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Physics 140 Winter 2006: Exam #1 Please print your name:___________________________________________________ Please list your discussion section number:___________________________________ Please list your discussion instructor:________________________________________ Form #1 Instructions 1. Fill in your name above 2. This will be a 1.5 hour, closed book exam. The exam includes 20 questions. 3. You may use a calculator, please do not share calculators 4. You may use one 3?x5? note card with notes and equations you think may be useful. You can write on both sides of the card if you like. 5. You will be asked to show your University student ID card when you turn in your exam. Discussion section reminder: Section 004 Svetlana Malinovskaia Section 005 Svetlana Malinovskaia Section 006 Svetlana Malinovskaia Section 007 Ctirad Uher Section 008 Ctirad Uher Section 009 Ctirad Uher Section 010 Fred Becchetti Section 011 Fred Becchetti Section 012 Fred Becchetti Section 013 Alan Krisch Section 014 Alan Krisch Section 015 Alan Krisch Section 016 Svetlana Malinovskaia Section 017 Svetlana Malinovskaia Section 018 Svetlana Malinovskaia Section 019 Konstantin Bobkov Section 020 Arthur Cole Section 021 Konstantin Bobkov Section 022 Dave Winn Section 023 Dave Winn Section 024 Dave Winn Section 025 Arthur Cole Section 026 Arthur Cole 1: A motorcycle travels due south 25 m/s relative to the highway. An SUV travels at the same speed relative to the highway, but at a heading of 30 degrees south of east. What is the speed of the SUV relative to the motorcycle? a) 37.5 m/s b) 12.5 m/s c) 3.3 m/s d) * 25 m/s e) 46.6 m/s 2: According to statistics compiled by zoo psychologists, suicidal elephants have a much larger terminal velocity than suicidal mice. Which of the following statements best explains this phenomenon? a) The air drag force acting on elephants is smaller than the air drag force acting on mice. b) Mice weigh less than elephants so gravity does not accelerate them downward as rapidly as it does for elephants. c) * Mice have a much larger surface area to mass ratio than elephants, so air drag has a relatively larger effect on them. d) Elephants fall faster because they are afraid of mice. e) The air drag force acting on elephants is much larger than the air drag force acting on mice due to the larger surface area of elephants. Thus, the zoo psychologists must have made a mistake in their measurements of terminal velocity. VSUV,Mot = VSUV ? Vmot These three vectors form an equilateral triangle, so the speed of the suv relative to the motorcycle must also be 25 m/s. ?F=?1/2?CA?v2?mg=0This can be solved for terminal velocity. Note that it depends upon the ratio of mass/area. Also note that the other answer choices can mostly be ruled out... a) The air drag force on elephants must be larger because elephants are much bigger. b) gravity accelerates all objects equally. d) this is just silly. e) This is tricky, refer to above equation. 3: The coefficients of friction between the incline and the block are as follows: ?s = 0.8, ?k =0.74 Which of the following statements about this situation is true? a) If the block is placed at rest on the incline it will begin to slide. b) * If the block is placed at rest on the incline it will remain motionless. If it is then given a shove down the incline it will begin to accelerate downward. c) If the block is placed at rest on the incline it will remain motionless. If it is then given a shove down the incline it will slow down and stop after the shove is over. d) If the block is placed at rest on the incline it will remain motionless. If it is given a shove down the incline it will move at a constant speed down the incline after the shove is over. e) The block will begin to slide downward after it is placed at rest on the incline, but it will quickly slow down and come to a stop after it travels a short distance. 4: ? ?x×?y??? ?y×?x? = a) ?z b) *-1 c) 1 d) 0 e) ?2 On incline with friction the forces are balanced (eg. net force is zero) if ?=tan??? For this incline tan???=0.75 So static friction is large enough to hold the block still. Kinetic friction is smaller than tan??? so the block accelerates down the incline if it is given a shove. ? ?x×?y?=?z , ? ?y×?x?=??z , and ?z????z?=?1 5: Shown below are velocity and acceleration vectors for objects undergoing different kinds of motion. Which one could (approximately) represent an object which is undergoing uniform circular motion? 