PHYSICS 221 SPRING 2013 EXAM 2: April 4, 2013 8:15-10:15pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions, each worth 3 points. Choose one answer only for each question. Choose the best answer to each question. Answer all questions. Allowed material: Before turning over this page, put away all materials except for pens, pencils, erasers, rulers and your calculator. There is a formula sheet attached at the end of the exam. Other copies of the formula sheet are not allowed. Calculator: In general, any calculator, including calculators that perform graphing, is permitted. Electronic devices that can store large amounts of text, data or equations (like laptops, palmtops, pocket computers, PDA or e-book readers) are NOT permitted. Wireless devices are NOT permitted. If you are unsure whether or not your calculator is allowed for the exam, ask your TA. How to fill in the bubble sheet: Use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. You will continue to use the same bubble sheet that you already used for the first midterm exam. Bubble answers 28-54 on the bubble sheet for this exam. If for some reason you have to use a new bubble sheet write and fill in the bubbles corresponding to: - Your last name, middle initial, and first name. - ««Your ID number (the middle 9 digits on your ISU card) ««- Special codes K to L are your recitation section. Always use two digits (e.g. 01, 09, 11, 13). Honors sections: H1 02; H2 13; H3 25. Please turn over your bubble sheet when you are not writing on it. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over. Best of luck, Drs. Kai-Ming Ho, Eli Rosenberg, and Kerry Whisnant
The following figure is used in Problems 28 and 29. 28. The potential energy for a particle of mass m is shown in the figure above. If the particle is released from rest at point G. Which regions of x are forbidden for the particle to be in? A) x B < x < x F B) x < x A and x > x G C) x C < x < x E D) Can t tell without knowing the value of m E) x < x B and x > x F Solution: The particle can only be where the total energy (indicated by the dashed line) is greater than the potential energy, so it must stay between A and G. Answer: B. 29. At which of the points labeled in the figure does the force on the particle act in the positive direction? A) F and G B) A, D, and G C) A and B D) C and E E) At no point is the force positive Solution: The force is the negative of the slope of U; a positive force requires a negative slope, which occurs at A and B. Answer: C.
30. Assume a potential energy is given by the equation U = x 2 y 2, where U is in J, and x and y are in m. What is the magnitude of the associated force, in N, at the point (x, y) = (3.0 m, 4.0 m)? A) 24 B) 96 C) 72 D) 168 E) 120 Solution: The force is minus the gradient of the potential energy, so! F =! "U î! "U ĵ =!2xy 2 î! 2x 2 yĵ "x "y =!2(3.0)(4.0)2 î! 2(3.0) 2 (4.0) ĵ =!96î! 72 ĵ. Then the magnitude is F = (96) 2 + (72) 2 =120 N. Answer: E. 31. An elastic spring of (k = 98 N/m) is attached to a rigid wall at one end and to a 500 g block at the other end. The system is placed on a horizontal table and the block is pushed against the spring compressing it by 10 cm from equilibrium. The coefficient of kinetic friction between the block and the table is 0.20. If the block is released, how far past the equilibrium point, in cm, will the block reach before it stops for the first time? A) 2.0 B) 4.0 C) 6.0 D) 8.0 E) 10 Solution: The work done by friction is equal to the change in mechanic energy. If we say that the block starts at negative x, then the work done by friction is W =! f "x =! f (x f! x i ). The friction force is f = µ k N = µ k mg, so!µ k mg(x f! x i ) = W = E f! E i = 1 2 kx2 f! 1 2 kx 2 i!2µ k mg(x f! x i ) = k(x 2 f! x i 2 ). This can be solved as a quadratic equation, or you can notice that (x 2 f! x i 2 ) = (x f! x i )(x f + x i ), so that you can divide both sides by (x f! x i ), which gives!2µ k mg = k(x f + x i ), or (putting all quantities in standard units) x f =!x i! 2µ k mg k =!(!0.10)! 2(0.20)(0.500)(9.8) 98 = 0.10! 0.02 = 0.08 m. Answer: D.
