* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Version 055 – Midterm 1
Survey
Document related concepts
Faster-than-light wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Coriolis force wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Equations of motion wikipedia , lookup
Mass versus weight wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Fictitious force wikipedia , lookup
Jerk (physics) wikipedia , lookup
Proper acceleration wikipedia , lookup
Transcript
Version 055 – Midterm 1 – OConnor (05141) This print-out should have 36 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. V1:1, V2:1, V3:3, V4:2, V5:1. 001 (part 1 of 1) 10 points The graph below shows the velocity v as a function of time t for an object moving in a straight line. 1 x t 7. 0 8. tQ tR tS tP x 0 t tQ tR tS tP x 9. t v 0 t tQ tR tS tP 0 Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval? x 1. t 0 correct tQ tR tS tP x 2. tQ tR tS tP Explanation: The displacement is the integral of the velocity with respect to time Z ~x = ~v dt . Because the velocity increases linearly from zero at first, then remains constant, then decreases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase proportional to negative time squared. From these facts, we can obtain the correct answer. x t 0 3. tQ tR tS tP x 0 0 t tQ tR tS tP x 4. t 0 tQ tR tS t tP tQ 5. t tQ tR tS 1. 2.304 m 3. 3.36 m tP 0 6. None of these graphs are correct. tS tP 002 (part 1 of 1) 10 points A man jogs at a speed of 1.5 m/s. His dog waits 1.6 s and then takes off running at a speed of 4 m/s to catch the man. How far will they have each traveled when the dog catches up with the man? 2. 2.94458 m x tR 4. 3.705 m 5. 3.84 m correct Version 055 – Midterm 1 – OConnor (05141) 6. 4.03636 m 2 y 7. 5.032 m t 3. 8. 6.5025 m 9. 7.26042 m y 10. 10.472 m Explanation: We calculate first the distance that separates the man and the dog when the dog takes off. It is d0 = vman ∗ twait = 2.4 m. Once the dog takes off, the position of the dog and the man at any time t is t 4. y t 5. xman = d0 + vman t xdog = vdog t. They meet when xman = xdog , so d0 +vman t = vdog t and the time is t= d0 = 0.96 s. vdog − vman y t 6. The distance is d = vdog ∗ t = 3.84 m. y 003 (part 1 of 1) 10 points An object was suspended in a fixed place (y = 0) and then allowed to drop in a free fall. Taking up as the positive vertical direction, which of the following graphs correctly represents its vertical motion as displacement vs time? t 7. y t 8. y t 1. y 9. correct y 2. t t Version 055 – Midterm 1 – OConnor (05141) 3 y hmax v0 Explanation: The object is undergoing a constant downward gravitational acceleration g. The slope of a position vs time curve represents the velocity. Given g 6= 0 m/s2 a < 0 m/s2 a = −g m/s2 v0 = 0 m/s y0 = 0 m . Since y = y0 + v 0 t + 1 2 at , 2 therefore 1 y = − g t2 . 2 This is a parabolic shaped curve starting a y = 0 with a continuously decreasing slope as time increases. Only the figure below is correct. y t 004 (part 1 of 1) 10 points A ball is thrown upward. Its initial vertical speed v0 , acceleration of gravity g, and maximum height hmax are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9.8 m/s2 . 