001 Version MIDTERM 02 REVIEW II turner (56910) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. SlidingBlockMITestQ 001 (part 1 of 2) 10.0 points Suppose a block of mass M is sliding down an inclined plane which makes an angle with the horizontal. The coefficient of kinetic friction between the plane and the block is (assume tan > ). The block is initially at rest. What is the work done by friction as the block travels a distance d along the plane? 1. M g tan d 2. M g cos d 3. M g sin d 4. - M g cos d correct 5. - M g tan d 6. - M g sin d Explanation: From the free-body diagram of the block, the normal reaction force N = M g cos . The frictional force Ff = M g cos . Since the block moves in a direction opposite to that of friction, Wf = -Ff d = - M g cos d. 002 (part 2 of 2) 10.0 points What is the speed of the block at this point? 1. 2. 3. 2 g d (sin - tan ) 2 g d (sin - cos ) 2 g d (sin + cos ) or
2 vf = 2 g d (sin - cos ).
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4. 5. 6.
2 g d (tan - cos ) 2 g d (sin - cos ) correct 2 g d (sin - cos )
Explanation: Ei = M g d sin . The final energy 1 2 Ef = M vf . 2 Therefore, energy conservation Ei + Wf = Ef implies, 1 2 M g d sin - M g cos d = M vf , 2
Therefore, vf = 2 g d (sin - cos ).
PlanetsMITestQ 003 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. 16 2. 1/4 3. 1/8 4. 2 correct 5. 1/16 6. 8 7. 4
Version 001 MIDTERM 02 REVIEW II turner (56910) 8. 1/2 Explanation: Since A has a circular orbit,
2 m vA GmM = , R R2
2
So the x component is -0.66 kg m/s . 005 (part 2 of 2) 10.0 points What is the change in kinetic energy of the fan cart over this 1.5 s time interval? Correct answer: -0.308314 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: v = p m -0.66, 0, 0 kg m/s = 0.84 kg = -0.785714, 0, 0 m/s .
which implies GmM 1 2 . KEA = m vA = 2 2R Similarly, for B,
2 m vB GM m = , 2R 4 R2
which implies GM m 1 2 . KEB = m vB = 2 4R Therefore, KEA = 2. KEB
So the final velocity is vf = vi + v = 0.0742857, 0, 0 m/s . Now, the change in kinetic energy is given by
FanCartMI06x046 004 (part 1 of 2) 10.0 points A fan cart of mass 0.84 kg initially has a velocity of vi = 0.86, 0, 0 m/s. Then the fan is turned on, and the air exerts a constant force of F = -0.44, 0, 0 N 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 along the x direction, the other components of momentum will not change.) Correct answer: -0.66 kg m/s. Explanation: The change in momentum is given by p = Fnet t = ( -0.44, 0, 0 N)(1.5 s) = -0.66, 0, 0 kg m/s .
KE = KEf - KEi 1 1 2 2 = m vf - m vi 2 2 1 = (0.84 kg) (0.0742857 m/s)2 2 1 - (0.84 kg) (0.86 m/s)2 2 = -0.308314 J . PullingBlockMI06x056 006 (part 1 of 2) 10.0 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?
Version 001 MIDTERM 02 REVIEW II turner (56910) 1. M , same momentum 2. Same kinetic energy, same momentum 3. m, M 4. Same kinetic energy, m 5. M , m 6. M , M 7. m, same momentum 8. Same kinetic energy, M correct 9. m, m 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 = p2 f 2m , 8. m, M 9. M , same momentum
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Explanation: If you pull the blocks for the same amount of time, then according to the Momentum Principle, p = Fnet t 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. AsteroidMI06p083 008 10.0 points The escape speed from an asteroid whose radius is 10 km is only 10 m/s. If you throw a rock away from the asteroid at a speed of 20 m/s, what will be its final speed? Use G = 6.7 10-11 N m2 /kg2 . Correct answer: 17.3205 m/s. Explanation: First use the escape speed to get the mass of the asteroid: 2GM R 2R 1v M = 2 G = 7.46269 1015 kg . vesc = Now, if vi = 20 m/s, then vf is found from Ei = Ef : Ui + K i = Uf + K f GM m 1 1 2 2 - + m vi = 0 + m vf ri 2 2 2GM 2 vf = vi - R = 17.3205 m/s .
