PHYSICS 221 SPRING 2016 EXAM 2: March 31, 2016; 8:15pm 10:15pm Name (printed): Recitation Instructor: Section # Student ID# 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. Fill in questions 28 through 54 for this exam. 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 front 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. 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. 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 24; H4 06; H5 42 - Fill in questions 28 through 54 for this exam 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 the table at the end of the exam and take them with you to compare with the posted answers. 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. Turn off your cell phone (don t just silence it) or leave it at home. Anyone using a cell phone must hand in their work; it will be considered academic dishonesty. Best of luck, Drs. Joseph Shinar, Kerry Whisnant, Kai-Ming Ho, and David Johnston
28. Which of these statements about the collision of two particles is false? A. Momentum is conserved during inelastic collisions. B. Momentum is conserved during elastic collisions. C. The total kinetic energy of the two particles is conserved during elastic collisions. D. The total kinetic energy of the two particles is not conserved during inelastic collisions. E. None of these statements is false. Solution: By definition, an elastic collision conserves total kinetic energy, and an inelastic collision does not. Also, momentum is conserved in either type of collision. Answer: E 29. A small block on a frictionless surface has a mass of 75 g. It is attached to a massless string passing through a hole in a horizontal surface (see diagram). The block is originally rotating in a circle of radius 50 cm with angular speed 0.90 rad/s. The force on the string is reduced until the radius of the circle in increases to 80 cm. What is the final angular speed, in rad/s? [You may treat the block as a point particle.] A. 0.35 B. 0.56 C. 0.90 D. 1.4 E. 2.3 Solution: The force of the string on the block is radial, so there is no torque on the block and angular momentum is conserved. Angular momentum is mvr = mωr 2, so mω i R i 2 = L i = L f = mω f R f 2 and ω f = ω i (R i / R f ) 2 = (0.90)(0.50 / 0.80) 2 = 0.35 rad/s. Answer: A 30. Four identical, uniform square boxes, each 2.0 m on a side, are attached as shown. Where is the center of mass of the system, (x CM, y CM ), when the coordinates are expressed in meters? A. (2.0, 1.0) B. (3.0, 2.0) C. (2.5, 1.5) D. (3.0, 2.5) E. (2.0, 1.5) Solution: The center of mass of each square is at its center. We can then treat the system as four point particles with their positions at the center of the squares. The center of mass is therefore at!! r CM = m i ri / m i = m(1.0,1.0)+ m(3.0,1.0)+ m(5.0,1.0)+ m(1.0,3.0) 4m = (2.5,1.5). Answer: C
31. A boat whose passengers are a cat and a dog is afloat on a lake (treat the water as a frictionless surface). The two animals are initially at the center of the boat but have a small scuffle and walk to opposite ends of the boat, the dog to the right and the cat to the left. If the dog is heavier than the cat, in which direction has the boat moved with respect to the shore? A. The boat hasn t moved B. To the left C. To the right D. It depends on the mass of the boat E. It is impossible to know Solution: There is no net force on the dog-cat-boat system, so the center of mass does not move. Since the dog is heavier, and the two animals move the same distance, the dog s motion affects the center of mass more when it moves. Therefore the boat must move opposite to the dog (in the same direction as the cat) to keep the center of mass of the system stationary. Answer: B 32. A baseball