32. Two stones, one of mass m and the other of mass 2m, are thrown directly upward with the same velocity at the same time from ground level and feel no air resistance. Which statement about these stones is true? A) The heavier stone will go twice as high as the lighter one because it initially had twice as much kinetic energy. B) At its highest point, the heavier stone will have twice as much gravitational potential energy as the lighter one because it is twice as heavy. C) At their highest point, both stones will have the same gravitational potential energy because they reach the same height. D) Both stones will reach the same height because they initially had the same amount of kinetic energy. E) The lighter stone will reach its maximum height sooner than the heavier one. Solution: With the same initial vertical velocity, the two stones will reach the same height. Since gravitational potential energy is mgh, the heavier stone will have twice as much since its mass is twice as big. Answer: B. 33. A roller coaster of mass 80 kg is moving with a speed of 30 m/s at position A as shown in the figure. The vertical height above ground level at point A is 200 m. Neglect friction. What is the speed of the coaster at point C, in m/s? A) 11 B) 23 C) 30 D) 41 E) 50 Solution: Total mechanical energy is conserved, so KE f + PE f = KE i + PE i. Then mgy f + 1 2 mv 2 f = mgy i + 1 2 mv i2 v f = v i 2! 2g(y f! y i ) = (30) 2! 2(9.8)(160! 200) = 41 m/s. Answer: D.
34. Martha throws a 0.15 kg rubber ball down onto the floor. The ball s velocity just before impact is 6.5 m/s, and just after impact it is 3.5 m/s. If the ball is in contact with the floor for 0.025 seconds, what is the magnitude of the average force applied by the floor on the ball, in N? A) 60 B) 18 C) 30 D) 120 E) 400 Solution: Note that the direction of the velocity changes; we will take the initial velocity to be negative. The average force is F avg =!p!t = m(v f " v i ) (0.15)[6.5" ("3.5)] = = 60 N.!t 0.025 Answer: A. 35. A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring between them. They are not attached to the spring. After they are released and have both moved free of the spring, which one of the following statements is true? A) The lighter block will have more kinetic energy than the heavier block. B) Both blocks will have the same amount of kinetic energy. C) The magnitude of the momentum of the heavier block will be greater than the magnitude of the momentum of the lighter block. D) Both blocks will have equal speeds. E) The heavier block will have more kinetic energy than the lighter block. Solution: Momentum is conserved, so the two momenta have equal magnitudes and opposite directions, and mv = MV. Then since M = 2m, v = 2V. The kinetic energy of the lighter block is therefore mv 2 /2, and the kinetic energy of the heavier block is MV 2 /2 = (2m)(v/2) 2 /2 = mv 2 /4. Thus the correct answer is A. 36. During a snowball fight two balls with masses 0.40 and 0.60 kg, respectively are thrown at each other in a manner such that they meet head-on and combine to form a single mass. The magnitude of the initial velocity for each is 15 m/s. What is the speed of the combined snowball after the collision, in m/s? A) 15 B) 0 C) 7.5 D) 9.8 E) 3.0