6: A catapult is mounted on top of the walls of a castle 20 meters above the ground. The catapult is designed to lob a stone projectile at a target 150 meters from the base of the wall. If the projectile arrives at its target 5 seconds after it is launched then what is the initial speed of the projectile? a) 22.3 m/s b) 28.4 m/s c) * 36.3 m/s d) 42.1 m/s e) 50.0 m/s v a v a v a v a v a a) b) c) d) e) a) Acceleration must be perpendicular to velocity for uniform circular motion. Find the x and y components of initial velocity separately, and then add them via Pythagorean theorem. x-dir: 150m=vxt y-dir: ?20m=vy0t??1/2?gt2 (note that the displacement is negative because the projectile lands beneath its starting point. 7: Which of the following relationships is correct for the tensions of the three ropes depicted in the diagram? a) T1 = T2 = T3 b) T1 = T2 > T3 c) T1 > T2 > T3 d) T2 > T3 > T1 e) cannot be determined without more information 8: Given the vectors below, calculate the magnitude of the vector B-A. The magnitude of vector A is 4.3 cm and the magnitude of B is 2.5 cm. (a) 3.15 cm (b)2.12 cm (c) ?3.15 cm (d)* 5.34 cm (e) -5.34 cm A B 30° 40° T1 T2 T3 Answer e was the original plan because angles are not given. Note that if rope 1 and 2 are nearly horizontal then they must have very large tensions in order to support T3 with their vertical components, and T3 << T1 and T2. If they are nearly vertical then (roughly) T1+T2=T3. However, if you estimate angles from the diagram (roughly 45 and 30 degrees below horizontal for rope 1 and 2 respectively) then you can conclude that T1 > T2 in order to match the x components and T3 > T1 >T2. Problem Credited. 9: The following graph shows the acceleration history of a particle moving along the x-axis during the time interval from 0 to 10 seconds. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6 7 8 9 10 11 time (s) ac ce ler ati on (m /s2 ) At the end of 10 seconds, how far did the particle travel from its original location? (a) 0 meters (b)10 meters (c) 50 meters (d)80 meters (e) *Need more information to determine displacement 10: A triangular block of mass m is placed on a frictionless inclined plane as shown in the figure below. A force, F, is applied in the horizontal direction. What is the smallest force F that will push the mass up the incline? (a) no solution exists (b)m g (c) m g sin(?) (d)m g cos(?) FF ? m To get velocity history you must integrate the acceleration history. However, initial velocity is not given, so only the shape of the velocity curve can be determined. Thus, more information is needed. Using vertical and horizontal axis... ?Fy=Ncos????mg=0 ?Fx=Nsin????F=0 Solve y equation for N and then substitute into x equation. (e) * m g tan(?) 11: Block M rests on a frictionless table. If the coefficient of static friction between block m and M is ? Then what is the minimum force needed to pull block m off of block M? a) * F=? mM ?m+M?g b) F=?mg c) F=?g d) F=??m+M?g e) Cannot be determined from the information given 12: A boat pilot wishes to cross a river along the dashed line in the diagram. At what angle relative to straight across the river must she point her boat? The speed of the boat relative to the water (Vb) is 10 m/s. Vr= 5m/s ? A B Vb a) 10.5 degrees b) 13.6 degrees c) * 30 degrees d) 60 degrees M m F If the two blocks stick together static friction must pull the bottom block to the right. ?FM=f ?static,max?=?mg=Ma a=?? mM ?g ?Fboth=F=?m+M?a=??m+M? mM g The horizontal component of Vb must be equal to 5 m/s to cancel the river velocity and allow the boat to cross from A to B. sin???= ?5m/s??10m/s?=0.5 e) Any angle will work provided the pilot aims upstream 13: A tennis ball is thrown upward at an angle from point A. It follows a parabolic trajectory and hits the ground at point D. At the instant shown, the ball is at point B. Point C represents the highest position of the ball above the ground. Point A is the location of the ball immediately after it is thrown and point D is the location of the ball immediately before it hits the ground. At which point is the velocity vector changing most rapidly with time? a) A b) B c) C d) D e) * The change in velocity with respect to time is the same for all points. 14: What is the tension in rope 1? You may assume that the table is frictionless and the ropes are massless. a) T1 = 3mg b) T1 = 2mg c) T1 = 1mg d) T1 = (2/3)mg e) *T1 = (1/3)mg m m m T1 T2 The change in velocity with respect to time is the acceleration due to gravity and it is the same everywhere. First, find the acceleration of the system. The only force accelerating the system is the force of gravity on the hanging mass: ?Fall=mg=3ma next look at the left mass ?F=T1=ma=m?13g? 15: A ?bicycle tire? shaped space station with a radius of 1000 meters rotates 0.95 times per minute. If an astronaut walks from the outer edge of the space station (1000 meters from the axis of rotation) along one of the spokes to a location 500 meters from the axis of rotation what happens to his apparent weight? a) His apparent weight quadruples b) His apparent weight doubles c) His apparent weight stays the same d) His apparent weight is reduced by a factor of 2 e) His apparent weight is reduced by a factor of 4. 