with a mass of 145 g is travelling with a speed of 40 m/s in the horizontal direction after being pitched. The batter hits the ball and it is now travelling in the opposite direction with a speed of 50 m/s. What is the magnitude of the impulse, in kg m/s, experienced by the ball when struck? A. 13 B. 65 C 10 D. 90 E. 0 Solution: The impulse is the change in momentum, so Δp = m(v f v i ) = (0.145)[(40) ( 50)] =13 kg-m/s. Answer A 33. Five forces of equal magnitude are applied within the plane of this page to a flat square plate that is also in the plane of this page as shown. Which one of the five forces exerts zero torque about the point O? A. F 1 B. F 2 C. F 3 D. F 4 E. F 5 Solution: A force that acts through the torque reference point gives no torque;! F 1 acts through point O. Answer: A 34. A sphere rolls without slipping along an inclined plane as shown. If the sphere is rolling up the incline, the direction of the static friction force of the inclined plane on the sphere is. A. directed down the plane B. directed into the page C. directed out of the page D. directed up the plane E. There is no static friction force on the sphere if the sphere is rolling up the plane. Solution: The ball is rotating in the CCW direction as it rolls without slipping up the hill. Since it is slowing down, there must be CW torque about the center of mass. Gravity and the normal force act through the center of mass and produce no torque. The only torque is due to static friction, which must act up the incline to give CW torque. [Another way to think about it is that
an object that is rolling has more kinetic energy than one that is just sliding, so it must go higher up the incline as the kinetic energy is converted to gravitational potential energy. The friction force is what helps push it higher up the incline compared to a sliding object.] Answer: D 35. A disk has a mass M = 1.00 kg and is spinning about a fixed vertical axis with angular speed ω 1 = 140 rad/s. A blob of clay of mass m = 0.200 kg drops vertically down and sticks to the rim of the disk as shown. The system consists of the clay and the disk. The final angular speed ω 2 of the disk plus clay is rad/s. [The moment of inertia of the disk about its rotation axis is I cm = (1/2)MR 2.] A. 50 B. 100 C. 150 D. 200 E. 250 Solution: There is no external torque, so angular momentum is conserved: Iω 1 = L 1 = L 2 = (I + mr 2 )ω 2 1! 2 MR2 ω 1 = 1 $ # 2 MR2 + mr 2 " % &ω =! # 1 2 " 2 M + m $ &R 2 ω 2 % M ω 2 = M + 2m ω = 1.00 1 140 =100 rad/s. 1.00 + 2(0.200) Answer: B 36. The power output of an electric motor is P = 500 W at an angular speed of ω = 955 rev/min. The magnitude τ of the torque produced by the motor is N m. A. 1.0 B. 2.0 C. 3.0 D. 4.0 E. 5.0 Solution: The power for rotational motion is P = τω. Converting the angular speed to rad/s:! ω = 955 rev $! 2π rad $! # &# & 1 min $ # & =100 rad/s. Then τ = P /ω = 500 /100 = 5.0 N-m. Answer: E " min %" rev %" 60 sec % 37. A non-uniform rod of mass M and length L is supported at its ends by two light wires, as shown in the figure. The tension in the left wire is one-third of the tension in the right wire. At what value of x is the center of mass of the rod? A. L/4 B. L/3 C. L/2 D. 2L/3 E. 3L/4 Solution: If the rod is stable, the torque about the CM is zero. Then if the CM is at x along the rod, we have T L x = T R (L x). Since T L = T R /3, then x = 3(L x) or x = 3L/4. [Note: the angle didn t affect the answer.] Answer: E
38. A uniform solid cylinder of mass m = 2.0 kg and radius r = 30 cm rotates about a fixed horizontal axis. A massless rope is wrapped around the cylinder and the end of the rope is attached to a box of mass M = 4.0 kg. If the rope does not slip on the cylinder as the mass falls, what is the magnitude of the angular acceleration of the cylinder, in rad/s? A. 14 B. 18 C. 22 D. 26 E. 30 Solution: Taking down to be positive, Newton s 2 nd law for the box is mg T = ma, where T is the tension in the rope. We also have for the rotation of the cylinder Tr = τ = Iα = (mr 2 / 2)α. Finally, if the rope doesn t slip, a = αr. Solving these three equations M g for the angular acceleration we get α = M + 1 2 m r = 4.0 9.8 4.0 + 1 = 26 rad/s. Answer: D 2 2.0 0.30 39. A 500 kg mass is to be hung vertically from a 1.5-m long steel cable. The cable has a circular cross section. If we require that the cable stretch no more than 0.10 mm, what is the minimum radius of the cable, in mm? [The Young s modulus of steel is 20 x 10 10 N/m 2. Ignore the mass of the cable.] A. 0.37 B. 1.8 C. 8.8 D. 11 E. 19 Solution: The definition of Young s modulus is Y = (F / A) / (ΔL / L o ) = (mg / π R 2 ) / (ΔL / L o ), so R = mgl 0 πyδl = (500)(9.8)(1.5) π (2.0 10 11 )(1.0 10 4 ) =1.1 10 2 m =11 mm. Answer: D 40. The comet Kobayashi has a perihelion distance to the Sun of 3.08 x 10 11 m and aphelion distance to the Sun of 2.27 x 10 12 m. What is its period as it revolves around the Sun, in years? [The mass of the Sun is 1.99 x 10 30 kg and one year is 3.16 x 10 7 s.] A. 2.95 B. 8.64 C. 25.3 D. 38.6 E. 59.0 Solution: The period is given by T = 2π the sum of the perihelion and aphelion. So T = 2π (1.289 10 12 ) 3 (6.67 10 11 )(1.99 10 30 ) = 7.98 108 s = 25.3 yr, where a is the semi-major axis, which is half and a 3 / GM a = (r P + r A ) / 2 =1.289 10 12 m. Answer: C
41. A particle is subjected to the force associated with the potential energy function U(x) shown below. No other forces are exerted on the particle. The particle s total mechanical energy E is given by the horizontal line marked E. Its turning point and point of maximal speed are and, respectively. A. 1, 3 B. 2,3 C. 1,5 D. 3,9 E. 3, 1 1 E 2 4 5 6 8 9 7 3 Solution: The total mechanical energy is E = K + U. The turning point occurs when the particle stops (K = 0), i.e., when U = E. This is at Point 1. The maximal speed occurs when K = E U is the largest, which occurs when U is the smallest, i.e., at Point 3. Answer: A 42. A particle is subject to the potential energy function U(x) = 5x 2 + 7x 12, where U is in J and x is in m. At what value of x, in m, is there equilibrium, and what kind of equilibrium is it? A. 1.0, stable B. 0, neutral C. 0.70, unstable D. -2.0, stable E. 2.4, unstable Solution: Equilibrium occurs when the force is zero. The force is F = du / dx, or F(x) = ( 10x + 7), which will be zero when x = 0.7. Also, d 2 U / dx 2 = 10 ; since this is negative (it s an inverted parabola), it s unstable. Answer: C
43. A block of mass m is pushed against a spring with constant k. The right end of the spring is initially compressed towards the left by a distance x from its equilibrium position x = 0 and then released from rest. The coefficient of kinetic friction between the block and the table is µ k. The speed of the block when its left side is at x = 0 is A.!!!!!!"!! B.!!!!!!"!! C.!!!! μ!gx D.!!!! 2μ!gx E.!!!! 4μ!gx Solution: The work done by friction (which is negative) is the change in total mechanical energy, so f k x = µ k Nx = E f E i = 1 2 mv2 1 2 kx2. The normal force is N = mg. Solving for v we get v =!!!! 