PHYSICS 221 SPRING 2013 EXAM 1: February 21, 2013; 8:15pm 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. 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 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. Eli Rosenberg, Kerry Whisnant, and Kai-Ming Ho
1. The speed limit on I-35 between Ames and Des Moines is 70 mph. What is this speed in m/s? A) 23 B) 27 C) 31 (Correct) D) 35 E) 39 Solution: We apply the appropriate conversions factors: 70 mi h 1609m mi 1 h = 31 m/s. 3600 s 2. A certain car traveling at 33.0 mph skids to a stop in 39.0 m from the point where the brakes were applied. In what distance, in m, would the car stop had it been going 66.0 mph? A) 39.0 B) 110 C) 55.0 D) 78.0 E) 156 (Correct) Solution: We can use the constant velocity equations; in particular, since we are given the velocities and distance, we can find the acceleration and apply it to the second situation. In fact, we don t even have to convert units if we take ratios:!v 2 = 0 " v 1 2 = 2a!x 1 and 0 " v 2 2 = 2a!x 2, so!x 2 = v 2 2 2 2a = v 2 = v 2 2 v 2 1 / 2!x 1 v!x 2 1 = (66)2 (39.0) =156 m. 2 1 (33)
3. The figure on the right shows two vectors! A and! B that lie in the x-y plane with magnitudes A and B, respectively. Their angles with respect to the positive x axis are 45 o and 30 o, respectively. What is the magnitude of the projection of vector! B on! A? A) A cos(45 o ) B) B cos(- 30 o ) C) A cos(15 o ) D) B cos(75 o ) (Correct) E) AB cos(75 o ) y 45 o! A x Solution: The projection of B onto A is B cos! AB = B cos75 o. 30 o 4. A ball is thrown directly upward off the roof of a 40 m building. For which situation are both the instantaneous velocity and the instantaneous acceleration zero?! B A) on the way up B) on the way down C) as it passes the roof of the building on its way to the ground D) when it reaches its highest point E) none of the above (Correct) Solution: Since gravity always acts, the acceleration is never zero. 5. Consider a particle moving along the y axis. The velocity v y of the particle as a function of time is given by v y = -10.0 m/s (3.00 m/s 2 )t + (4.00 m/s 3 )t 2. If at t = 0 the particle is at y = 2.00 m, what is its position, in m, at t = 6.00 s? A) 47.0 B) 116 C) 118 D) 174 E) 176 (Correct) Solution: To find position from velocity, we integrate: t x(t)! x(0) = " v(t)dt =!10.0t! 3.00 0 2 t 2 + 4.00 3 t3, or x(6.00) = 2.00!10.0(6.00)!1.50(6.00) 2 +1.33(6.00) 3 =176 m
6. A ball is projected directly upward at time t = 0.0 s from a point on a roof 80 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 38 m/s. Consider all quantities as positive in the upward direction. At time t = 5.0 s, what is the velocity of the ball, in m/s? A) 49 B) 11 (Correct) C) 0 D) 38 E) 87 Solution: The velocity is given by v(t) = v(0)! gt = 38! (9.8)(5.0) =!11 m/s. 7. Which of the following situations is impossible? A) An object has constant non-zero acceleration and changing velocity. B) An object has constant non-zero velocity and changing acceleration. (B is Correct) C) An object has velocity directed east and acceleration directed east. D) An object has velocity directed east and acceleration directed west. E) An object has zero velocity but non-zero acceleration. Solution: If the acceleration is changing, it is not always zero, and so the velocity must be changing.! 8. Vectors A, B,! and satisfy the vector equation A! + B! = C!. Their magnitudes are related by A! C!! B! = C!. Which of the following is an accurate statement? A) B! and C! are perpendicular vectors.! B) B and!! C are antiparallel vectors. (Correct) C) A, B,! and C! form the sides of an equilateral triangle.! D) A, B,! and C! form the sides of a right triangle. E) The angle between A! and B! can have any value, in view of the limited information. Solution: The only case when the length of the sum of two vectors is equal to the difference of their respective lengths is when the two vectors are in opposite directions.
