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Version 073 – midterm 1 v1 – shih – (58505) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. may the force be with you 001 10.0 points A rope has a cross section A = 8.18 m2 and density ρ = 2270 kg/m3 . The linear density of the rope µ, defined to be the mass per unit length, can be written in the form µ = ρx Ay . Based on dimensional analysis, determine the powers x and y by choosing an expression below. A2 ρ2 ρ 2. µ = A 1 3. µ = ρA 1. µ = 6. µ = 7. µ = 8. µ = 9. µ = 2y − 3x = −1 2y − 3 = −1 y = 1. Thus µ = ρ1 A1 = ρ A . 002 10.0 points Despite a very strong wind, a tennis player manages to hit a tennis ball with her racquet so that the ball passes over the net and lands in her opponent’s court. Consider the following forces: 1. A downward force of gravity, 2. A force by the hit, and 3. A force exerted by the air. Which of the above forces is (are) acting on the tennis ball after it has left contact with the racquet and before it touches the ground? 1. 1 and 3. correct 2. 1, 2, and 3. 4. µ = ρ A2 5. µ = 1 A2 ρ ρ A2 A ρ2 A ρ 1 ρ A2 3. 1 only. 4. 2 and 3. 5. 1 and 2. Explanation: The forces acting on the tennis ball after it has left contact with the racquet are gravity and air resistance. A force by the hit is no longer acting on the ball. 10. µ = ρ A correct Explanation: [x] means “the units of x”. [µ] = [ρx Ay ] x y [M] M = L2 3 [L] L +1 −1 [M] [L] = Mx L−3x L2y = Mx L2y−3x Equating the exponents, x=1 003 10.0 points The diagram describes the acceleration vs time behavior for a car moving in the xdirection. a Pb Q b 0 At the point Q, the car is moving t Version 073 – midterm 1 v1 – shih – (58505) dv >0 . which means positive derivative dt In other words, the car has positive acceleradv tion a = > 0 . Since v > 0, this means the dt car is accelerating. 1. with a decreasing speed. 2. with an increasing speed. correct 3. with a constant speed. Explanation: dv a = . As long as the acceleration is dt positive the velocity is always increasing. 004 10.0 points A car moves along a straight line. The graph below shows its velocity in the x-direction vx as a function of time. 6 vx (m/s) 5 P 4 3 2 1 0 0 1 2 3 4 5 time (s) 2 6 7 8 9 005 10.0 points A jet airliner moving initially at 786 mph to the east where there is no wind moves into a region where the wind is blowing at 414 mph in a direction 79◦ north of east. What is the new speed of the aircraft with respect to the ground? 1. 635.595 2. 865.317 3. 605.287 4. 1432.06 5. 1030.55 6. 737.237 7. 1538.03 8. 1004.33 9. 955.705 10. 1157.37 Correct answer: 955.705 mph. Explanation: Which statement correctly describes the car’s motion at point P on the graph? Vwind φ 1. The car is stationary. 2. The car is accelerating forward. correct 3. The car has zero acceleration. 6. The car’s direction is about 30◦ off the x axis. Explanation: The graph depicts the velocity v(t) as a function of time and NOT the position x(t). At point P on the graph, the velocity is positive (indeed, v ≥ 0 at all times depicted on the graph) and the curve v(t) slopes upward, θ Vjet The new velocity of the aircraft is the sum of the initial velocity and the velocity of the wind: 4. The car is decelerating. 5. The car is climbing a hill. V ~vf = ~vi + ~vwind = vx ı̂ + vy ̂ where vx = vi + vwind cos φ vy = vwind sin φ with φ being the angle between vwind and vjet . So the speed is q |~v | = vx2 + vy2 . 006 10.0 points Version 073 – midterm 1 v1 – shih – (58505) 10 m y A student rolls a 5 kg ball off the horizontal roof of the building in the direction of the target. v 10 m 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. 