PHYSICS 221 SPRING 2015

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1 PHYSICS 221 SPRING 2015 EXAM 2: April 2, :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 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, H4 06, H5 49. 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. Lekha Adhikari, Rana Biswas, Kai-Ming Ho, and Kerry Whisnant

2 28. An object of mass M is subject to the force associated with the U(X) shown below, and is released from rest at point B. The object has maximum speed at A) point A B) point B C) point C D) point D E) point E U(x) B C D A E Position (x) Solution: The kinetic energy (and hence the speed) is largest when the potential energy is the smallest. Answer: C 29. A massless rod of length, L = 1.0 m, is used to support a cylinder of weight 8.0 N, as shown in the figure. In each of three different cases, a force of different magnitude is holding the rod and the cylinder at rest. The rod is fixed at its left end using a pin and the 1.00 m long rod is equally divided into four equal segments, each of length 0.25 m. Rank the magnitude of the force acting on the rod in each of the three cases. A) F B > F A > F C B) F A = F B = F C C) F B = F A > F C D) F B < F A < F C E) None of the above F A F B F C Solution: The torque due to the tension must cancel the torque due to the weight. The torque is equal to the force times the moment arm, which in this case is the distance from the left end. The torque due to the weight is largest for Case B since that has the longest moment arm, and the moment arm for the force is smallest, so Case B must have the largest force. The moment arm for F is the same for A and C, and the torque due to the weight is the smallest for C, so the force in Case C is the smallest. Answer: A 30. A baseball of mass 0.20 kg travelling at 50 m/s is struck by a baseball bat and travels back in the opposite direction at a speed of 40 m/s. If the baseball was in contact with the bat for 2.0 ms what is the average force exerted on the bat, in kn? A) 9.0 B) 8.0 C) 6.0 D) 3.0 E) 1.0

Solution: The average force is the change in momentum divided by the time over which the force acts, so F ave =!p!t = m(v f " v i ) (0.20)[("40)" 50] = = 9000 N. Answer: A!t 0.002 31.

A) The momentum of the ball is conserved B) The kinetic energy of the ball is conserved C) The total mechanical energy of the ball is conserved D) Both the momentum of the ball and the total

3 Solution: The average force is the change in momentum divided by the time over which the force acts, so F ave =!p!t = m(v f " v i ) (0.20)[("40)" 50] = = 9000 N. Answer: A!t A ball falls down vertically from a ledge on top on a building. Ignore air resistance. Which one of the following statements is true? A) The momentum of the ball is conserved B) The kinetic energy of the ball is conserved C) The total mechanical energy of the ball is conserved D) Both the momentum of the ball and the total mechanical energy are conserved E) The weight of the ball does no work on the ball during this process Solution: The speed of the ball is increasing as it falls, so both kinetic energy and momentum are increasing in magnitude, but without frictional losses the total mechanical energy is conserved. Answer: C 32. A horizontal, uniform disk of mass m and radius r rotates without friction about its axis of symmetry with angular velocity ω. Four weights, each with mass m, are added to the edge of the disk, as shown in the figure, without introducing any external torque. What is the new angular velocity ω of the system? A) ω/9 B) ω/5 C) ω/3 D) 6ω/5 E) ω/7 Before masses added: After masses added: ω ω ' r r Solution: Total angular momentum of the system will be conserved since the only forces are frictional forces, which by Newton s third law are equal and opposite and therefore have equal and opposite torques, giving no net torque on the system. The moment of inertia of the disk is mr 2 /2. The moment of inertia of the added masses is 4mr 2 since they are all placed at the outer rim. Therefore