9.8 m/s2 t 10. What is its maximum height, hmax (in terms of the initial speed v0 )? √ 2 3v 1. hmax = √ 0 2 2g 5 v02 2. hmax = 8g v2 3. hmax = 0 g 3 v02 4. hmax = 4g √ 2 5v 5. hmax = √ 0 2 2g √ 2 3 v0 6. hmax = 2g v2 7. hmax = 0 4g v2 8. hmax = 0 correct 2g v2 9. hmax = √ 0 2g Explanation: Basic Concept: For constant acceleration, we have (1) v 2 = v02 + 2 a (y − y0 ) . Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v, a, y0 , and y (no time) is the easiest one to use. Choose the positive direction to be up; then a = −g and 0 = v02 + 2 (−g) (hmax − 0) Version 055 – Midterm 1 – OConnor (05141) or Explanation: Basic Concept: For constant acceleration, we have v2 = 0 . 2g hmax 005 (part 1 of 1) 10 points A ball is thrown upward. Its initial vertical speed v0 , acceleration of gravity g, and maximum height hmax are shown in the figure below. Given: g = 9.8 m/s2 . Neglect: Air resistance. v = v0 + a t . 0 = v0 + (−g) tup v0 or v0 . g tup = 9.8 m/s2 hmax (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v, a, and t . Choose the positive direction to be up; then a = −g and 4 006 (part 1 of 1) 10 points A ball is thrown upward. Its initial vertical speed v0 , acceleration of gravity g, and maximum height hmax are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9.8 m/s2 . 2. tup = 3. tup = 4. tup = 5. tup = 6. tup = 7. tup = 8. tup = 9. tup = v0 2g 4 v0 g √ 2 v0 g √ 3 v0 g 2 v0 g v √0 2g √ 2v √ 0 3g v0 correct g v0 4g v0 1. tup = 9.8 m/s2 Whatis its time interval, tup (in terms of the initial speed v0 ), between the release of the ball and the time it reaches its maximum height? hmax What is its time interval, tup (in terms of the maximum height hmax ), between the release of the ball and the time it reaches its maximum height? √ 2g 1. tup = hmax √ 4g 2. tup = hmax √ 2 g 3. tup = hmax Version 055 – Midterm 1 – OConnor (05141) 5. tup 6. tup 7. tup 8. tup 9. tup 10. tup while it is in Michael’s hand and after he lets it go, assuming it has not yet hit the ground? s 4 hmax g √ g = 2h s max 2 hmax = correct g s hmax = 2g g = hmax hmax = g s hmax =2 g 4. tup = a 1 2 at . 2 t 1. correct a t 2. a 3. Explanation: Basic Concept: For constant acceleration, we have y = y0 + v 0 t + 5 a (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing hmax = y − y0 , a, and t . Choose the positive direction to be up, then a = −g and 1 2 gt 2 1 = g t 2 − g t2 2 1 2 = gt , 2 t t 4. a 5. t hmax = v0 t − where v0 = g t . Therefore r 2 ymax tup = . g 007 (part 1 of 1) 10 points Michael stands motionless holding a baseball in his hand. After a while he tosses it upwards, and it travels up for a while before turning about and heading towards the ground. Define upwards to be positive. Which of the following diagrams can describe the vertical acceleration of the ball, a 6. t Explanation: The ball first experiences a period of zero acceleration when Michael is just holding the ball. Then, as the ball is thrown upward, it feels an upward acceleration. Quickly the ball is let loose. Once the ball leaves Michael’s hand it is in free-fall. In free-fall all objects feel a downward (here negative) constant gravitational acceleration. This scenario is shown below. Version 055 – Midterm 1 – OConnor (05141) 6 Basic Concepts: a ~a = t ∆~v ∆t For constant acceleration 008 (part 1 of 3) 10 