because as we move to the right from r1 to r2 , U increases. Statement II is true because the kinetic energy is given as the total energy minus the potential energy, and since the total energy is A, and U (r2 ) = B, the kinetic energy is A − B. Statement III is true because the total energy A is negative, which indicates a bound system. Statements IV-VI can be understood from the previous three explanations. 016 10.0 points Given that the y axis represents energy and the x axis represents separation, which graph correctly shows the potential energy U for two interacting electrons? I
II
III
IV
1. Statements I, IV, V, VI 2. Statements I, II, III, VI 1. II and III 3. Statements II, III, VI correct 2. I 4. Statements III, V, VI 3. II and IV
Version 100/ABCBA – Exam 3 – haley – (56465) 4. II correct 5. I and IV 6. IV 7. I and II 8. III 9. None 10. III and IV Explanation: Electrons interact repulsively, so the correct potential energy graph will decrease to zero at large separation. Only graph B has this feature. 017 10.0 points In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. If the volume flow rate is V˙ and the height of the falls is h, what is the amount of electrical power produced, assuming the plant’s efficiency is k? Let the density of water be notated by ρ. ρgh2 k V˙ ρV˙ gh 2. P = k ρV˙ gh2 3. P = k 4. P = kρV˙ gh2
1. P =
kρgh V˙ 6. P = kρV˙ gh correct 5. P =
ρgh k V˙ kρgh2 8. P = V˙ Explanation: The electrical power of a hydroelectric plant
8
is the potential energy converted to electrical energy per second. The mass of water flowing ˙ = ρV˙ . The through the plant per second is M potential energy change of this mass is: ˙ gh = −ρV˙ gh. ∆U = −M Therefore, the electrical energy produced per second is this potential energy times the conversion efficiency: P = k(−∆U ) = kρV˙ gh. 018 10.0 points In a star, a secondary fusion process involves helium-3 and helium-4 fusing together into beryllium-7. Helium-3 has a mass of 3.01603 u, helium-4 has a mass of 4.0026 u, and beryllium-7 has a mass of 7.01693 u. The atomic mass unit u is 1.66054 × 10−27 kg. Each one of these fusion reactions will convert rest mass energy into kinetic energy. If you want to have a total of E = 87 J of energy that has been converted from rest mass energy, how many of these reactions must take place? Keep six significant figures throughout this problem, and use c = 2.99792 × 108 m/s. 1. 1.73426e+14 2. 90654500000000.0 3. 3.42911e+14 4. 2.40432e+14 5. 3.90209e+14 6. 19707500000000.0 7. 2.089e+14 8. 3.19262e+14 9. 1.34011e+14 10. 1.18245e+14 Correct answer: 3.42911 × 1014 . Explanation: The rest mass energy released by one of these reactions is
7. P =
(mHe−3 + mHe−4 − mBe ) c2 = 2.5371 × 10−13 J. In order to get the number of reactions to produce energy E, we divide E by the energy per reaction that is found above:
Version 100/ABCBA – Exam 3 – haley – (56465) E 87 J = −13 2.5371 × 10 J 2.5371 × 10−13 J = 3.42911 × 1014 .
9
1. M , m 2. M , M 3. m, same momentum correct
019 (part 1 of 2) 5.0 points You pull a block of mass m across across a frictionless table with a constant force. You also pull with an equal constant force a block of larger mass M . The blocks are initially at rest. If you pull the blocks through the same distance, which block has the greater kinetic energy, and which block has the greater momentum, respectively?
4. m, M 5. M , same momentum 6. m, m 7. Same kinetic energy, m 8. Same kinetic energy, same momentum
1. M , same momentum 9. Same kinetic energy, M 2. M , m 3. m, same momentum 4. Same kinetic energy, m 5. m, m 6. M , M 7. Same kinetic energy, same momentum 8. m, M 9. Same kinetic energy, M correct Explanation: You do the same work on each block. According to the Energy Principle, each block will have the same ∆K. Since they both start from rest, Kf is the same for each block. Kf =
p2f
, 2m so the larger mass block will have the larger final momentum. 020 (part 2 of 2) 5.0 points If instead you pull the blocks for the same amount of time, which block has the greater kinetic energy, and which block has the greater momentum, respectively?
Explanation: If you pull the blocks for the same amount of time, then according to the Momentum Principle, ~ net ∆t ∆~p = F will be the same for the two blocks. Since they both start from rest, ~pf will be the same. Since ~pf ~vf = , m we know that ~vf is greater for the smaller block and the smaller block will have the greater final kinetic energy. 021 (part 1 of 2) 5.0 points A fan cart of mass 0.88 kg initially has a velocity of ~vi = h0.81, 0, 0i m/s. Then the fan is turned on, and the air exerts a constant force of ~ = h−0.47, 0, 0i N F on the cart for 1.5 s. What is the change in the x component of momentum of the fan cart over this 1.5 s time interval? (Since the force is only applied
Version 100/ABCBA – Exam 3 – haley – (56465) along the x direction, the other components of momentum will not change.) 1. -0.555 2. -0.435 3. -0.795 4. -0.345 5. -0.36 6. -0.42 7. -0.705 8. -0.585 9. -0.81 10. -0.465 Correct answer: −0.705 kg · m/s. Explanation: The change in momentum is given by
~ net ∆t ∆~p = F = (h−0.47, 0, 0i N)(1.5 s) = h−0.705, 0, 0i kg · m/s .
So the x component is −0.705 kg · m/s . 022 (part 2 of 2) 5.0 points What is the change in kinetic energy of the fan cart over this 1.5 s time interval? 1. -0.3132 2. -0.367048 3. -0.3885 4. -0.23865 5. -0.288649 6. -0.300888 7. -0.2352 8. -0.380739 9. -0.225273 10. -0.1953 Correct answer: −0.288649 J. Explanation: To find the change in kinetic energy, we need to know the initial and final velocities. We know the initial velocity; to get the final, we first find ∆~v :
10
∆~p m h−0.705, 0, 0i kg · m/s = 0.88 kg = h−0.801136, 0, 0i m/s . So the final velocity is ∆~v =
~vf = ~vi + ∆~v = h0.00886364, 0, 0i m/s . Now, the change in kinetic energy is given by ∆KE = KEf − KEi 1 1 = m vf2 − m vi2 2 2 1 = (0.88 kg) (0.00886364 m/s)2 2 1 − (0.88 kg) (0.81 m/s)2 2 = −0.288649 J . 023 10.0 points Starting from rest, a woman lifts a barbell of mass mbb with a constant force F through a distance h, at which point she is still lifting, and the barbell has acquired a speed v. Let Ewoman stand for the following energy terms associated with the woman: Ewoman = Echemical,woman + Kwoman + Ugrav,woman+Earth + Ethermal,woman The change in the kinetic energy of the barbell is 1 1 mbb v 2 − 0 = mbb v 2 . 2 2 The general statement of the energy principle is: ∆Esys = Wsurr For which of the following systems will the left hand side of this equation have ONLY the 1 terms +mbb gh and mbb v 2 ? 2 1. woman + barbell + Earth
Version 100/ABCBA – Exam 3 – haley – (56465) 2. woman only 3. there is no such system 4. woman + barbell 5. barbell + Earth correct 6. barbell only 7. Earth only 8. woman + Earth Explanation: We may first note that +mbb gh on the left hand side of the energy principle equation represents a potential energy change, which only multibody systems can exhibit. Thus the single body answer choices cannot be correct. Of the multibody choices, any containing “woman” must have the term ∆Ewoman on the left side of the energy principle. The only remaining choice is barbell + Earth. This is correct because the change in potential energy of this system is +mbb gh due to the barbell rising, and the change in kinetic energy of this 1 system is mbb v 2 due to the barbell gaining 2 speed v.
11