9. A dentist s drill has an initial angular speed of 1.05 x 10 4 rad/s. It undergoes constant angular acceleration for 3.00 s and reaches an angular speed of 8.85 x 10 4 rad/s. How many revolutions does the drill bit make during this period? A) 1.49 x 10 5 B) 2.99 x 10 5 C) 4.72 x 10 4 D) 2.36 x 10 4 (Correct) E) 7.80 x 10 4 Solution: The easiest way to solve this is by using ω avg = (ω i +ω f )/2 for constant acceleration, or ω avg = (1.05 x 10 4 + 8.85 x 10 4 )/2 = 4.95 x 10 4. Then θ = ω avg t = (4.95 x 10 4 )(3.00) = 1.49 x 10 5 rad. To convert to revolutions, we must divide θ by 2π, to get 1.49 x 10 5 rad/(2π) = 2.36 x 10 4 revolutions. You could also find the angular acceleration α from Δω/Δt and then use Δθ = ω 0 t + αt 2 /2. 10. A ball is thrown at a 60.0 angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the ground with a speed of 29.9 m/s and experiences no appreciable air resistance. What is the time the ball remains in the air before striking the ground, in s? A) 3.21 B) 4.28 C) 5.36 (Correct) D) 6.43 E) 7.50 Solution: Hitting the ground involves only the vertical motion, so we use 0 = y = y 0 + v 0 y t! 1 2 gt 2 = y 0 + v 0 sin! t! 1 2 gt 2 = 2.00 + (29.9)sin60 o t! 4.9t 2. This is a quadratic equation, which has two solutions: t = -0.076 or 5.36 s. The positive answer makes physical sense.
11. The Ferris wheel in the figure rotates counter-clockwise with uniform angular speed. What is the direction of the average acceleration of a gondola as it goes from the top to the bottom of its trajectory? A) Up. B) Down. C) To the left. D) To the right. (Correct) E) Zero, because the motion is uniform. Solution: The velocity at the top is to the left, and at the bottom it is to the right, so the average acceleration is! v! /!t, which is proportional to! v bottom! v " top = v! bottom! (! v! bottom ) = 2 v! bottom, or to the right. 12. Shown in Fig. 3.4 are the trajectories of four artillery shells. Each was fired with the same initial speed. Which was in the air the longest? A) A (Correct) B) B C) C D) D E) All were in the air for the same time. Solution: The shell that goes the highest must be in the air the longest. Remember that the time to the top is given by 0 = v 0y gt, or v 0y = gt, and the height is then y = v 0y t gt 2 /2 = gt 2 /2. Thus the shell with the highest y will also be in the air the longest.
13. A boat must cross a 260-m-wide river and arrive at a point 110 m upstream from where it starts (see figure). To do so, the pilot must head the boat at a 45º upstream angle. The current has speed 2.7 m/s. The trip across the river takes 55 s. What is the speed of the boat in still water, in m/s? A) 3.1 B) 4.3 C) 5.6 D) 6.7 (Correct) E) 8.2 Solution: This situation is similar to a written homework problem, although more information is given so there are two ways to solve this. The simplest is to use the distance across and the time to find the component of the boat across the river, v y = (260 m)/(55 s) = 4.73 m/s. Since we know the angle at which the boat aims, this is 4.73 m/s = v b cos 45 o, so v b = 6.7 m/s. We could also solve this in the same way the homework was solved. The net component of the velocity upstream is v b sin 45 o v w, the component downstream is v b cos 45 o, and the ratio of these two must be 110/260, since that is the resulting direction. Therefore v b sin45 o! v w = 110 v b cos45 o 260, or v = v w b = 6.6 m/s. sin45 o o! 0.423cos45 This is a little different from the first answer due to round-off error. 14. A ball sits at rest on a horizontal tabletop. The weight of the ball is one half of a Newton s third law force pair. Which force is the other half? A) The force of the Earth s gravity on the ball B) The upward force that the table top exerts on the ball C) The upward force that the ball exerts on the Earth (Correct) D) The downward force that the ball exerts on the table top E) The frictional force between the ball and the table top Solution: The weight of the ball is the Earth s upward force on the ball, so the other half of this 3 rd law pair is the ball s upward force on the Earth.