points Given: The acceleration of gravity on Earth is 9.8 m/s2 . Consider a ball thrown up from the ground (the point O). It passes a window (the segment AB) in the time interval 0.314 s (see the figure). The points in the figure represent the sequential order, and are not drawn to scale. The distance AB = 1.42 m. C B y vi + v f ∆s = ∆t 2 v = v0 + a t . v̄ = For the acceleration of gravity (a = −g) v = v0 − g t . Solution: The average velocity is given by |AB| ∆s = ∆t ∆t 1.42 m = 0.314 s = 4.52229 m/s . v̄ = 1.42 m Alternative Part 1: x A d = yB − yA = v A t − 1 2 gt 2 O Find the average speed as the ball passes the window. 1. 2.59936 m/s d+ vA = 1 2 gt 2 t 1 (9.8 m/s2 ) (0.314 s)2 2 = (0.314 s) = 6.06089 m/s 1.42 m + 2. 2.67491 m/s 3. 2.80435 m/s 4. 3.20225 m/s 5. 3.47692 m/s 6. 3.68421 m/s 7. 4.52229 m/s correct 8. 5.05576 m/s 9. 5.2459 m/s 10. 5.45455 m/s Explanation: vB = v A − g t = 6.06089 m/s − (9.8 m/s2 ) (0.314 s) = 2.98369 m/s vA + v B 2 6.06089 m/s + 2.98369 m/s = 2 = 4.52229 m/s . v= 009 (part 2 of 3) 10 points What is the magnitude of the decrease of the velocity from A to B? 1. 2.156 m/s Version 055 – Midterm 1 – OConnor (05141) 2. 2.3912 m/s 5. 0.21444 s 3. 2.6362 m/s 6. 0.304458 s correct 4. 2.7048 m/s 7. 0.313938 s 5. 2.7734 m/s 8. 0.381394 s 6. 2.9498 m/s 9. 0.413296 s 7. 2.989 m/s 8. 3.0772 m/s correct 10. 0.446586 s Explanation: The velocity change is vA − vB = ∆v 9. 3.1654 m/s 10. 3.4888 m/s Explanation: For a constant acceleration ~a = =⇒ vA = vB + ∆v and the average velocity is ∆~v , ∆t =⇒| ∆~v |=| ~a ∆t |= g ∆t . v̄ = 2 vB + ∆v vA + v B = . 2 2 Thus Thus the decrease in velocity is ∆v = g ∆t = (9.8 m/s2 )(0.314 s) = 3.0772 m/s . Alternative Part 2: ∆v = vA − vB = 6.06089 m/s − 2.98369 m/s = 3.0772 m/s . 010 (part 3 of 3) 10 points If the ball continues its path upward without obstruction, find the travel time between B and C, where point C is at the ball’s maximum height. 1. 0.109241 s 2. 0.13145 s 3. 0.148158 s 4. 0.192288 s 7 vB = v̄ − ∆v 2 = 4.52229 m/s − 3.0772 m/s 2 = 2.98369 m/s . The ball reaches its maximum height when its velocity is zero. Based on v = v0 + a t vC = 0 = v B − g t vB g 2.98369 m/s = 9.8 m/s2 t= = 0.304458 s . 011 (part 1 of 3) 10 points The position of a softball tossed vertically upward is described by the equation y = c 1 t − c 2 t2 , Version 055 – Midterm 1 – OConnor (05141) where y is in meters, t in seconds, c1 = 8.59 m/s, and c2 = 7.4 m/s2 . Find the ball’s initial speed v0 at t0 = 0 s. 1. 3.15 m/s 5. −5.7912 m/s 6. −3.0588 m/s 2. 5.69 m/s 7. −1.63 m/s 3. 5.89 m/s 8. −0.5417 m/s 4. 6.04 m/s 9. 3.64864 m/s 5. 6.61 m/s 8 10. 6.75688 m/s Explanation: Substituting t = 1.08 s into the above formula for v, we obtain 6. 6.93 m/s 7. 8.4 m/s 8. 8.59 m/s correct v = 8.59 m/s − 2(7.4 m/s2 )(1.08 s) = −7.394 m/s . 9. 10.3 m/s 013 (part 3 of 3) 10 points Find its acceleration at t = 1.08 s. 1. −15.4 m/s2 10. 10.8 m/s Explanation: Basic Concepts: dx v= dt a= dv d2 x = 2 dt dt Solution: The velocity is simply the derivative of y with respect to t: v= dy = 8.59 m/s − 2(7.4 m/s2 )t, dt which at t = 0 is v0 = 8.59 m/s . 