3 θ′ 45◦ x 5m R/2 R The time of flight tR of the cannonball for this maximum range R is given by 2 v0 3 g 1 v0 2. tR = √ 3 g √ v0 3. tR = 3 g 1 v0 4. tR = 2 g 1 v0 5. tR = 4 g 1 v0 6. tR = √ 2 g v0 7. tR = 4 g v0 8. tR = g v0 9. tR = 2 g √ v0 correct 10. tR = 2 g Explanation: The cannonball’s time of flight is 2 v0y 2 v0 sin 45◦ √ v0 tR = = = 2 . g g g 1. tR = 007 10.0 points A target lies flat on the ground 5 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m/s2 . Air resistance is negligible. The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly √ 1. v = 2 5 m/s. √ 2. v = 5 2 m/s. √ 5 3 3. v = m/s. 3 √ 5 m/s. 4. v = 5 √ 2 m/s. 5. v = 5 √ 6. v = 5 3 m/s. √ 7. v = 3 5 m/s. √ 5 2 8. v = m/s. correct 2 √ 3 9. v = m/s. 5 √ 10. v = 5 5 m/s. Explanation: m = 5 kg , not required h = 10 m , x = 5 m , and g = 10 m/s2 . Observe the motion in the vertical direction only and it is a purely 1-dimension movement Version 073 – midterm 1 v1 – shih – (58505) with a constant acceleration. So the time need for the ball to hit the ground is s 2h t= g and the horizontal speed should be v= x t for the ball to hit the target. Therefore r g v=x 2h s 10 m/s2 = (5 m) 2 (10 m) 5 = √ m/s 2 √ 5 2 = m/s . 2 008 10.0 points A block of mass 9 m can move without friction on a horizontal table. This block is attached to another block of mass 6 m by a cord that passes over a frictionless pulley, as shown. 4 2 g correct 5 13 g 6. k~ak = 20 2 7. k~ak = g 9 18 8. k~ak = g 23 6 9. k~ak = g 11 13 10. k~ak = g 30 Explanation: 5. k~ak = Let : m1 = 9 m , m2 = 6 m , and m1 + m2 = 15 m . a m1 g T m2 9m 6m Let the tension of the cord be T . Apply Newton’s second law to the two masses: T = m1 a . (1) m2 g − T = m2 a , If the masses of the cord and the pulley are negligible, what is the magnitude of the acceleration of the descending block? 14 g 17 8 2. k~ak = g 11 17 3. k~ak = g 19 4 g 4. k~ak = 11 so (2) m2 g = m1 a + m2 a so m2 a= g m1 + m2 2 6m g = g. = 15 m 5 1. k~ak = keywords: 009 10.0 points A toy cannon fires a 0.0714 kg shell with initial Version 073 – midterm 1 v1 – shih – (58505) velocity vi = 9.3 m/s in the direction θ = 44 ◦ above the horizontal. The shell’s trajectory curves downward because of gravity, so at the time t = 0.535 s the shell is below the straight line by some vertical distance ∆h. 5 1 Thus, x = x̂ but y = ŷ − gt2 , or in other 2 words, the shell deviates from the straight-line path by the vertical distance ∆h = ŷ − y = 9. 3 m ∆h /s ∆y ◦ 44 y g t2 . 2 Note: This result is completely independent on the initial velocity vi or angle θ of the shell. It is a simple function of the flight time t and nothing else (besides the constant g = 9.8 m/s2 ). g t2 2 (9.8 m/s2 ) (0.535 s)2 = 2 = 1.4025 m . ∆h = ∆x Find this distance ∆h in the absence of air resistance. The acceleration of gravity is 9.8 m/s2 . 1. 0.586608 2. 3.01181 3. 1.4025 4. 1.63701 5. 0.696432 6. 0.34151 7. 0.252492 8. 0.935748 9. 0.889232 10. 2.88262 Correct answer: 1.4025 m. Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: x̂ = t vi cos θ , ŷ = t vi sin θ . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate at constant downward acceleration ay = −g, hence x = t vi cos θ, g t2 . y = t vi sin θ − 2 010 10.0 points A car makes a 310 km trip at an average speed of 35.7 km/h. A second car starting 1 h later arrives at their mutual destination at the same time. What was the average speed of the second car for the period that it was in motion? 1. 37.7808 2. 55.5063 3. 59.6118 4. 58.5872 5. 47.9567 6. 42.5061 7. 40.3463 8. 35.481 9. 41.9242 10. 34.226 Correct answer: 40.3463 km/h. Explanation: Let : d = 310 km , v = 35.7 km/h , ∆t = 1 h . Average velocity is v = d . t and Version 073 – midterm 1 v1 – shih – (58505) ∆t = t1 − t2 = v = 2 m/s , r = 8 m. d d − v1 v2 d d = − ∆t v2 v1 v1 v2 1 · = d d v − ∆t 1 v1 d v1 v2 = d − v1 ∆t (310 km)(35.7 km/h) = 310 km − (35.7 km/h)(1 h) v2 r (2 m/s)2 = (8 m) = 0.5 m/s2 . 012 10.0 points The mass of the worker m1 = 50 kg. The mass of the block at the end of the rope, m2 = 100 kg. 10.0 points b A student is swinging a 6 kg ball in a circular path in the vertical plane. Consider the instant when the string is horizontal as the ball is on its way up as shown in the figure. T T m2 a 8m 2 m/s a y m1 x M and ac = = 40.3463 km/h . 