123 Version Midterm 03 florin (56930) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points 7. Ic,IIa 8. Ib,IIa 9. Ib,IIb
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Two spheres, A and B, have the same mass and radius. However, sphere B is made of a dense core and a less dense shell around it. How does the moment of inertia of sphere A about its center of mass compare to the moment of inertia of sphere B about its center of mass? Ia. IA > IB Ib. IA < IB Ic. IA = IB If the two spheres are rolled down an incline from the same height simultaneously, IIa. sphere A reaches the bottom first. IIb. sphere B reaches the bottom first. IIc. spheres A and B reach the bottom simultaneously. Choose the correct pair of statements. 1. Ia,IIb correct 2. Ic,IIc 3. Ib,IIc 4. Ic,IIb 5. Ia,IIa 6. Ia,IIc
Explanation: Since the two spheres share the same mass and radius, a comparison of their moment of inertias depends on how their mass is distributed relative to an axis through the center of mass. Sphere B has a dense core, implying that its mass is on average closer to the axis than A's mass. Thus, A has a greater moment of inertia than B. Let be the unitless parameter in the moment of inertia formula I = M R2 . Sphere B has a smaller from the preceding argument. As a consequence of energy conservation, it can be shown that vCM = 2gh 1+
Therefore, B reaches the bottom first. 002 10.0 points Find the moment of inertia of a solid sphere of mass M = 3.5 kg and radius R = 0.6 m about an axis that is tangent to the sphere. m r
1. 3.087 2. 1.372 3. 4.032 4. 1.715 5. 2.772 6. 0.875 7. 1.764 8. 0.252 9. 1.232 10. 0.693 Correct answer: 1.764 kg m2 . Explanation:
Version 123 Midterm 03 florin (56930) The moment of inertia of a solid about a diameter is Icm = 2 M R2 . 5 Thus, the power expended will be P = -Ecar t 1 mcar (vcar )2 = 2 t 1 (750 kg)(7.5 m/s)2 = 2 0.35 s = 60267.9 W .
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Using the parallel-axis theorem, the moment of inertia about an axis that is tangent to the sphere is I = Icm + M R2 2 = M R2 + M R2 5 7 = M R2 5 7 = (3.5 kg)(0.6 m)2 5 = 1.764 kg m2
004 10.0 points Consider the spacing of vibrational energy levels of Pb and Al based on the quantum harmonic oscillator model for the interatomic bound. Pb has a stiffness of ks 5N/m and an atomic mass of 207 mN (where mN is the mass of a nucleon). For Al, the stiffness is ks 17N/m, and the atomic mass is 27 mN . Determine the ratio of the energy level spacEAl . Choose one : ings, EP b 1. 5.11 correct 2. 2 3. 3.33 4. 6 5. 1.33 6. 0.5 7. 4 8. 1 Explanation: A quantum harmonic oscillator has energy level spacing E N = h 0 N + E 0 . So the level spacing is given by E = h 0 = h Thus the ratio is EAl h Al 0 = EP b h P 0 b ks m
003 10.0 points A test car of mass 750 kg is moving at a speed of 7.5 m/s when it crashes into a wall to test its bumper. If the car comes to rest in 0.35 s, how much average power is expended in the process? 1. 59189.1 2. 44252.9 3. 47940.0 4. 60267.9 5. 42820.9 6. 58285.9 7. 41605.4 8. 45615.1 9. 55151.1 10. 51201.9 Correct answer: 60267.9 W. Explanation: The power expended is the energy lost per unit time. The energy that is lost in the process of stopping the car is -Ecar = -Kcar 1 = mcar (vcar )2 . 2
Version 123 Midterm 03 florin (56930) =
Al ks mAl P ks b mP b
3
=
Al ks mP b P ks b mAl
=
(17N/m)(207 mN) (5 N/m)(27 mN) = 5.11 .