012 (part 2 of 3) 10 points Find its velocity at t = 1.08 s. 1. −20.4 m/s 2. −12.989 m/s 3. −7.394 m/s correct 4. −6.6046 m/s 2. −14.8 m/s2 correct 3. −13.9 m/s2 4. −13.14 m/s2 5. −12.98 m/s2 6. −11.3 m/s2 7. −10.1 m/s2 8. −6.46 m/s2 9. −4.56 m/s2 10. −3.96 m/s2 Explanation: The acceleration is the derivative of velocity with respect to time: a= dv = −2(7.4 m/s2 ) = −14.8 m/s2 . dt 014 (part 1 of 2) 10 points Version 055 – Midterm 1 – OConnor (05141) A commuter airplane starts from an airport and takes the route shown in the figure. It first flies to city A located at 198 km in a direction 29 ◦ north of east. Next, it flies 135 km 16◦ west of north to city B. Finally, it flies 180 km due west to city C. y (km) N 250 C 200 180 km km 135 150 R100 50 BW 198 29◦ 50 km 16◦ E S A x (km) 100 150 200 How far away from the starting point is city C? 1. 212.275 km 2. 219.693 km 3. 230.017 km correct 4. 234.482 km β b a α The x-component of the resultant is rx = a x + bx + c x = a cos α − b sin β − c = (198 km) cos 29◦ − (135 km) sin 16◦ − 180 km = −44.0363 km . The y-component of the resultant is ry = a y + by + c y = a sin α + b cos β + 0 = (198 km) sin 29◦ + (135 km) cos 16◦ = 225.763 km . and the resultant is q R = rx 2 + r y 2 q = (−44.0363 km)2 + (225.763 km)2 = 230.017 km . 5. 239.753 km 6. 246.565 km 015 (part 2 of 2) 10 points What is the direction of the final position vector r, measured from North? Use counterclockwise as the positive angular direction, between the limits of −180◦ and +180◦ . 1. 9.37141 ◦ 7. 249.497 km 8. 258.343 km 9. 269.981 km 2. 11.0373 ◦ correct 10. 283.602 km Explanation: 3. 13.8464 ◦ 4. 15.0734 ◦ 5. 16.0841 ◦ Given : 9 a = 198 km , α = 29◦ , b = 135 km , β = 16◦ , and c = 180 km . 6. 16.5333 ◦ 7. 19.6103 ◦ 8. 21.2758 ◦ Version 055 – Midterm 1 – OConnor (05141) 9. 21.8767 ◦ 6. tmax rx 7. tmax 8. tmax r ry γ 9. tmax 10. tmax Since γ is the angle between r and the yaxis, for r is in the second quadrant tan γ = Thus |rx | . |ry | Explanation: The cannonball’s time of flight is t= ¶ |rx | γ = tan |ry | ¶ µ −1 44.0363 km = tan 225.763 km −1 µ 1. hmax 016 (part 1 of 3) 10 points Denote the initial speed of a cannon ball fired from a battleship as v0 . When the initial projectile angle is 45◦ with respect to the horizontal, it gives a maximum range R. 2. hmax 3. hmax v0 4. hmax ◦ 5. hmax 45 x R The time of flight tmax of the cannonball for this maximum range R is given by 1 v0 4 g √ v0 = 3 g √ v0 = 2 correct g v0 =4 g 6. hmax 7. hmax 1. tmax = 8. hmax 2. tmax 9. hmax 4. tmax 2 v0y 2 v0 sin 45◦ √ v0 = = 2 . g g g 017 (part 2 of 3) 10 points The maximum height hmax of the cannonball is given by = 11.0373◦ . 3. tmax 2 v0 3 g 1 v0 = 2 g √ v0 = 3 g v0 =2 g v0 = g 1 v0 =√ 3 g 5. tmax = 10. 22.687 ◦ Explanation: y 10 10. hmax v02 =4 g 1 v2 =√ 0 2 g 2 v02 = 3 g 1 v02 = 2 g √ v02 = 2 g √ v02 = 3 g 2 1 v0 correct = 4 g v02 = g v2 =2 0 g 1 v2 =√ 0 3 g Explanation: Version 055 – Midterm 1 – OConnor (05141) oo k Use the equation 11 B 2 vy2 = vy0 − 2gh. At the top of its trajectory vy = 0. Solving for h yields h= 2 vy0 2g v 2 sin2 45◦ = 0 2g 2 1 v0 = . 4 g 018 (part 3 of 3) 10 points The speed vhmax of the cannonball at its maximum height is given by 1. vhmax Which free body diagram best describes this system? normal 1. friction gravitational 1 = v0 2 2. vhmax = v0 √ 3. vhmax = 3 v0 2. normal gravitational 4. vhmax = 2 v0 5. vhmax = 4 v0 √ 6. vhmax = 2 v0 1 7. vhmax = √ v0 correct 2 2 8. vhmax = v0 3 1 9. vhmax = √ v0 3 1 10. vhmax = v0 4 Explanation: At the top of the cannonball’s trajectory, vy = 0. Hence the speed is equal to vx . 