4 L i = L f! I i! = I f!"! and! =! / 9. Answer: A 1 # 2 mr2! = 1 & % 2 mr2 + 4mr 2 (!" = 9 $ ' 2 mr2!" 33. Consider the in-class demonstration with 2 gliders on an air track. Initially glider A is moving with speed v in the +x direction toward another stationary glider B. After the collision gliders A and B stick together in a completely inelastic collision and move with a common speed v in the +x direction. Which one of the following statements is true? A) The kinetic energy is conserved B) The linear momentum is conserved C) Both the linear momentum and kinetic energy are conserved D) The final speed v can be equal to the initial speed v E) The velocity of the center of mass changes before and after the collision Solution: When objects stick together, kinetic energy will not be conserved. Linear momentum will always be conserved in any collision (and therefore the velocity of the center of mass remains constant). Answer: B 34. An electric motor exerts a constant 10 N-m torque on a grindstone, which has a moment of inertia of 2.0 kg-m 2 about its shaft. The system starts from rest and rotates about its shaft. Find the average power P av, in W, delivered by the motor in 10 s. A) 2.5 B) 5.0 C) 125 D) 250 E) 500 Solution: P =!", so the average power will be P =!". We have! = (! i +! f ) / 2 = (0 +! f ) / 2 =! f / 2,! f =! 0 +"t = "t, and! = " / I. Putting it all together we get P =! 2 t / (2I) = (10) 2 (10) /[2(2.0)] = 250 W. Answer: D 35. A uniform sheet of metal is cut into the shape shown in the figure. Find the x component, in m, of the position of the center of mass of this object. A) 2.5 B) 0.50 C) 1.3 D) 2.3 E) 2.4 Solution: We can treat this as a small 1.0 m by 1.0 m square centered at (0.50, 0.50)

5 plus a 3.0 m by 3.0 m square at (2.5, 1.5). The big square has nine times the area and therefore nine times the mass of the small square. The CM in the x direction is x CM = m(0.5)+ 9m(2.5) m + 9m = 23 = 2.3 m. Answer: D A uniform rod of length L, and mass M = 3.0 kg, is supported on its left end by a massless hinge fixed to the wall. The other end of the rod is connected to a cable, as shown in the figure. Find the tension T in the cable, in N, if the rod does not move. A) 46 B) 30 C) 23 D) 19 E) 15 L T 40 Solution: The net torque must be zero. Taking the torque about the left end we get T sin!(l)! mg(l / 2) = 0 or T = mg / (2sin!) = (3.0)(9.8) /[2(sin 40 )] = 23 N. Answer: C Note that by taking the torque about the left end, the unknown force acting at the left end dropped out of the torque equation and the tension was easy to find. 37. A block of mass m = 3.0 kg hangs at the end of a massless rope that is wrapped around a pulley (see figure). The pulley is a uniform disc of mass M = 7.0 kg and radius R = 0.50 m that rotates without friction about an axis through its center. The rope does not slip on the pulley. Calculate the linear acceleration, in m/s 2, when the mass m starts moving downward. A) 4.5 B) 9.8 C) 6.0 D) 9.0 E) 3.3 Solution: Defining down as positive, the equation of motion (Newton s second law) for the mass is mg!t = ma. The equation of motion for the rotating pulley is TR =! = I". The linear and angular acceleration are related by! = a / R if the rope does not slip on the pulley. Eliminating T and α from these equations, we get a = g / (1+ I / mr 2 ). In this case I = MR 2 / 2, where M is the mass of the pulley, so a = g /[1+ (MR 2 / 2) / mr 2 ] = g /[1+ M / (2m)] = 9.8 /[1+ 7 / (2!3)] = 4.5 m/s 2. Answer: A 38. An object, initially at rest at! = 1.0 m, is subject to the potential energy given by!! =!! 3!! 4! + 6 where x is in meters and U is in Joules. When the object is released, it A) will accelerate in +x direction B) will accelerate in x direction C) is in stable equilibrium

6 D) is in neutral equilibrium E) is in unstable equilibrium Solution: The force is F x =!du / dx =!(3x 2! 6x! 4) = 3x 2 + 6x + 4. Evaluating at x = 1.0 we get F x = +7 N. Answer: A 39. If the net torque on an object is zero, then A) the forces on it also add up to zero B) it can be accelerating linearly but it cannot be rotating about its center of mass C) the object cannot be rotating about its center of mass D) the object is at rest E) the object can be both accelerating linearly and rotating about its center of mass Solution: If the net torque is zero, there is no angular acceleration but it could have constant nonzero angular velocity (this is the rotational equivalent of Newton s first law an object can remain in motion at constant velocity if there is no net force acting on it). So it can be turning. The zero torque condition does not say anything about whether the net force is zero, so the object can be accelerating. Answer: E 40. A uniform thin rod (mass M = 10.0 kg, length L = 1.00 m), initially at rest, is allowed to rotate about its left end on a horizontal frictionless plane. A bullet of mass m = 10.0 gm, is fired horizontally with the velocity v in the direction perpendicular to the rod, as shown in the figure. The bullet remains embedded in the rod and the combined object (the rod and the bullet) rotates about the fixed left end of the rod with an angular velocity of magnitude 2.00 rad/s. What is L the magnitude of the velocity v of O the bullet, in m/s? Axis of rotation A) 167 B) 169 C) 669 D) 969 E) 1000 Velocity of bullet= v Solution: This is a rotational equivalent of a perfectly inelastic collision; angular momentum about the axis of rotation will be conserved. The initial angular momentum is mvl. After the bullet becomes embedded in the rod, the two rotate together. Then