16: A remote controlled car starts from rest and begins to accelerate as follows: ?a=?5m/s3?t?x??3m/s4?t2 ?y At what time will the velocity vector make a 45 degree angle with the x-axis? a) 2.0 seconds b) *2.5 seconds c) 3.0 seconds d) 3.5 seconds e) 4.0 seconds The apparent weight of the astronaut is the normal force acting on him. This normal force is present because the station is rotating and the astronaut undergoes uniform circular motion. ?F=N=mv2r However, note that velocity is proportional to radius... v=?2?r?T Where T is the time for one complete rotation. When the astronaut moves inward his apparent weight changes as follows: ?F=N?1/2?=m?2?0.5r/T? 2 0.5r =0.5m?2?r/T ? 2/r=0.5N So his weight is cut in half. To get the change in the velocity vector integrate acceleration from time 0 to time t. Add the change in velocity to the initial velocity (0 m/s) to get velocity as a function of time. ?v=?52m/s3?t2 ?x??1m/s4?t3 ?y Set x and y components equal to each other to find time when velocity makes 45 degree angle with horizontal. 17: During an episode of Mythbusters Jamie and Adam test the urban legend that a box of Kleenex placed in the back seat of a car can become a deadly projectile if the car stops suddenly. During their test they cause a car to decelerate from 27 m/s (60 mph) to 0 m/s very rapidly. The Kleenex box then flies forward (relative to the car) and rams a wall of ballistic gel. Assume that the Kleenex box weighs 100 grams and that it penetrates 0.5 cm into the gel. What is the average force exerted by the box on the gel? a) 634 N b) 5320 N c) 6980 N d) * 7290 N e) 1.03 x 104 N 18: A box is sliding up a rough incline, as shown. What is the correct free body diagram for the box? A B C D E v ?vf ?2=?vi?2?2xa where x = 0.005 m, vi is 27 m/s and the final velocity is zero... Solve for a (72900 m/s^2) Next use F=ma to get force. The correct answer is A). The force of friction resists the motion of the block, so it points down the inclince. Velocity is not a force, so there are no force vectors pointing up the incline. 19: Hector and Achilles compete with each other in a game of tug-of-war. The two heroes stand on opposite sides of a mud puddle and attempt to pull the other into the mud. Eventually, Hector pulls Achilles into the puddle. Which of the following is an accurate description of the reasons behind Hector's victory? a) Hector wins because he has a greater mass than Achilles. b) Hector and Achilles both exert the same force on the rope. Hector wins because the rope exerts a greater force on Achilles than it exerts on Hector. c) * Hector and Achilles both exert the same force on the rope and the rope exerts the same force on both heroes. Hector wins because he exerts a greater frictional force on the ground than does Achilles. d) The rope exerts the same force on Hector that Hector exerts on the rope. The rope also exerts the same force on Achilles that Achilles exerts on the rope. Hector wins because the force that he exerts on the rope is greater than the force that Achilles exerts on the rope. e) Hector wins because his victory was ordained by Zeus. 20: A navy frigate fires projectiles simultaneously at ship A and ship B. Which of the following statements about this situation is false? a) The projectile fired at ship B must have a larger initial speed. b) Ship A gets hit after ship B gets hit. c) The y component of the velocity of projectile A must be larger than the y component of the velocity of projectile B. d) The x component of the velocity of projectile A must be smaller than the x component of projectile B. e) The crew of the frigate must have fantastic aim to hit two targets with a single volley of shots. My original notion was that a) is the correct answer since launch angles are not given and neither are the heights or ranges. However, if you assume that the scale in the drawing is correct then you can conclude (roughly) that projectile A reaches a height 1.5 times B's height and B goes three times as far. This leads to VAy=??1.5?VBy and VBx?3VAx This means that B does have a larger launch speed. Since we did not want you guys to waste 20 to 30 minutes figuring this out, we just removed the problem. mckay Physics 140 Fall 2004: Exam #1

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About this note

Author: Arun G.

Textbook: University Physics Vol 1 (Chapters 1-20) (12th Edition) (Chapters 1-20 v. 1)

Created: 2008-06-08

Updated: 2008-06-08

File Size: 11 page(s)

Views: 58

Textbook: University Physics Vol 1 (Chapters 1-20) (12th Edition) (Chapters 1-20 v. 1)

Created: 2008-06-08

Updated: 2008-06-08

File Size: 11 page(s)

Views: 58

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