1 |v| = vx = v0 cos 45◦ = √ v0 . 2 019 (part 1 of 3) 10 points Consider a book that remains at rest on an incline. friction 3. normal gravitational normal 4. friction gravitational 5. normal gravitational correct Version 055 – Midterm 1 – OConnor (05141) 12 021 (part 3 of 3) 10 points Are they opposite in direction? 6. friction 1. Yes, the normal force acts opposite the weight force because the book is stationary. gravitational 2. Yes, the normal force always acts opposite the weight force. Otherwise, the book would fall through the inclined plane. 7. friction normal 3. No, the normal force acts perpendicular to the surface of the inclined plane. correct gravitational 4. No, the normal force acts up the incline so that the book does not slide down. 8. friction gravitational Explanation: The frictional force keeps the book from sliding down. The normal force acts perpendicular to the inclined surface. The gravitational force acts down. 020 (part 2 of 3) 10 points Compare the normal force exerted on the book by the inclined plane and the weight force exerted on the book by the earth. Are they equal in magnitude? Explanation: The normal force acts perpendicular to the surface of the inclined plane. Because the plane is inclined, this is not opposite in direction to the weight force. 022 (part 1 of 1) 10 points A man stands in an elevator in the university’s administration building and is accelerating upwards. (During peak hours, this does not happen very often.) Elevator Cable 1. Yes 2. No correct 3. Their magnitudes cannot be determined since the forces are not in the same direction. Explanation: They are not equal in magnitude. The normal force given by N = m g cos θ, with θ the angle the incline makes with the ground. Since | cos θ| is less than 1 as long as the incline is not horizontal, the magnitude of the normal force, N , will be less than the magnitude of the weight, m g. Choose the correct free body diagram for the man, where Fi,j is the force on the object i, from the object j. 1. Fman, elevator Felevator, cable Version 055 – Midterm 1 – OConnor (05141) 2. Felevator, cable 13 shown below. The left-hand cable had tension T2 and makes an angle of 44 ◦ with the ceiling. The right-hand cable had tension T1 and makes an angle of 48◦ with the ceiling. Fman, earth T 3. 44 ◦ 48◦ T 1 2 Fman, cable 679 N a) What is the tension in the cable labeled T1 slanted at an angle of 48◦ ? 1. 381.046 N 4. Felevator, cable Fman, floor 2. 385.203 N 3. 392.64 N Fman, earth 4. 417.991 N 5. Fman, floor 5. 429.023 N 6. 438.515 N Fman, earth 7. 452.597 N 8. 457.621 N correct 6. Fman, acceleration 9. 463.535 N Explanation: Only the forces acting directly on the man are to be in the free body diagram. Therefore, the force from the cable should be omitted, while those from gravity and from the floor’s normal force should be included. 023 (part 1 of 2) 10 points Consider the 679 N weight held by two cables F 2 F θ2 1 10. 