! mvl = I f! f = 1 $ # 3 ML2 + ml 2 " % &! ' v =! # 1 f " 3 M + m $ & L! f % m =! 1 $ # 10.0 + 0.010& (1.00)(2.00) = 668 m/s. " 3 % 0.01 Answer: C 41.

7 ! mvl = I f! f = 1 $ # 3 ML2 + ml 2 " % &! ' v =! # 1 f " 3 M + m $ & L! f % m =! 1 $ # & (1.00)(2.00) = 668 m/s. " 3 % 0.01 Answer: C 41. The acceleration due to gravity on the surface of the dwarf planet Pluto is 0.66 m/s 2, and the radius of Pluto is 1.15 x 10 6 m. What is the density of Pluto, in kg/m 3? A) 6.7 B) 4.3 x 10 2 C) 2.1 x 10 3 D) 5.5 x 10 3 E) 1.6 x 10 4 Answer: Surface gravity is given by g = GM / R 2 and the mass is M =!V = 4" R 3! / 3, so g = 4! R"G / 3 and therefore! = 3g / (4" RG) = 3(9.8) /[4(3.14)(1.15!10 6 )(6.67!10 "11 )] = 2.1!10 3 kg/m 3. Answer: C 42. Four objects, each with mass M and radius R, roll without slipping up an inclined plane that makes a 30 o angle with the horizontal. If they all start at the bottom of the incline with the same linear speed v, which object is highest when it stops? A) They all reach the same height B) Hollow spherical shell C) Uniform, solid cylinder D) Uniform, solid sphere E) Hollow cylindrical shell Solution: The objects have the same translational kinetic energy Mv 2 /2 and the same angular velocity ω = v/r. The object with the biggest moment of inertia will have the most initial rotational kinetic energy Iω 2 /2 and therefore the most initial total mechanical energy, and therefore will go the highest before stopping. The hollow cylindrical shell has I = MR 2, while the others are 2MR 2 /5, MR 2 /2, 2MR 2 /3, respectively, Answer: E 43. A uniform, horizontal bench seat with length 4.00 m and weight 100 N has supports 1.00 m from each end. A large 500 N marble ball is placed at the extreme right edge of the bench, as shown. What is the force on the seat at the left support, assuming the ball does not move? A) 500 N, up B) 600 N, up C) 300 N, up D) 400 N, down E) 200 N, down

8 Solution: Let the force acting up at the left support be F 1 and the force acting at the right support be F 2. For static equilibrium, the sum of the vertical forces must add to zero, so F 1 + F 2!100! 500 = 0, or F 1 + F 2 = 600 N. The sum of torques about any point must also be zero; if we take torque about the right support and call CCW positive we get!f 1 (2.0)+100(1.0)! 500(1.0) = 0, or F 1 =!200 N. The minus sign tells us that the force at the left support is actually down in this case. Answer: E 44. A satellite that will explore the outer planets launches from the surface of the Earth (mass 5.97 x kg and radius 6.38 x 10 6 m) with speed 15 km/s. How fast is it going, in km/s, when it reaches the Moon (3.84 x 10 8 m from Earth)? Consider only the effects of the Earth s gravity. A) 0.50 B) 10 C) 3.4 D) 6.3 E) 12 Solution: Using conservation of total mechanical energy 1 2 mv 2 i! GmM r i = 1 2 mv 2 f! GmM, r f or solving for v f " v 2 f = v 2 i + 2GM 1! 1 % $ # r f r ' = " (1.5(104 ) 2 + 2(6.67(10!11 )(5.97( ) i & 3.84 (10! 1 % $ ' =1.0 (10 8 # (10 6 & or v f = 1.0 x 10 4 m/s = 10 km/s. Answer: B 45. A block of mass M 1 = 1.0 kg, moving to the right (+x direction) with a speed of 5.0 m/s, hits a second block of mass M 2 = 2.0 kg, moving with a speed of 2.0 m/s to the right (+x direction). If the collision between the blocks is elastic and one-dimensional, what is the final velocity of the block of mass M 2? Assume there is no friction. A) 5.0 m/s to the right B) 4.0 m/s to the right C) 2.0 m/s to the right D) 1.0 m/s to the right E) 7.0 m/s to the right V1 =5.0 m/s M1 V2 =2.0 m/s M2 +x direction Solution: Momentum and energy are conserved, which gives m 1 v 1i + m 2 v 2i = m 1 v 1 f + m 2 v 2 f and v 2 f! v 1 f =!(v 2i! v 1i ). Thus v 1 f + 2v 2 f = (1)(5)+ (2)(2) = 9 and v 2 f! v 1 f =!(2! 5) = 3, and solving for v 2f we get +4.0 m/s. Answer: C