011 6 Determine the acceleration. At this instant, the instantaneous tangential speed is 2 m/s , and the radius is 8 m. What is the centripetal acceleration? 1. 10.125 2. 5.33333 3. 40.5 4. 1.33333 5. 6.0 6. 1.0 7. 2.77778 8. 0.5 9. 9.14286 10. 4.0 Correct answer: 0.5. m2 g = 2g m1 m2 − m1 2. a = g=g m1 m2 − m1 1 3. a = g = g correct m1 + m2 3 Explanation: Apply “F = m a”on the m1 + m2 mass system. The net force is 1. a = m2 g − m1 g = (m1 + m2 ) a This leads to a= 1 m2 − m1 g= g m1 + m2 3 Explanation: 013 Let : M = 6 kg , 10.0 points Version 073 – midterm 1 v1 – shih – (58505) Consider a man standing on a scale which is placed in an elevator. When the elevator is stationary, the scale reading is Ss . 7 ~ 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 Ss − m g = 0 , Scale Find Sup , the scale reading when the elevator is moving upward with acceleration 1 ~a = g ̂, in terms of Ss . 3 5 1. Sup = Ss 4 2. Sup = 2 Ss 3 Ss 2 1 = Ss 2 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 mg Equation 1 in the upward direction reads Sup − m g = m a , 3. Sup = 4. Sup 5. Sup = 0 m/s2 6. Sup = 7. Sup = 8. Sup = 9. Sup = 10. Sup = 1 Ss 3 5 Ss 3 3 Ss 4 4 Ss correct 3 2 Ss 3 Explanation: 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 and since a = Sup − m g = 1 g at this particular instant, 3 1 mg 3 Sup = m g + 4 4 1 m g = m g = Ss . 3 3 3 Thus the scale gives a higher reading when the elevator is accelerating upwards. Comments for Matter Interaction readers: Apply the derivative form of the momentum dp principle = Fnet . The left hand side is dt 1 m a = m g, and the right hand side is Sup − 6 1 m g. This leads to Sup − m g = m g. 6 014 10.0 points An object was suspended in a fixed place (y = 0) and then allowed to drop in a free fall. Version 073 – midterm 1 v1 – shih – (58505) Taking up as the positive vertical direction, which of the following graphs correctly represents its vertical motion as displacement vs time? 8 Explanation: The object is undergoing a constant downward gravitational acceleration g. The slope of a position vs time curve represents the velocity. y t 1. Let : y t 2. a = −g , v0 = 0 m/s , y0 = 0 m . y = y0 + v0 t + y t 3. t 015 10.0 points Two 30 N forces and a 60 N force act on a hanging box as shown. y 30 N t 5. 1 2 1 a t = − g t2 . 2 2 This is a parabolic shaped curve starting at y = 0 with a continuously decreasing slope as time increases. y 4. and 30 N y t 6. correct y t 7. 60 N Will the box experience acceleration? y t 8. 1. Yes; upward. 2. Yes; downward. correct y t 9. 3. Unable to determine without the angle. 4. No; it is balanced. y 10. t Explanation: The horizontal components of the two 30 N forces cancel, leaving an upward force that is less than 60 N. Thus, the net force on the box is down, causing it to accelerate downward. Version 073 – midterm 1 v1 – shih – (58505) 016 10.0 points A particle undergoes two displacements. The first has a magnitude of 161 cm and makes an angle of 138.1◦ with the positive x axis. The resultant of the two displacements is 105 cm directed at an angle of 33.6◦ to the positive x axis. Find the magnitude of the second displacement. 1. 213.099 2. 178.491 3. 226.149 4. 198.204 5. 161.148 6. 245.214 7. 201.354 8. 189.272 9. 174.295 10. 176.24 Correct answer: 213.099 cm. Explanation: Let : |A| = 161 cm , θa = 138.1◦ , |R| = 105 cm , θR = 33.6◦ . and A 138.1◦ R 33.6◦ x B R=A+B B = R−A. The components of the second displacement B are Bx = Rx − Ax = R cos θR − A cos θA = (105 cm) cos 33.6◦ − (161 cm) cos 138.1◦ = 207.291 cm and By = Ry − Ax = R sin θR − A sin θA = (105 cm) sin 33.6◦ − (161 cm) sin 138.1◦ = −49.4149 cm , so the magnitude of B is q |B| = Bx2 + By2 q = (207.291 cm)2 + (−49.4149 cm)2 = 213.099 cm . 9