488.729 N correct Explanation: Observe the free-body diagram below. Wg θ1 Version 055 – Midterm 1 – OConnor (05141) Note: The sum of the x- and y-components of F1 , F2 , and Wg are equal to zero. Given : Wg = 679 N , θ1 = 48◦ , and θ2 = 44◦ . Basic Concept: Vertically and Horizontally, we have x Fnet y Fnet F1x − F2x = =0 = F1 cos θ1 − F2 cos θ2 = 0 (1) y y = F1 + F2 − Wg = 0 = F1 sin θ1 + F2 sin θ2 − Wg = 0 (2) Solution: Using Eq. 1, we have F2 = F 1 cos θ1 . cos θ2 14 7. 468.891 N 8. 488.842 N 9. 502.789 N 10. 512.395 N Explanation: Using Eq. 1, we have F2 = F 1 cos θ1 cos θ2 = (488.729 N) cos 48◦ cos 44◦ = 454.617 N . 025 (part 1 of 3) 10 points Consider a man standing on a scale which is placed in an elevator. When the elevator is stationary, the scale reading is Ss . Substituting F2 from Eq. 2 into Eq. 1, we have F1 sin θ1 + F2 sin θ2 = Wg cos θ1 sin θ2 = Wg F1 sin θ1 + F1 cos θ2 F1 sin θ1 + F1 cos θ1 tan θ2 = Wg Wg F1 = sin θ1 + cos θ1 tan θ2 679 N F1 = ◦ sin 48 + cos 48◦ tan 44◦ = 488.729 N 024 (part 2 of 2) 10 points a) What is the tension in the cable labeled T2 slanted at an angle of 44◦ ? 1. 397.788 N 2. 414.851 N 3. 421.94 N 4. 428.768 N 5. 447.713 N 6. 454.617 N correct Scale Find Sup , the scale reading when the elevator is moving upward with acceleration 1 ~a = g ̂, in terms of Ss . 5 7 1. Sup = Ss 6 2 2. Sup = Ss 3 4 3. Sup = Ss 5 4. Sup = 0 m/s2 Version 055 – Midterm 1 – OConnor (05141) 3 5. Sup = Ss 5 6 6. Sup = Ss correct 5 3 7. Sup = Ss 2 5 8. Sup = Ss 4 7 9. Sup = Ss 5 5 10. Sup = Ss 6 Explanation: Basic Concepts: Newton’s 2nd law X ~ = m ~a . F Equation 1 in the upward direction reads Sup − m g = m a , and since a = 1 mg 5 Sup = m g + 6 mg 5 6 = Ss . 5 1 mg 5 = Thus the scale gives a higher reading when the elevator is accelerating upwards. (1) 026 (part 2 of 3) 10 points Find Sdown , the scale reading when the elevator is moving downward with constant velocity, in terms of Ss . 1. Sdown = 2. Sdown = 3. Sdown = 4. Sdown = 5. Sdown = Ss − m g = 0 , or Ss = m g as we would expect. Call the scale reading for this part Sup (in units of Newtons). Consider the free body diagram for each the case where the elevator is accelerating down (left) and up (right). The man is represented as a sphere and the scale reading is represented as S. Sup a Sdown a mg 1 g at this particular instant, 5 Sup − m g = Solution We consider the forces acting on the man. Taking up (̂) as positive, we know that m g acts on the man in the downward (−̂) direction. The only other force acting on the ~ s from the scale. By man is the normal force S the law of action and reaction, the force on the scale exerted by the man (i.e., the scale reading) is equal in magnitude but opposite ~ s vector. Initially, the elin direction to the S evator is moving upward with constant speed (no acceleration) so mg 15 6. Sdown = 7. Sdown = 7 Ss 5 5 Ss 6 4 Ss 5 5 Ss 4 3 Ss 2 3 Ss 5 6 Ss 5 8. Sdown = Ss correct 2 Ss 3 7 10. Sdown = Ss 6 Explanation: At constant velocity the acceleration is zero. Equation (1) in the upward direction reads 9. Sdown = Sdown − m g = 0 , Version 055 – Midterm 1 – OConnor (05141) 16 since a = 0 at this particular instant 027 (part 3 of 3) 10 points Suppose all of a sudden the rope breaks when the elevator is moving upward. What is the reading of the scale Sb after the rope is broken? 