15. A block is on a frictionless horizontal table, on Earth. This block accelerates at 8.0 m/s 2 when a 50 N horizontal force is applied to it. The block and table are then set up on the Moon where the acceleration due to gravity is 1.62 m/s 2. A horizontal force of 25 N is applied to the block when it is on the Moon. What acceleration, in m/s 2, does this force impart to the block? A) 3.6 B) 4.0 (Correct) C) 4.4 D) 4.8 E) 5.2 Solution: Since there is no friction and the table is horizontal, the normal force cancels gravity in the vertical direction. In the horizontal direction we have F = ma, or 50 N = m(8.0 m/s 2 ), so m = 6.25 kg. On the Moon, the normal force and gravity still cancel, although they are about a factor of 6 less than on Earth. In the horizontal direction, a = F/m = (25 N)/(6.25 kg) = 4.0 m/s 2. 16. A block of mass m 1 = 3.0 kg is pushed by a force F against another block of mass m 2 = 2.0 kg, as shown. If the mass m 2 is accelerating at a = 1.0 m/s 2, how much force, in N, is being exerted on mass m 1 by the hand? A) 4.0 B) 1.0 C) 3.0 D) 2.0 E) 5.0 (Correct) F m1 m2 Solution: For m 2 we have the force F 21 (1 acting on 2). This must give the acceleration F 21 = m 2 a 2 = (2.0 kg)(1.0 m/s 2 ) = 2.0 N. By Newton s 3 rd law there is an equal an opposite force on 1 due to 2, F 12 = 2.0 N. Then for m 1 we have F net = F F 12 = m 1 a, so F = F 12 + m 1 a = 2.0 N + (3.0 kg)(1.0 m/s 2 ) = 5.0 N. You can also treat the system as a whole, F = (m 1 + m 2 )a = (3.0 kg + 2.0 kg) x (1.0 m/s 2 ) = 5.0 N.
17. A student who weighs 750 N stands on a scale in an elevator. If the scale reads 650 N, only one of the following statements can be true. Which one is it? A) The velocity is down and the speed is increasing. (Correct) B) The velocity is up and the speed is increasing. C) The velocity is down and the speed is decreasing. D) The velocity is constant. E) None of the above statements is true. Solution: For the vertical forces acting on the student we have N mg = F net = ma, so N = m(g+a). The scale reading is N. For N to be less than the actual weight mg, a must be negative. If the velocity is down and the speed is increasing, the acceleration is also down, i.e., negative. The acceleration in B and C is positive (going up and speed increasing and going down and speed decreasing, respectively). 18. A 0.30 kg rock is swung in a circular path in a vertical plane on a 0.25 m length string. At the top of the path, the angular velocity is 13.0 rad/s. What is the tension in the string at that point? A) 9.7 B) 2.9 C) 16 D) 13 E) 0 Solution: For circular motion, at the top both the tension and gravity act down, and must combine to give the centripetal acceleration. Thus T + mg = mω 2 R, so T = m(ω 2 R-g) = (0.30)[(13.0) 2 (0.25) 9.8] = 9.7 N. 19. A car weighing 8900 N is parked on a hill as shown in the figure. What is the minimum coefficient of static friction between the tires and the ground that will prevent the car from sliding down the hill? A) 0 B) 0.60 C) 0.15 D) 1.3 E) 0.75 (Correct) Solution: The component of acceleration down the incline is mg sinθ, and the friction is µ s N = µ s mg cosθ, so if the car is not sliding, mg sinθ = µ s mg cosθ, or µ s = tanθ = tan37 o = 0.75.
20. It is amateur night at the tightrope walkers convention, and a 600-N performer finds himself in the awkward position shown in the figure. If the angle between the ropes and the horizontal is 8.00 o, what is the tension in either rope, in N? A) 2.16 x 10 3 (Correct) B) 6.00 x 10 2 C) 3.00 x 10 2 D) 2.20 x 10 2 E) 4.21 x 10 3 Solution: Because the situation is symmetric, the two tensions both have vertical component T sin8 o, and gravity acts down, so the net force in the vertical direction is F y,net = 2T sin8 o mg = 0, or T = mg/(2 sin8 o ) = (600 N)/(2 sin8 o ) = 2.16 x 10 3 N. 21. In the figure, a person pulls horizontally on block B and both blocks A and B move horizontally as one unit. There is friction between block B and the horizontal table. Air resistance is negligible. If the two blocks are moving to the right at constant velocity, which one of the following statements is true? A) The horizontal force that B exerts on A points to the left. B) The horizontal force that B exerts on A points to the right. C) B exerts no horizontal force on A. (Correct) D) The pull force is greater than the friction force. E) The pull force is less than the friction force. Solution: Since block B is moving at constant speed, it is not being accelerated, and the force of block A acting on block B must be zero because that is the only possible horizontal force on B. Due to Newton s 3 rd law, the horizontal force of block B acting on block A is also zero.
22. Consider the block on the incline pictured below. When the block is released from rest (at the top of the incline) it takes time t to reach the bottom, traveling a distance x. What is the coefficient of kinetic friction between the block and the incline? x A) µ k = 2x / t 2 B) µ k = tan! C) µ k = 2x / (gt 2 cos!) θ D) µ k = tan!! 2x / (gt 2 cos!) (Correct) E) It is impossible to calculate this without the mass of the block. Solution: The net force down the incline is mg sinθ f k, where the kinetic friction force is f k = µ k N and the normal force is N = mg cosθ. Therefore the net force is F net = mg(sinθ µ k cosθ) = ma. For the block sliding down the incline, x = at 2 /2, or a = 2x/t 2. Substituting in for a, we can solve for the coefficient of kinetic friction: µ k = [g sinθ 2x/(gt 2 )]/ cosθ = g tanθ 2x/(gt 2 cosθ). Questions 23 and 24 refer to the diagram at the right, with m 2 = 2.4 kg. The coefficient of kinetic friction between m 1 and the table is µ k = 0.60. The friction and mass of the pulley may be neglected and the rope is ideal. When the system is set into downward motion, m 2 descends with a constant speed of 1.1 m/s for 2.0 s. 23. What is the mass of m 1, in kg? A) 1.2 B) 2.4 C) 3.0 D) 4.0 E) 6.0 Solution: The forces acting on the system are gravity acting down on m 2 and friction acting to the left on m 1. The speed is constant so the acceleration must be zero, and therefore the net force is zero. Therefore 0 = F net = m 2 g µ k m 1 g, or m 1 = m 2 /µ k = (2.4 kg)/(0.60) = 4.0 kg.
24. What is the work done by gravity on the system, in J, during the 2.0 s the masses are moving? A) 25 B) 39 C) 52 Correct) D) 69 E) 98 Solution: Gravity does no work on m 1 since that mass does not move up or down. The work done by gravity on m 2 is m 2 g Δy = mgv y Δt = (2.4 kg )(9.8 m/s 2 )(1.1 m/s)(2.0 s) = 52 N. 25. In gym class, a 50 kg student climbs a 5.0 m rope and stops at the top. If she does this with constant speed of 0.40 m/s, what is her power output during the climb, in W? A) 70 B) 100 C) 140 D) 200 (Correct) E) 300 Solution: Power is force times velocity, or Fv = mgv = (50)(9.8)(0.40) = 196 W, or 2.0x10 2 W.
26. Three blocks are connected as shown. The ropes and pulleys are of negligible mass. When released, block C moves downward, block B moves up the ramp, and block A moves to the right. After each block has moved a distance d, the force of gravity has done A) positive work on A, B, and C. B) zero work on A, positive work on B, and negative work on C. C) zero work on A, negative work on B, and positive work on C. (Correct) D) zero work on A and B, and negative work on C. E) negative work on A, B, and C. Solution: Gravity acts down. Since the motion of A is horizontal, there is n work done on A. B s motion has an upward component, so the work done on it by gravity is negative. C s motion is down, so the work done by gravity on it is positive. 27. A 0.10 kg golf ball hits a wall with a velocity of 10 m/s and bounces back with a velocity of 7.5 m/s. How much work does the wall do on the ball, in J? A) 0.13 B) 2.2 (Correct) C) +7.8 D) +2.2 E) +0.13 Solution: By the Work/Kinetic Energy theorem, the work done on the ball is just the change in kinetic energy, so W = 1 2 mv 2 f! 1 2 mv 2 i = 1 2 m(v 2 f! v 2 i ) = 1 2 (0.10) " # (7.5)2! (10) 2 $ % =!2.2 J.
Physics 221 midterm exam 1 - KEY 1 C 11 D 21 C 2 E 12 A 22 D 3 D 13 D 23 D 4 E 14 C 24 C 5 E 15 B 25 D 6 B 16 E 26 C 7 B 17 A 27 B 8 B 18 A 9 D 19 E 10 C 20 A