Solution: Note that the initial velocities are in the opposite direction, and we will take the velocity of the 0.40-kg mass to be positive. Linear momentum is conserved, so m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v f v f = m 1v 1 + m 2 v 2 0.40(15)+ 0.60(!15) = =!3.0 m/s, m 1 + m 2 0.40 + 0.60 and the final speed is 3.0 m/s. Answer: E. 37. In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is A) perfectly elastic B) partially inelastic C) completely inelastic D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved Solution: The total momentum before and after is 6.0 kg-m/s, so momentum is conserved. Kinetic energy is also conserved, which you can find by calculating it directly or by noting that (v 2f v 1f ) = -(v 2i v 1i ), which is true if kinetic energy is conserved. Answer: A. 38. The cross-section of a uniform piece of plastic is depicted below. What is the x-position of the center of mass? A) 0.33 B) 0.75 C) 0.60 D) 0.50 E) 0.45
Solution: The formula for CM is x CM =! m i x i / M, where M is the total mass. We don t i know the masses of the individual pieces, but each square is the same; call it m. There are two squares with their centers at x = 2.5, one square at x = 1.5, three squares at x = 0.5, and four squares at x = -0.5. Then 2m(2.5)+ m(1.5)+ 3m(0.5)+ 4m(!0.5) x CM = = 0.6. Note that the mass of one square 10m canceled out. Answer: C. 39. In the figure, a weightlifter's barbell consists of two identical uniform solid spherical masses each with radius 0.17 m and mass of 50 kg. The weights are connected by a 0.96-m uniform steel rod with a mass of 12 kg. Find the moment of inertia of the barbell, in kg-m 2, about an axis through the center and perpendicular to the rod (see figure). A) 25 B) 36 C) 44 D) 47 E) 55 Solution: For each sphere we can use the parallel axis theorem, where the distance from the axis to the CM of the sphere is 0.48 + 0.17 = 0.65 m: I = I CM + md 2 = 2 5 mr2 + md 2 = 2 5 (50)(0.17)2 + 50(0.65) 2 = 21.7 kg-m 2. For the rod we just have I CM = 1 12 ml2 = 1 12 (12)(0.96)2 = 0.92 kg-m 2. Adding two spheres and the rod we get I = 21.7 + 21.7 + 0.96 = 44 kg-m 2. Answer: C. 40. A wheel is acted on by three forces as shown in the figure. The force F 1 = 10 N acts at r = 15 cm from the center of the wheel, while F 2 = 20 N and F 3 = 15 N act at R = 30 cm from the center of the wheel. What is the magnitude of the net torque acting on the wheel, in N-m? (You can ignore any possible friction.) A) 3.0 B) 0 C) 14 D) 12 E) 6.8
Solution: In each case, the forces are tangential, i.e., perpendicular to the radial direction, so that the torque in each case is just the force times the radial distance. Calling torque in the CCW direction positive,! =!F 1 r + F 2 R! F 3 r =!(10)(0.15)+ (20)(0.30)! (15)(0.30) = 0. Answer: B. 41. A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw? Assume that both people are small enough compared to the length of the seesaw to be thought of as point masses. A) The side the girl is sitting on will tilt downward B) It is impossible to say without knowing the distances C) The side the boy is sitting on will tilt downward D) It is impossible to say without knowing the masses E) Nothing will happen, the seesaw will still be balanced Solution: Initially the torques are balanced, so m i gr 1 = m 2 gr 2. If both distances are halved, the torques will still balance. Answer: E. 42. A 4.6-m massless rod is loosely pinned to a frictionless pivot at point P. A 2.3-kg ball is attached to the other end of the rod (see figure). The ball is held at point A, where the rod makes a 30-degree angle above the horizontal, and is released from rest. The ball-rod assembly then swings freely in a vertical circle. What is the tension in the rod, in N, when the ball reaches the lowest point at B? A) 29 B) 107 C) 68 D) 45 E) 90 Solution: We can use conservation of mechanical energy to find the speed at the bottom: KE i + PE i = KE f + PE f 0 + mgh = 1 2 mv2 + 0, or v 2 = 2gh. In this case h = R + Rsin! = R(1+ sin30 o ) =1.5R. Then the net force on the ball at point B must give the proper centripetal force: T! mg = mv2 R = m(2gh) R = 3mg, or T = 4mg = 4(2.3)(9.8) = 90 N. Answer: E.