so the larger mass block will have the larger final momentum. 007 (part 2 of 2) 10.0 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? 1. Same kinetic energy, m 2. M , m 3. M , M 4. Same kinetic energy, same momentum 5. m, m 6. m, same momentum correct 7. Same kinetic energy, M
Version 001 MIDTERM 02 REVIEW II turner (56910) PendulumMI06p107 009 (part 1 of 2) 10.0 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, , measured from the vertical. Set U = 0 at the location of the mass when the pendulum is hanging straight down. 1. U = m g L sin() 2. U = -m g L cos() 3. U = m g L cos() 4. U = m g L 5. U = m g L(1 - sin()) 6. U = m g L(1 - cos()) correct 7. U = -m g L sin() 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() from L. The first value is the vertical distance of the mass from the pivot, while hanging at an angle of , 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()) U =m g L(1 - cos()) . 010 (part 2 of 2) 10.0 points Suppose that you hit the stationary hanging mass so it has an initial speed vi . What is the minimum initial speed needed the for pendulum to go over the top ( = 180 )? 1. vi = 2. vi = 3. vi = 4 4. vi = 5. vi = 2 6. vi = 8 7. vi = 1 2 gL gL 2 gL gL g L correct gL 2gL
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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 2 m vi + 0 =0 + m g L(1 - cos(180 )) 2 2 vi =2 g L(2) vi =2 gL
FusionRxnMITestQ 011 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 = 16 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.
Version 001 MIDTERM 02 REVIEW II turner (56910) Correct answer: 6.3064 10 . Explanation: The rest mass energy released by one of these reactions is
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energy per second. The mass of water flowing through the plant per second is M = V . The potential energy change of this mass is: U = -M gh = -V gh.
Therefore, the electrical energy produced per (mHe-3 + mHe-4 - mBe ) c2 = 2.5371 10-13 J. second is this potential energy times the conversion efficiency: In order to get the number of reactions to P = k(-U ) = kV gh. produce energy E, we divide E by the energy per reaction that is found above: E 16 J = 2.5371 10-13 J 2.5371 10-13 J = 6.3064 1013 . NiagaraFallsAlgebraicMI07X11 012 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 . 1. P = kV gh2 2. P = 3. P 4. P 5. P 6. P 7. P gh2 kV V gh2 = k V gh = k kgh = V kgh2 = V = kV gh correct gh kV NonLinearSpringWorkMI06 013 10.0 points The force due to a deflected cantilever beam can be modeled using a linear and cubic term, i.e. Fbeam = -k1 - k3 3 where is the deflection at the end of the beam. What work must be done on the beam to deflect it by 10 cm if k1 = 92 N/m and k3 = 15000 N/m3 ? Correct answer: 0.835 J. Explanation: The work done on the beam is given by Won beam = -Wby beam
max
=-
0 max
Fbeam d k1 + k3 3 d
=
0
=
1 1 k1 2 + k3 4 2 4
max 0
= 0.835 J HeavyNucleiMI06p102s 014 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
8. P =
Explanation: The electrical power of a hydroelectric plant is the potential energy converted to electrical
Version 001 MIDTERM 02 REVIEW II turner (56910) fewer protons and two fewer neutrons than 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. 1.26367 10-12 J 2. 9.88962 10-13 J 3. 1.09885 10-12 J correct 4. 9.3402 10-13 J 5. 1.15379 10-12 J 6. 1.31862 10-12 J 7. 1.37356 10-12 J 8. 8.79078 10-13 J 9. 1.20873 10-12 J 10. 1.0439 10-12 J Explanation: We begin by using conservation of energy to write Ei = Ef Erest,Ra = Erest,Rn + Erest, + KRn + K = Erest,Rn + Erest, + = Erest,Rn + Erest, + p2 2 1 1 + mRn m p2 p2 Rn + 2mRn 2m K + Ug
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Note that p2 = p2 since the original par Rn 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 p2 1 1 . + 2 mRn m Now we solve for p : = Erest,Ra - Erest,Rn - Erest, = mRa c2 - mRn c2 - m c2 = 1.11921 10-12 J p2 = 2(1.11921 10-12 J) 1 1 + mRn m p = 1.20834 10-19 kg m/s .
Don't forget to make the appropriate conversions. Now that we know the momentum, the kinetic energy is just KE = p2 2 m
= 1.09885 10-12 J . GravitationalPotentialPlanetIM6x078mc 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 . 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?
Version 001 MIDTERM 02 REVIEW II turner (56910) 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 . 1. Statements I, III, IV 2. Statements I, II, III, VI 3. Statements I, IV, V, VI 4. Statements II, III, IV 5. Statements I, III, IV, V 6. Statements III, V, VI 7. Statements II, III, VI correct 8. Statements I, II, III, IV 9. Statements II, IV, V, VI 10. Statements I, II, III 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.
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