1. Sb = 2. Sb = 3. Sb = 4. Sb = 5. Sb = 6. Sb = 7. Sb = 8. Sb = 9. Sb = 2 Ss 3 3 Ss 2 7 Ss 6 4 Ss 5 5 Ss 6 7 Ss 5 3 Ss 5 5 Ss 4 6 Ss 5 θ2 42 0 θ1 0N Thus the scale gives the same reading as when stationary if the elevator is moving downward at constant velocity. 028 (part 1 of 2) 10 points Hint: sin2 θ + cos2 θ = 1 . Consider the 674 N weight held by two cables shown below. The left-hand cable had tension 420 N and makes an angle of θ2 with the ceiling. The right-hand cable had tension 500 N and makes an angle of θ1 with the ceiling. N 50 Sdown − m g = 0 Sdown = m g − 0 = mg = Ss . 674 N a) What is the angle θ1 which the righthand cable makes with respect to the ceiling? 1. 35.9706 ◦ 2. 36.5949 ◦ 3. 37.0824 ◦ 4. 37.9641 ◦ 5. 39.627 ◦ 6. 41.9197 ◦ 7. 48.4217 ◦ 8. 49.1665 ◦ 10. Sb = 0 N correct Explanation: When the rope breaks, the whole system moves under the influence of gravity with the acceleration ~a = −g ̂, so Sb − m g = −m g Sb = 0 N . Intuitively this implies that there is no force between the person and the scale. Note: There are four different presentations of this problem. 9. 50.9586 ◦ 10. 51.5544 ◦ correct Explanation: Observe the free-body diagram below. Version 055 – Midterm 1 – OConnor (05141) 2 θ1 1 F F θ2 (420 N)2 − 2 (500 N) (674 N) = 51.5544◦ . ¸ 029 (part 2 of 2) 10 points b) What is the angle θ2 which the left-hand cable makes with respect to the ceiling? 1. 41.1969 ◦ Wg Note: The sum of the x- and y-components of F1 , F2 , and Wg are equal to zero. Given : Wg = 674 N , F1 = 500 N , F2 = 420 N . Basic Concepts: X 17 2. 42.211 ◦ correct 3. 43.6346 ◦ 4. 44.1286 ◦ and 5. 46.0622 ◦ 6. 46.9763 ◦ 7. 47.6435 ◦ x F =0 F1x F1 cos θ1 F12 cos2 θ1 X and Fy = F2x = F2 cos θ2 = F22 cos2 θ2 8. 51.5378 ◦ (1) (2) =0 F1y + F2y + F3y = 0 F1 sin θ1 + F2 sin θ2 − F3 = 0 F1 sin θ1 = −F2 sin θ2 + F3 F12 sin2 θ1 = F22 sin2 θ2 −2 F2 F3 sin θ2 + F32 , since (3) F3 sin θ3 = F3 sin 270◦ = −F3 , and F3 cos θ3 = F3 cos 270◦ = 0 . Solution: Since sin2 θ + cos2 θ = 1 and adding Eqs. 2 and 3, we have F22 = F12 − 2 F1 F3 sin θ1 + F32 F 2 + F32 − F22 sin θ1 = 1 2 F 1 F3 µ 2 ¶ F3 + F12 − F22 θ1 = arcsin 2 F 1 F3 · (674 N)2 + (500 N)2 = arcsin 2 (500 N) (674 N) 9. 53.4745 ◦ 10. 54.7926 ◦ Explanation: Using Eq. 1, we have F1 cos θ1 F2 µ ¶ F1 θ2 = arccos cos θ1 F2 ¶ µ 500 N ◦ cos 51.5544 = arccos 420 N = 42.211◦ . cos θ2 = 030 (part 1 of 3) 10 points A 3.14 kg block slides down a smooth, frictionless plane having an inclination of 32◦ . The acceleration of gravity is 9.8 m/s2 . Version 055 – Midterm 1 – OConnor (05141) 18 W|| = m g sin θ = m a 5m 4 3.1 3. 8 Motion has a constant acceleration. Recall the kinematics of motion with constant acceleration. Solution: Because the block slides down along the plane of the ramp, it seems logical to choose the x-axis in this direction. Then the y-axis must emerge perpendicular to the ramp, as shown. Let us now examine the forces in the xdirection. Only the weight has a component along that axis. So, by Newton’s second law, kg µ= 0 32◦ Find the acceleration of the block. 1. 3.3518 m/s2 2. 3.67115 m/s2 X 3. 3.98602 m/s2 Fx = m a = m g sin θ Thus 4. 4.14166 m/s2 a = g sin θ 5. 4.29604 m/s2 With our particular value of θ, 6. 4.60082 m/s2 a = (9.8 m/s2 ) sin 32◦ = 5.19321 m/s2 7. 5.04737 m/s2 8. 