43. A tire is rolling along a road, without slipping, with a velocity v. A piece of tape is attached to the tire. When the tape is opposite the road (at the top of the tire), its velocity with respect to the road is A) v B) 2v C) 0 D) 1.5v E) undeterminable without knowing the radius of the tire Solution: The velocity of a point on the tire is the sum of the CM velocity plus the velocity of the point relative to the CM. For a round object rolling without slipping, v = ωr. The tangential speed of a point on the edge relative to the CM is also ωr, and for a point on the top it also points forward, so the velocity of a point on the top is v + v = 2v. Answer: B. 44. A ball moving horizontally at a constant speed strikes a pendulum hanging on a string from the ceiling. If the ball sticks to the pendulum, what quantities are conserved for the ball-pendulum system in this collision? A) Mechanical energy B) Total linear momentum C) Total angular momentum about the ceiling anchor D) All of the above E) None of the above Solution: Mechanical energy is not conserved because it is an inelastic collision. Linear momentum of the ball-pendulum system is not conserved because of the force acting on the pendulum at the pivot that holds the top end of the string in place. Angular momentum of the ball-pendulum system about the pivot is conserved since the force at the pivot acts at the pivot and there is no torque. Answer: C. 45. A wheel of radius 1.00 m is spun with a constant tangential force of 2.50 N acting at the rim (see figure). What angular displacement, in degrees, is required for this torque to do 5.90 J of work? A) 45.0 B) 75.0 C) 105 D) 135 E) 165
Solution: The work done is W =!!" = FR!", so!! = W FR = 5.90 = 2.36 rad = 2.36 rad180o (2.50)(1.00) 2" =135o. Answer: D. 46. A uniform solid disk of radius 1.6 m and mass 2.3 kg rolls without slipping to the bottom of an inclined plane. The inclined plane makes an angle of 10 o with the horizontal. What is the force of friction acting in the disk, in N? You may assume that the coefficient of static friction is large enough to keep the disk from slipping. A) 1.3 B) 2.6 C) 3.9 D) 5.2 E) 6.5 Solution: The force equation in the direction down the incline is mgsin!! f s = ma. The only force acting on the disk that exerts a torque is friction, with f s R = I!, and the rolling without slipping condition is a = Rα. Then eliminating α, we get f s R =! = a mr2, or ma = f s. Substituting this into the first equation we get I R I mgsin!! f s = f s, or f s = mgsin!. Since I = mr 2 /2 for a solid disk, 1+ mr2 I f s = 1 3 mgsin! = 1 3 (2.3)(9.8)sin10o =1.3 N. Answer: A. 47. When you ride a bicycle in the forward direction, in what direction is the angular momentum of the wheels? A) To your left B) Backwards C) Up D) To your right E) Forwards Solution: Viewed from the left, the rotation is CCW, so, using the right hand rule, the angular velocity vector (and therefore also the angular momentum vector) is towards the viewer, which is to the left of the rider. Answer: A.
48. A carousel has a radius of 3.0 m and a moment of inertia of 3000 kg-m 2. The carousel is rotating unpowered and without friction with an angular velocity of 1.2 rad/s. An 80-kg man with a velocity of 5.0 m/s, on a line tangent to the rim of the carousel, as shown in the figure. When the man reaches the carousel he grabs onto the edge and hangs on. What is the final angular velocity of the carousel after the man jumps on, in rad/s? A) 0.97 B) 1.2 C) 1.3 D) 1.4 E) 5.0 Solution: Angular momentum is conserved. The initial angular momentum comes from both the carousel [L = I! = (3000)(1.2) = 3600 kg-m 2 / s] and from the man[l = mrv = (80)(3.0)(5.0) =1200 kg-m 2 / s], for a total of 4800 kg-m 2 /s. Once the man is on the carousel, he has moment of inertia mr 2 = (80)(3.0) 2 = 720 kg-m 2, so the total moment of inertia is 3000 + 720 = 3720 kg-m 2. Then since angular momentum is conserved, L = I f! f, or! f = L / I f = (4800) / (3720) =1.3 rad/s. Note that this is faster than the initial angular velocity of the wheel; that s because when the man grabs the carousel he has an effective angular velocity of v/r = (5.0 m/s)/(3.0 m) = 1.67 rad/s, faster than the carousel s initial angular velocity. Answer: C. 49. If the net torque on an object is zero, which one of the following statements is true? A) The object could be both rotating and accelerating linearly. B) The object cannot be rotating. C) The object is at rest. D) The forces on it also add up to zero. E) The object could be accelerating linearly but it could not be rotating. Solution: When the net torque is zero, there is no angular acceleration, but it could still be rotating with constant angular velocity. There is no constraint on the net force, so it could be accelerating linearly. Answer: A.