2μ!gx. Answer: D 44. Consider a pendulum of length l and mass m that is released from rest at an angle θ0. The maximum angular and centripetal accelerations are at the angles and, respectively. A. θ0, θ0 B. θ0, 0 C. 0, θ0 D. 0, 0 E. θ0/2, θ0/2 Solution: The angular acceleration is α = τ / I = mgsinθ / I, so the maximum acceleration occurs at the largest angle, i.e., θ 0. The maximum kinetic energy, and hence maximum speed, occurs when the gravitational potential energy is the smallest, which occurs at the bottom. Answer: B 45. An electric clock is hanging on a vertical wall. The hands of the clock are moving in the clockwise direction. Its battery is losing power and the clock is slowing down. What is the direction of the angular acceleration vector of the minute hand? A. To the right, B. To the left, C. Up, D. Into the wall, E. Out from the wall, Solution: The angular velocity, using the right-hand rule, is into the wall. Since it is decreasing in magnitude, and the angular acceleration is related to the change of the angular velocity, the angular acceleration is directed out from the wall. Answer: E
46. A car and a truck collide at an intersection and stick together. The truck, traveling north, has mass 1200 kg and the car, traveling east, has mass 800 kg. The final velocity of the pair is 16 m/s in a direction 24 o east of north, as shown in the figure. What was the speed of the car before the collision, in m/s? A. 12 B. 16 C. 24 D. 36 E. 40 Solution: Momentum is conserved. In the x direction (east- west) we have m c v c = (m c + m t )v f sin24, or v c = (m c + m t )v f sin24 / m c =16 m/s. [Note that we didn t need the y equation it could have been used to find the initial speed of the truck.] Answer: B 47. Planet Z is known to have a radius that is 2.2 times that of Earth. A future explorer lands on Planet Z and finds that the acceleration due to gravity at the surface is 8.0 m/s 2. What is the average mass density of Planet Z compared to that of Earth (i.e., ρ Z / ρ E )? You may assume that the Earth and Planet Z are both spherically symmetric. A. 0.37 B. 0.56 C. 1.8 D. 2.7 E. 3.6 Solution: Surface gravity is g = GM / R 2, so = GM Z = R 2 E (4π R 3 Z )ρ Z = ρ Z R Z, or 2 2 g E R Z GM E R Z (4π R 3 E )ρ E ρ E ρ Z = g Z R E = 8.0 R E = 0.37. Answer: A ρ E g E R Z 9.8 2.2R E g Z R E 2 R E 48. A mass m 1 is travelling with velocity v when it strikes another mass m 2 which is initially at rest. If the collision is completely inelastic, what is the ratio of the final kinetic energy to the initial kinetic energy? A. 2 B. (m 1 + 2m 2 )/m 1 C. m 1 /(m 1 + m 2 ) D. m 2 /m 1 E. m 1 /m 2 Solution: Momentum is conserved in the collision, so m 1 v = (m 1 + m 2 )v f. The ratio of final to = (m 1 + m 2 )v 2 f / 2 = m + m 1 2 m 1 v 2 / 2 initial kinetic energy is therefore K f m # 1 v& K i m 1 " m 1 + m 2 % that this is always less than one, so kinetic energy is necessarily lost.] Answer: C! $ 2 1 v 2 = m 1 m 1 + m 2. [Note
49. A uniform solid sphere of radius R = 2.236 m and mass M = 0.143 kg is welded to a horizontal massless rod of length L = 2.236 m. The assembly rotates about a vertical axis at the left end of the rod as shown in the figure with an angular speed of ω = 4.00 rad/s. The moment of inertia of the sphere about the axis of rotation is kg m 2. [The moment of inertia of a uniform solid sphere about an axis through its center of mass is (2/5)MR 2.] A. 0.286 B. 1.00 C. 1.72 D. 3.14 E. 6.72 Solution: Using the parallel axis theorem, the moment of inertia about an axis that goes through the edge of the sphere is I = I CM + Md 2 = 2MR2 + M(R + L) 2 = 2 5 5 (0.143)(2.236)2 + (0.143)(4.472) 2 = 3.14 kg-m 2. Answer: D 50. A 1.0 kg mass slides up a 35 degree incline with an initial speed of 6.0 m/s. The coefficient of kinetic friction is 0.040. How much work does friction do on the mass by the time it comes to a stop 1.74 m above the initial position? A. 0.97 J B. 2.06 J C. 1.15 J D. 1.50 J E. 1.25 J Solution: The change in total mechanical energy is equal to the (negative) work done by friction. So W = E f E i = mgh mv 0 2 / 2 = (1.0)(9.8)(1.74) (1.0)(6.0) 2 / 2 = 0.95 J. You could also solve