Ephoton = |E1,3 | = |E1 - E3 | = |-9 eV + 1.8 eV| = 7.2 eV Ephoton = |E2,3 | = |E2 - E3 | = |-4.8 eV + 1.8 eV| = 3 eV 006 10.0 points Consider the head-on collision of two masses, m1 = 8 kg moving to the right with speed 3 m/s and m2 = 16 kg moving to the left with speed 1.5 m/s. Given that they stick together after the collision, what is the increase in internal energy of the system during the collision? Assume no energy is lost to the surroundings and neglect any external forces acting on the two masses. 1. 112.5 2. 54.0 3. 192.0 4. 450.0 5. 216.0 6. 131.25 7. 84.0 8. 303.75 9. 60.0 10. 600.0 Correct answer: 54 J. Explanation: We know that the velocity of the center of mass is given by vCM = m1 v1 + m2 v2 m1 + m2
005 10.0 points The energy levels of a particular quantum object are -9 eV, -4.8 eV, and -1.8 eV. If a collection of these objects is bombarded by an electron beam so that there are some objects in each excited state, what are the energies of the photons that will be emitted? 1. -4.2 eV, -7.2 eV, -3 eV 2. 4.2 eV, 6.6 eV, 13.8 eV 3. 7.2 eV, -4.8 eV, 1.8 eV 4. 10.8 eV, -4.2 eV, 7.2 eV 5. 13.8 eV, 3 eV, 10.8 eV 6. 13.8 eV, 10.8 eV, 6.6 eV 7. 4.2 eV, 13.8 eV, 3 eV 8. 4.2 eV, 7.2 eV, 3 eV correct 9. -9 eV, -4.8 eV, -1.8 eV 10. 9 eV, -4.8 eV, -1.8 eV Explanation: 4.2 eV, 7.2 eV, 3 eV is the correct answer. The 3 emitted photons correspond to 3 transitions between the states. Label them as E1 = -9 eV E2 = -4.8 eV E3 = -1.8 eV Then the possible transitions are given by Ephoton = |E1,2 | = |E1 - E2 | = |-9 eV + 4.8 eV| = 4.2 eV
By substituting the values of m1 , m2 , v1 and v2 we have (m)(v0 ) + (2m)(- vCM = m + 2m v0 ) 2
Version 123 Midterm 03 florin (56930) vCM = 0 Thus, the velocity of the center of mass is zero. We then apply the energy principle to this problem E = K + Eint = 0 Eint = - K = Ki - Kf But, the final kinetic energy is zero since the velocity of the center of mass is zero. Thus, we have Eint = Ki Ki = 1 1 2 m v0 + (2m) 2 2 1 = 2 1 1+ 2 3 2 m v0 4 v0 2
2 m v0 2
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Initially, the coffee filter's energy is Einitial = Uinitial + Kinitial = mgh + 0 = (1.2 g)(9.8 m/s2 )(2 m) = 0.02352 J . When the coffee filter reaches the ground, it's energy is Ef inal = Uf inal + Kf inal 1 = 0 + m v2 2 1 1kg = (1.2 g)(0.6 m/s)2 2 1000g = 0.000216 J . Since these two are not equal, energy must be lost to the air molecules, meaning that they gain kinetic energy. Thus, Kair = Kinitial - Kf inal = 0.02352 J - 0.000216 J = 0.023304 J . 008 10.0 points A spring of stiffness k and relaxed length L stands vertically on a table. A person takes a mass M and very slowly lets the mass down onto the spring. On letting go, he finds that the mass does not move and the spring is compressed by a length d1 . The same experiment is now repeated by letting the mass go all of a sudden when the spring is still unstretched. The spring is found to compress by a length d2 when the mass momentarily comes to rest. What is the ratio d1 /d2 ? 1. 2 2. 4 3. 1 4 4. kL Mg 5. Mg
kL
1kg 1000g
Eint
Eint = Eint
3 = (8 kg) (3 m/s)2 4
Eint = 54 J 007 10.0 points A coffee filter of mass 1.2 g dropped from a height of 2 m reaches the ground with a speed of 0.6 m/s. How much kinetic energy Kair did the air molecules gain from the falling coffee filter? 1. 0.050726 2. 0.0761215 3. 0.053548 4. 0.05848 5. 0.04868 6. 0.023304 7. 0.019355 8. 0.02922 9. 0.070176 10. 0.068425 Correct answer: 0.023304 J. Explanation:
6. 1
Version 123 Midterm 03 florin (56930) 7. 1 correct 2 Explanation: When the mass is very gradually let down onto the spring, the net work done on the mass is 0 because at every instant while it is being let down, the net force acting on the mass is 0 (since the man always provides just enough force to balance all the forces). The final spring compression is determined by the fact that there is no net force acting on the mass even when the man has let it down, i. e., the spring force and the force of gravity cancel. Hence we can write kd1 - M g = 0. From this, we get d1 = Mg . k
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From these two expressions, we can see that the required ratio is 1/2. 009 10.0 points
When the mass is let go all of a sudden, it is easy to see that when the spring has been compressed by d1 , there will still be a downward momentum for the mass due to the fact that it has been pulled down by the force of gravity which was always greater in magnitude than the spring force (the work done till that instant would be non-zero). Hence, the mass goes down further, until all its Kinetic energy has been converted into potential energy. Eventually, the mass oscillates about the equilibrium position conserving its total energy all the time. By using conservation of energy between the topmost point of oscillation (the point at which the man just let the mass go) and the bottommost point of oscillation (when the mass just comes to stop momentarily), we get KE1+GP E1 +SP E1 = KE2 +GP E2 +SP E2 . This translates to 1 0 + 0 + 0 = 0 - M gd2 + kd2 . 2 2 From this, we get 2M g . d2 = k