5.19321 m/s2 correct 031 (part 2 of 3) 10 points What is the block’s speed when, starting from rest, it has traveled a distance of 3.85 m along the incline. 1. 4.03606 m/s 9. 5.7603 m/s2 10. 5.89779 m/s2 Explanation: 2. 4.15573 m/s Given : m = 3.14 kg , θ = 32◦ , and µs = 0 . 3. 4.93036 m/s 4. 5.11916 m/s Consider the free body diagram for the block N m in gs θ N os µN g c m = W = mg Basic Concepts: Fx,net = F cos θ − W|| = 0 θ 5. 5.3329 m/s 6. 5.8349 m/s 7. 6.0893 m/s 8. 6.2011 m/s 9. 6.32358 m/s correct 10. 6.74584 m/s Explanation: Version 055 – Midterm 1 – OConnor (05141) Since v0 = 0, a = 5.19321 m/s2 and L = 3.85 m, vf2 = v02 + 2 a (x − x0 ) = 2 (5.19321 m/s2 ) (3.85 m) vf = 6.32358 m/s . It is very important to note at this point that neither of these values depended on the mass of the block. This may seem odd at first, but recall what Galileo discovered 300 years ago – objects of differing mass fall at the same rate. 032 (part 3 of 3) 10 points What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 3.85 m down the incline? 1. 18.6733 N 2. 19.6422 N 3. 20.138 N 4. 24.5779 N 5. 24.9669 N 6. 26.0961 N correct 19 Two blocks on a frictionless horizontal surface are connected by a light string. The acceleration of gravity is 9.8 m/s2 . 12.4 kg T 20.7 kg 49.8 N µ Find the acceleration of the system. 1. 1.45455 m/s2 2. 1.50453 m/s2 correct 3. 1.63548 m/s2 4. 1.67869 m/s2 5. 1.93723 m/s2 6. 1.97674 m/s2 7. 2.04829 m/s2 8. 2.14503 m/s2 9. 2.17932 m/s2 10. 2.25905 m/s2 Explanation: 7. 32.7944 N Let : 8. 36.7439 N 9. 38.4731 N 10. 39.3247 N Explanation: By examining the free body diagram again, we see that the force in the y direction is given by X F = −m g cos θ = FN . =⇒ | FN | = m g cos θ = 3.14 kg (9.8 m/s2 ) cos 32◦ = 26.0961 N. 033 (part 1 of 4) 10 points m1 = 12.4 kg , m2 = 20.7 kg , F = 49.8 N , and µ = 0 , Parts 1 and 2 µ = 0.109 , Parts 3 and 4. Apply Newton’s second law to the each block F1net = m1 a = T , (1) F2net = m2 a = F − T . Adding these equations gives us (m1 + m2 ) a = F a= F . m1 + m 2 (2) Version 055 – Midterm 1 – OConnor (05141) 034 (part 2 of 4) 10 points What is the tension in the string between the blocks? 1. 7.21356 N 20 10. 1.31433 m/s2 Explanation: The friction force on each block is Ff r = µ m g . 2. 8.64446 N 3. 9.12074 N 4. 9.31809 N 5. 9.78168 N 6. 11.6628 N 7. 12.9232 N 8. 18.6562 N correct so apply Newton’s second law to each block F1net = m1 a1 = T1 − µ m1 g (3) F2net = m2 a1 = F − T1 − µ m2 g (4) Adding these equations gives (m1 + m2 ) a1 = F − µ m1 g − µ m2 g a1 = a1 = 9. 19.0545 N 10. 19.893 N Explanation: From equation 1, T = m1 a . F − µ (m1 + m2 ) g m1 + m 2 F − µg. m1 + m 2 036 (part 4 of 4) 10 points What would be the new tension in the string between the blocks? 1. 7.21356 N 2. 8.64446 N 035 (part 3 of 4) 10 points If the surface were frictional, and the coefficient of kinetic friction between each block and the surface is 0.109, what would be the new acceleration? 1. 0.436332 m/s2 correct 3. 9.12074 N 4. 9.31809 N 5. 9.78168 N 2. 0.552573 m/s2 6. 11.6628 N 3. 0.714369 m/s2 7. 12.9232 N 4. 0.750544 m/s2 8. 18.6562 N correct 5. 0.870439 m/s2 9. 19.0545 N 6. 0.898744 m/s2 7. 0.927834 m/s2 8. 1.09643 m/s2 9. 1.18841 m/s2 10. 19.893 N Explanation: From equation 3, T1 = µ m 1 g + m 1 a 1 .