this by using Newton s 2 nd law; the frictional force is µ N = µ k mgcosθ and the distance is d = h / sinθ, so W = f k d = µ k mghcotθ = (0.04)(1.0)(9.8)(1.74)cot35 = 0.97 N. The difference in the two answers is due to round-off error. Answer: A 51. A yo-yo is placed on a horizontal surface as shown. There is sufficient friction for the yo-yo to roll without slipping. If the string is pulled, then 1 2 3 A. 1, 2 and 3 all roll to the right B. 1 and 3 roll to the right, and 2 rolls to the left C. 1 and 2 roll to the right, and 3 rolls to the left D. The answer depends on the magnitude of the applied force relative to the force of friction E. 1, 2 and 3 all roll to the left
Solution: All of the forces except F (normal, gravity, and friction) act through the contact point, so the only torque about the contact point is from F. In each case, F tends to create CW rotation about the contact point, which will result in rolling to the right if it rolls without slipping. Answer: A 52. Five objects with the same mass and radius are rolling across a horizontal surface with the same center of mass speed. They then roll up an incline that makes an angle of 20 o with respect to the horizontal. Which object moves the farthest up the ramp? Assume each object rolls without slipping. A. A solid sphere B. A hollow sphere C. A solid cylinder D. A hollow cylinder E. They all go the same distance up the ramp Solution: Total mechanical energy is conserved, where the kinetic energy has both translational and rotational parts. The initial K at the bottom is changed into U at the top. So 1 2 mv2 + 1 2 Iω 2 = mgh. If it rolls without slipping, ω = v / r ; substituting in and multiplying by! 2/m we get v 2 1+ I $ # & = 2gh. Clearly whichever object has the largest value of I/mR 2 will go " mr 2 % the highest if they start with the same speed; that is a hollow cylinder, with I = MR 2. [Note that this does not depend on the exact values of m and R, only on the ratio I/mR 2, which is determined by the shape. It also doesn t depend on the angle.] Answer: D 53. A uniform beam of mass 50 kg is attached to the ground at the pivot point P and supported by a wire attached to the top end of the beam (see diagram). If the system is in static equilibrium, what is the tension in the wire, in N? A. 180 B. 470 C. 670 D. 760 E. 980 Solution: In static equilibrium the net force is zero and the net torque is zero. The angle between the wire and the beam can be determined to be 15 o. Taking the torque about point P and setting it equal to zero, we get TLsin15 = mg(l / 2)sin 45, so
T = mgsin 45 / (2sin15 ) = (50)(9.8)sin 45 / (2sin15 ) = 670 N Note that we don t need the length of the beam. Answer: C 54. In a Newton s cradle experiment, a steel ball of mass 2m has an elastic collision with a stationary hanging ball of mass m. If the ball of mass m rises to a height of h after the collision, what is the speed of the ball of mass 2m just before the collision? A. 2gh B. 9gh / 8 C. 3gh / 2 D. 2gh / 3 E. 3gh Solution: The collision is elastic, so we can use both conservation of momentum [ (2m)v 0 = (2m)v 2 + (m)v 1, where v 2 and v 1 are the velocities of 2m and m right after the collision] and the 1-D conservation of energy equation [ v 0 0 = (v 2 v 1 ) ] in which the relative velocity changes sign. Solving for the final velocities we get v 1 = 4v 0 /3 and v 2 = v 0 /3 [note: both are moving to the right after the collision]. The kinetic energy of m gets converted to gravitational potential energy, so mv 2 1 / 2 = mgh, or v 1 = 2gh. Then v 0 = 3v 1 /4, or v 0 = 3. Answer: B 4 2gh = 9 16 2gh = 9 8 gh
31 B 41 A 51 A 32 A 42 C 52 D 33 A 43 D 53 C 34 D 44 B 54 B 35 B 45 E 36 E 46 B 37 E 47 A 28 E 38 D 48 C 29 A 39 D 49 D 30 C 40 C 50 A