9 46. A hockey player of mass m 1 = 80 kg collides with another player of mass m 2 = 80 kg that is initially at rest. The final speed of the player 1 is v 1f = 6.0 m/s. He comes out at an angle θ 1 = 30 with respect to his original direction. Player 2 comes out at an angle θ 2 = 60 with respect to Player 1 s original direction. What is the final speed of player 2, in m/s? A) 6.0 B) 4.0 C) 3.5 D) 3.0 E) 2.5 Solution: We can take the initial direction of the moving hockey player as the +x direction. Then the initial total momentum in the y direction is zero. By conservation of linear momentum 0 = m 1 v 1 f sin! 1! m 2 v 2 f sin! 2, or v 2 f = (m 1 v 1 f sin! 1 ) / (m 2 sin! 2 ) = (v 1 f sin! 1 ) / sin! 2 = (6.0)sin30 / sin60 = 3.5 m/s. Answer: C 47. A 15-kg box slides along a frictionless horizontal surface with speed 5.0 m/s. It collides with a 10-kg box that is initially at rest and attached to a spring. After the collision the boxes stick together and the spring compresses by 3.0 m. What is the spring constant, in N/m? A) 10 B) 15 C) 20 D) 25 E) 30 Solution: This is the spring equivalent of a ballistic pendulum you might call it a ballistic spring. It is a two-step process: first there is a completely inelastic collision, which conserves linear momentum, and then the spring compresses, which conserves total mechanic energy. If M is the mass of the bigger box and m the mass of the smaller box, then conservation of momentum in the collision gives Mv 0 = (m + M )v, or v = v 0 M / (m + M ) = (5.0)(15) / (10 +15) = 3.0 m/s. When the spring compresses the kinetic energy of the two masses converts to elastic potential energy of the spring: 1 2 (m + M )v2 = 1 2 kx2, or k = (m + M )v 2 / x 2 = (10 +15)(3.0) 2 / (3.0) 2 = 25 N/m. Answer: D 48. A 40-kg boy sits on the left end of a 10-kg boat that is at rest in still water. The length of the boat is 2.5 m. If the boy walks to the right end of the boat, how far has he walked relative to the fixed ground, in m? Assume no friction between the boat and the water. A) 0.75 B) 0.50 C) 1.00 D) 0.60 E) 0.80 Solution: The center of mass of the boy-boat system will not move. Using the initial position of the boy as x = 0, the CM of the boat is at L/2 = 1.25 m. Therefore the CM of the system is at x CM = (m 1 x 1 + m 2 x 2 ) / (m 1 + m 2 ) = [(40)(0)+ (10)(1.25)] / (40 +10) = 0.25 m.