In the figure, a block sitting on a frictionless horizontal surface is attached to a rigid wall on the right through a spring (whose axis is horizontal). A bullet is shot at the block from the left and gets embedded in it, causing the block to move to the right, thus compressing the spring. (Assume the bullet is travelling perfectly horizontally, along the axis of the spring, before hitting the block). Which of the following are true? A. The initial kinetic energy of the bullet is completely converted to spring potential energy when the spring reaches its maximal compression. B. The initial momentum of the bullet is equal to the momentum of the bullet+block system just after the bullet enters the block. C. Part of the momentum of the bullet+block system is lost during the collision (i.e. before the spring-compression starts). D. Part of the energy of the bullet+block system is "lost" (no longer present as macroscopic kinetic energy) during the collision, before the spring-compression starts. E. If we are given the masses of the block and the bullet, the initial speed of the bullet and the spring constant, it is possible to find the maximum compression of the spring. 1. A, B 2. D 3. B, D
Version 123 Midterm 03 florin (56930) 4. A, E 5. A, C 6. A, B, E 7. A, C, E 8. B, D, E correct 9. A, B, D Explanation: Since the bullet gets embedded in the block, the collision is inelastic. Hence, some the the Kinetic energy of the bullet is converted into internal energy, heat, etc during the collision and does not get stored up in the spring. Therefore, A is false and D is true. However, momentum is conserved during the collision, since forces acting on the bullet-block system during the short time period are negligible. Therefore, B is true and C is false. If we are given initial velocity and masses, we can use momentum conservation to find velocity (and hence kinetic energy) of bullet-block system just after collision. We can then use energy principle to get maximum compression of the spring. Hence, E is true. 010 10.0 points A 300 g block of aluminum at a temperature of 610 K is placed in intimate contact with a 600 g block of iron at 300 K. The blocks are contained within an insulated enclosure. What is the final temperature of the two blocks? Given: The specific heat capacity of aluminum is 1 J/K/g and the specific heat capacity of iron is 0.4 J/K/g. 1. 527.8 2. 327.8 3. 494.4 4. 666.7 5. 372.2 6. 650.0 7. 588.9 8. 627.8 9. 638.9 10. 472.2 Correct answer: 472.2 K. Explanation: : Let TAl mAl CAl TF e mF e CF e = 610 K , = 300 g , = 1 J/K/g , = 300 K , = 600 g , and = 0.4 J/K/g .
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Let the final temperature be Tf . Heat will flow from the hot object to the cold object subject to the constraint that the net thermal energy will be conserved. Ethermal = C m T Ethermal (Al) + Ethermal (F e) = 0 CAl mAl TAl + CF e mF e TF e = 0 (1 J/K/g)(300 g)(Tf - 610 K) + (0.4 J/K/g)(600 g)(Tf - 300 K) = 0 Thus, the final temperature Tf is Tf = 472.2 K 011 10.0 points A beam of high-energy - (negative pions) is shot at a flask of liquid hydrogen, and sometimes a pion interacts through the strong interaction with a proton in the hydrogen, in the reaction - + p+ - + X + , where X + is a positively charged particle of unknown mass. Before After - - p+ X+
45
Version 123 Midterm 03 florin (56930) The incoming pion momentum is 2920 MeV/c. The pion is scattered through 45 , and its momentum is measured to be 1515 MeV/c. The X + particle is scattered through an unknown angle with an unknown momentum. A pion has a rest energy of 140 MeV, and a proton has a rest energy of 938 MeV. Find the scattering angle of the X + particle. 1. 30.0906 2. 27.4524 3. 29.752 4. 28.1837 5. 25.5265 6. 25.8799 7. 26.6746 8. 26.2914 9. 29.0747 10. 24.9826 Correct answer: 30.0906 Degrees. Explanation: We conserve momentum in the x and y directions: p,i =p,f cos 45 + pX cos 0 =p,f sin 45 - pX sin We rearrange the second expression to solve for pX , and then plug it into the first expression and solve for . p,i = p,f cos 45 + pX cos sin 45 p,i = p,f cos 45 + p,f tan p,f sin 45 tan = p,i - p,f cos 45 1515 MeV/c sin 45 = 2920 MeV/c - 1515 MeV/c cos 45 = 30.0906 .
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I. Vibrational states of a diatomic molecule such as O2 II. Idealized quantized spring - mass oscillator III. Electronic, vibrational, and rotational states of a diatomic molecule such as O2 IV. Electronic states of a single atom such as hydrogen
1. I-B, II-C, III-A, IV-D 2. I-C, II-B, III-D, IV-A 3. I-C, II-A, III-B, IV-C 4. I-A, II-B, III-D, IV-C correct 5. I-D, II-B, III-A, IV-C 6. I-D, II-C, III-A, IV-B 7. I-A, II-C, III-D, IV-B 8. I-A, II-D, III-A, IV-C 9. I-B, II-A, III-C, IV-D 10. I-B, II-D, III-C, IV-A Explanation: I-A, II-B, III-D, IV-C is the correct answer. Vibrational states of diatomic molecules have the typical potential curve, and energy levels become closely spaced as energy goes
012 10.0 points Which energy diagram in the figure below is appropriate for each of the following situations?
Version 123 Midterm 03 florin (56930) up to zero - (A) Quantum oscillator has equally spaced energy levels - (B) Electronic states have more wide separation than vibrational states, and vibrational states have more spacing than rotational states; so they combine to give (D) Electronic states of hydrogen are well known, which is also the only remaining option - (C) 013 10.0 points A projectile of mass m1 moving with a speed v1 in the +x direction strikes a stationary target of mass m2 head-on in an elastic collision. Find the final velocity of the projectile m1 . Hint: You can use the energy and momentum principles, OR you can solve the problem in the center of mass reference frame. 1. 2. 3. 4. 5. 6. 7. 8. m1 + m2 v1 m1 - m2 m2 v1 m1 m2 m1 + m2 m2 v1 2m1 2m1 v1 m1 + m2 m1 - m2 v1 correct m1 + m2 m1 v1 m2 m1 v1 m1 + m2
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Plugging in the expression for v1,f in terms of v2,f in the above expression, one can solve for v2,f and obtain, v2,f = v1,f 2m1 v1 m1 + m2 m1 - m2 = v1 . m1 + m2
Alternate Solution: The problem can also be solved by changing to the center of mass reference frame. vcm = m1 v1 m1 + m2 m2 v1 m1 + m2
v1i = v1 - vcm =
v1f = -v1i = -
m2 v1 m1 + m2
v1f = v1f + vcm =
(m1 - m2 ) v1 m1 + m2
014 10.0 points Which of the following forces are conservative forces? A) Gravitational force between the Earth and the Sun B) Spring force C) Air resistance D) A constant force E) Friction force
Explanation: From momentum conservation, m1 v1 = m1 v1,f + m2 v2,f which implies v1,f = v1 - m2 v2,f /m1 . From energy conservation,
2 2 2 m1 v1 = m1 v1,f + m2 v2,f .
1. B 2. A,B 3. A,B,C,D,E 4. A,B,D,E
Version 123 Midterm 03 florin (56930) 5. A 6. A,B,C,E 7. A,D,E 8. C,D,E 9. C,E 10. A,B,D correct 4. = 5. = 6. = 7. = 8. = 9. = 10. = 2. = 3. = 1 M R2 1 M R2 2 3M R2 4 M R2 4 M R2 2 3M R2 2 M R2 4 3M R2 2 M R2 FR Fl FR 1 F R - M v2 2 1 F l - M v2 2 Fl 1 F R - M v2 2 FR 1 F l - M v2 2
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correct
Explanation: The work due to a conservative force only depends on the initial and final positions, being completely independent of the path taken. This is true of the gravitational force, the spring force, and any constant force. The directions of the forces of friction and air resistance both depend on the direction in which the object in question is moving, and hence the particular path taken from the initial position to the final position is important. These two are therefore not conservative forces. 015 10.0 points As seen from above in the image, a string is wrapped around the edge of a uniform cylinder of radius R and mass M which is initially resting motionless on a frictionless table. F M R
Explanation: By the energy principle, K = W . The moment of inertia of the disk is I = (1/2)M R2. 1 1 M v 2 + I 2 = F l 2 2 1 1 1 M v 2 + ( M R2 ) 2 = F l 2 2 2 So, 4 M R2 1 F l - M v2 2
=
The end of the string is pulled with a force of F over a total distance l. If the linear speed of the cylinder is found to be v after pulling this distance, what is the angular speed of the cylinder? (Note that v = R in this case.) 1. = 4 3M R2 Fl
Note that there is a way to eliminate v from the final answer using the angular momentum principle, but in that case the answer is = 8 Fl 3M R2
016 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
Version 123 Midterm 03 florin (56930) 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. Consider the following statements regarding energy. I Ktrans + Einternal = F d II. Ktrans + Einternal = -F d III. Ktrans + Einternal = 0 IV. Ktrans = F d V. Ktrans = -F d What is the correct form of the energy principle for the skater as a real system and as a point particle (PP) system?
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Translational kinetic energy is the only type of energy a point particle system can have. IV is correct. 017 10.0 points Consider the following diagram. E3 E2 E1 I E0 III II IV
1. Real: III, 2. Real: IV, 3. Real: II, 4. Real: I, 5. Real: II, 6. Real: I, 7. Real: V, 8. Real: V, 9. Real: IV, 10. Real: III,
PP: IV correct PP: III PP: V PP: V PP: IV PP: IV PP: I PP: IV PP: IV PP: V
Say which arrows in the above diagram correspond to the following processes respectively: Absorption of a photon from the ground state to the first excited state; Emission of a photon from an excited state to another excited state. 1. Transition I; Transition IV 2. Transition IV; Transition II 3. Transition III; Transition II 4. Transition I; Transition III 5. Transition IV; Transition III 6. Transition II; Transition IV 7. Transition II; Transition III 8. Transition III; Transition IV 9. Transition III; Transition I 10. Transition I; Transition II correct Explanation: Absorption sends the atom into a higher energy state, so for the first process the arrow will point upward. Given that the amount of energy is represented by the difference between E1 and E0 for the first process, it should
Explanation: As a real system, the wall does no work on the skater since the contact point of the skater's hand and the wall does not move. Also, the wall does not transfer energy to the skater, i.e. by cooling down. The right hand side of the energy principle must therefore be zero for the real system. Furthermore, the left hand side must contain the skater's change in internal energy for a real system. III is correct. As a point particle system, the skater's center of mass moves a distance d while a force F is applied. Thus, the work done is F d.
Version 123 Midterm 03 florin (56930) be clear that Transition I is the correct choice for the first process. On the other hand, emission sends the atom into a lower energy state, so for the second process the arrow will point downward. The phrase "to another excited state" tells us that the final state should be either E2 or E1 . Given the possibilities on the diagram, we find that Transition II is the correct choice for the second process. 018 10.0 points In a certain time interval, natural gas with energy content 14000 J was piped into a house during a winter day. In the same time interval sunshine coming through the windows delivered 1000 J of energy into the house. The temperature of the house didn't change. What was Ethermal of the house? Ia. 14000 J Ib. 0 J Ic. 15000 J Id. 1000 J For the system of the house, what was Q, the energy transfer between the house and the air? IIa. -1000 J IIb. 15000 J IIc. 1000 J IId. -15000 J
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Explanation: Since the temperature doesn't change, the amount of thermal energy cannot change, so Ethermal = 0. Since we have no change in the thermal energy, the energy that is being added by the natural gas and the sunshine must transfer out to the surrounding outside air. Hence, Q = -15000 J, where the sign is negative since the energy flows from the system (the house) to the surroundings (the outside air). 019 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. there is no such system 2. Earth only 3. barbell only 4. woman only 5. barbell + Earth correct
1. Ic, IIc 2. Ib, IIb 3. Id, IIb 4. Ib, IId correct 5. Ic, IId 6. Ia, IIa 7. Ia, IIc 8. Id, IIa
Version 123 Midterm 03 florin (56930) 6. woman + barbell 7. woman + Earth 8. woman + barbell + 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.
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