50. A 70.0 kg diver stands on the edge of a uniform diving board of mass 10.0 kg and length 3.80 m. The diving board is supported at two points as shown. What is the magnitude and direction of the force at point A, in N? A) 2800 N, up B) 2010 N, up C) 784 N, up D) 2010 N, down E) 2800 N, down Solution: The net force must be zero and the net torque must also be zero for the board to be stationary. Let m be the mass of the board, and M the mass of the diver. For the torque, since we want to find F A, we will choose point B as the point about which to take torque; that way F B drops out and we can solve for F A. We will choose up to be positive for the forces at A and B. The CM of the board is 0.90 m from B and the diver is 2.8 m from B, so (choosing CCW torque to be positive), 0 =! =!(1.00)F A! (0.90)mg! (2.80)Mg, or F A =!(0.90)mg! (2.80)Mg =!(0.90)(10.0)9.8! (2.80)(70.0)9.8 =!2010 N. The minus sign indicates the force is down. Answer: D. 51. In the figure, a 10-m long bar is attached by a frictionless hinge to a wall and held horizontal by a rope that makes an angle θ of 53 with the bar. The bar is uniform and weighs 40 N. How far from the hinge, in m, should a 10-kg mass be suspended for the tension T in the rope to be 125 N? A) 2.3 B) 4.7 C) 6.4 D) 8.1 E) 9.5 Solution: We take the torque about the hinge, with CCW taken as positive; it must be zero to be in static equilibrium.
LT sin!! xw 1! L 2 w 2 = 0, or T sin!! 1 x = L 2 w 2 125 sin53 o! 1 =10 2 40 = 8.1 m. w 1 10(9.8) Answer: D. 52. A solid sphere of Lead has a radius of 50 cm. When it is placed at the deepest part of the ocean, the pressure is 1.1x10 8 Pa, and the radius of the sphere decreases by 0.045 cm. What is the Bulk Modulus, in Pa? A) 1.1x10 8 B) 4.1x10 10 C) 1.2x10 11 D) 2.4x10 9 E) You can t determine the Bulk Modulus without knowing the density of Lead. Solution: The bulk modulus is B = P/ ΔV/V. The fractional change in volume is 4!V V = V f "V i = 3! R 3 f V i 4 "1= R 3 # f 49.955& 3 "1= 3! R 3 % ( 3 R i $ 50.000 ' "1= "0.0027, so B = 1.1x1010 0.0027 = 4.1x1010 Pa. i Answer: B. 53. Which one of the following statements is false? A) If two objects move closer together, the gravitational force between them increases. B) The center of gravity of an object always lies within the object. C) The moon pulls on the earth with the same magnitude force with which the earth pulls on the moon. D) The gravitational force between two objects is always attractive. E) The planets move in elliptical orbits with the Sun at one focus of the ellipse. Solution: A symmetrical object with a hole in the middle, such as a donut, will have the CM at the center, which is not within object. Answer: B. 54. From what height above the surface of the earth, in km, should an object be dropped to initially experience an acceleration of 0.920 g? The radius of the earth is 6.38 10 6 m. A) 510 B) 160 C) 272 D) 554
E) 385 Solution: The acceleration due to gravity is a = GM the Earth a g = GM / r 2 E GM / R = R 2 E 2 E r, so r = R g 2 E a = (6.38x103 km) Then r! R E = 6652! 6380 = 272 km. Answer: C. r 2, so compared to g at the surface of g 0.92g = 6.652x103 km.
28 B 37 A 46 A 29 C 38 C 47 A 30 E 39 C 48 C 31 D 40 B 49 A 32 B 41 E 50 D 33 D 42 E 51 D 34 A 43 B 52 B 35 A 44 C 53 B 36 E 45 D 54 C