10 Afterwards the boy is at x and the CM of the boat is at x L/2, so 0.25 = x CM = [m 1 x + m 2 (x!1.25)] / (m 1 + m 2 ) = (50x!12.5) / 50, or x = 0.50 m. Answer: B 49. A uniform solid cylinder of mass, M = 2.0 kg, and radius R = 20 cm is attached to the right edge of a uniform thin rod of length L = 1.0 m and mass m = 3.0 kg. The combined system (the rod and the cylinder) is allowed to rotate about an axis through the left edge of the rod that is parallel to the height of the cylinder, as shown in the figure. If the system rotates about the axis with an angular speed of 5.0 rad/s, find the rotational kinetic energy of the system, in J. A) 49 B) 31 C) 29 D) 13 E) 3.6 m=3.0 kg L=1.0 m R =20 cm M =2.0 kg axis of rotation Solution: The rotational kinetic energy is KE = I! 2 / 2. To find the total moment of inertia we add the two contributions together I tot = 1! 3 ml2 + 1 $ # 2 MR2 + Md 2 &, where d = L + R is the " % distance from the axis of symmetry of the cylinder to the axis of rotation. Then I tot = 1! 3 (3.0)(1.0)2 + 1 $ # 2 (2.0)(0.20)2 + (2.0)(1.2) 2 & = 3.92 kg-m 2, and KE = (3.92)(5.0) 2 / 2 = 49 J. " % Answer: A 50. A block of mass 0.50 kg with initial speed of 8.0 m/s, is sliding up an incline plane, as shown in the figure. If the final speed of the block after rising by 2.0 m in the vertical direction is 2.0 m/s, how much work was done by the force of friction, in J? A) 2.4 B) 9.8 C) 0.0 D) 5.2 E) 9.8 V 1 = 8.0 m/s V 2 = 2.0 m/s 2.0 m Solution: The work done by non-conservative forces like friction is

11 W nc =!E =!KE +!U = 1 2 m(v 2 f " v 2 i )+ mg(y f " y i ) = 1 2 (0.50)[(2.0)2 " (8.0) 2 ]+ (0.50)(9.8)(2.0) = "5.2 J. Answer: D 51. A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case, and assume that air resistance is negligible. Case A: Thrown straight up Case B: Thrown straight down Case C: Thrown straight out horizontally Case D: Thrown out at an angle of 45 o above the horizontal In which case is the speed of the stone the greatest when it hits the water below? A) Case A B) Case B C) Case C D) Case D E) The speed will be the same in all cases Solution: From conservation of total mechanical energy, since the change in potential energy is the same in each case, the change in mechanical energy must be the same in each case, and so the final speeds are all the same if the initial speeds are the same. Answer: E 52. How deep, in m, would a submarine have to dive in water (density 1000 kg/m 3 ) in order to have an increase in pressure of 10.0 atmospheres? A) 70.8 B) 103 C) 2.56 D) 2.41x10 3 E) 13.2 Solution: The increase in pressure is given by!p =!g!y, so the height required for a pressure change is!y =!P / (!g) = (10 "1.01"10 5 ) /[(1000)(9.8)] =103 m. Answer: B 53. A 20-m long steel cable supporting a load of 500 kg is being used in a situation where the length cannot change by more than 2.5 mm when the load is attached to the cable. What is the minimum cross sectional area required for the cable, in cm 2? The Young s modulus of steel is 2.0 x N/m 2. Ignore the mass of the cable. A) B) C) 1.0 D) 2.0 E) 3.0 Solution: Using the elasticity formula Y = (F / A) / (!L / L), and since the force is the weight of the load, the area required is A = (mgl) / (Y!L) = (500)(9.8)(20) /[(2.0 "10 11 )(0.0025)] = 2.0 "10 #4 m 2, or 2.0 cm 2. Answer: D 54. A 3.2-kg mass sliding on a frictionless, horizontal surface is attached to a spring. It has speed 20 cm/s when the mass is 40 cm from the equilibrium position of the spring,

12 and speed 80 cm/s when it is at the equilibrium position. What is the spring constant, in N/m? A) 76 B) 52 C) 38 D) 20 E) 12 Solution: We can use conservation of total mechanical energy: 1 2 mv 2 i kx 2 i = 1 2 mv 2 f kx 2 f, or solving for k, k = m(v 2 f! v 2 i ) (x 2 i! x 2 f ) = 3.2[(0.80)2! (0.20) 2 ] =12 N/m. Answer: E [(0.40) 2! (0) 2

13 31 C 41 C 51 E 32 A 42 E 52 B 33 B 43 E 53 D 34 D 44 B 54 E 35 D 45 B 36 C 46 C 37 A 47 D 28 C 38 A 48 B 29 A 39 E 49 A 30 A 40 C 50 D

PHYSICS 221 SPRING 2015

PHYSICS 221 SPRING 2015 EXAM 2: April 2, 2015 8:15-10:15pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions,