Quick QuizzesAnswers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 3 Vectors and Two-Dimensional Motion Quick Quizzes 1. (c). The largest possible magnitude of the resultant occurs when the two vectors are in the s me direction. In this case, the magnitude of the resultant is the sum of the magnitudes of JG and BJ: R = A + B = 20 units. The smallest possible magnitude of the resultant occurs when the two vectors are in opposite directions, and the magnitude is the difference of the magnitudes of A and B : R = |A – B|= 4 units. aAGJG JG2. (b). The resultant has magnitude A + B when AJG is oriented in the same direction as BJG. 3. Vector x component y component AJG – + BJG + – AJG + BJG– – 4. (b). If velocity is constant, the acceleration (rate of change in velocity) is zero. An object may have constant speed (magnitude of velocity) but still be accelerating due to a change in direction of the velocity. If an object is following a curved path, it is accelerating because the velocity is changing in direction. 5. (a). Any change in the magnitude and/or direction of the velocity is an acceleration. The gas pedal and the brake produce accelerations by altering the magnitude of the velocity. The steering wheel produces accelerations by altering the direction of the velocity. 6. (c). A projectile has constant horizontal velocity. Thus, if the runner throws the ball straight up into the air, the ball maintains the horizontal velocity it had before it was thrown (that is, the velocity of the runner). In the runner’s frame of reference, the ball appears to go straight upward and come straight downward. To a stationary observer, the ball follows a parabolic trajectory, moving with the same horizontal velocity as the runner and staying above the runner’s hand. 7. (b). The velocity is always tangential to the path while the acceleration is always directed vertically downward. Thus, the velocity and acceleration are perpendicular only where the path is horizontal. This only occurs at the peak of the path. 5556 CHAPTER 3 Answers to Even Numbered Conceptual Questions 2. The magnitudes add when and AJGBJG are in the same direction. The resultant will be zero when the two vectors are equal in magnitude and opposite in direction. 4. The minimum sum for two vectors occurs when the two vectors are opposite in direction. If they are unequal, their sum cannot add to zero. 6. The component of a vector can only be equal to or less than the vector itself. It can never be greater than the vector. 8. The components of a vector will be equal in magnitude if the vector lies at a 45° angle with the two axes along which the components lie. 10. They both start from rest in the downward direction and accelerate alike in the vertical direction. Thus, they reach the ground with the same vertical speed. However, the ball thrown horizontally had an initial horizontal component of velocity which is maintained throughout the motion. Thus, the ball thrown horizontally moves with the greater speed. 12. The car can round a turn at a constant speed of 90 miles per hour. Its velocity will be changing, however, because it is changing in direction. 14. The balls will be closest at the instant the second ball is projected. The first ball will always be going faster than the second ball. There will be a one second time interval between their collisions with the ground. The two move with the same acceleration in the vertical direction. Thus, changing their horizontal velocity can never make them hit at the same time. 16. Let v and 0x 0yv represent its original velocity components. We know that the vertical component of velocity is zero at the top of the trajectory. Thus, 00yvgt=− and the time at the top of the trajectory is 0yg=tv. (a) 00yxvxvg= and 202yvyg= (b) Its velocity is horizontal and equal to . 0xv (c) Its acceleration is vertically downward, –g. With air resistance, the answers to (a) and (b) would be smaller. As for (c) the magnitude would be somewhat larger because the total acceleration would have a component horizontally backward in addition to the vertical component of –g. 18. The equations of projectile motion are only valid for objects moving freely under the influence of gravity. The only acceleration such an object has is the free-fall acceleration, g, directed vertically downward. Of the objects listed, only a and d meet this requirement. 20. The passenger sees the ball go into the air and come back in the same way he would if he were at rest on Earth. An observer by the tracks would see the ball follow the path of a projectile. If the train were accelerating, the ball would fall behind the position it would reach in the absence of the acceleration.Vectors and Two-Dimensional Motion 57 Answers to Even Numbered Problems 2. (a) Approximately 484 km (b) Approximately 18.1° N of W 4. Approximately 83 m at 33° N of W 6. (a) Approximately 6.1 units at 113° (b) Approximately 15 units at 23° 8. (a) Approximately 5.2 m at + 60° (b) Approximately 3.0 m at -30° (c) Approximately 3.0 m at + 150° (d) Approximately 5.20 m at -60° 10. 1.31 km north, 2.81 km east 12. 358 m at 2.00° S of E 14. 42.7 yards 16. 788 mi at 48.1° N of E 18. (a) 185 N at 77.8° (b) 185 N at 258° 20. (a) 74.6° N of E (b) 470 km 22. 2.68 ft (0.817 m) 24. 3.19 s, 36 at 60.1° below the horizontal .1 m s26. 2.8 m from base of table; 5.0 m s, 5.4 m sxyvv==− 28. 337.23 10 m, 1.68 10 mxy=× =×30. (a) clears the bar by 0.85 m (b) falling, 13.4 m syv=− 32. 18.7 m 34. 9.91 m s 36. 61 s 38. (a) 10.1 m s at 8.53° E of N (b) 45.0 m 40. (1. ), Yes 81 m s, 5.43 m s42. 15.3 m 44. 7.87 N at 97.8° counterclockwise from a horizontal line to the right58 CHAPTER 3 46. (a) 57.4m, 1.22 s (b) 57.4 m, 1.22 s 48. 0.344 m, 2.34 m 50. (a) 0.85 m s (b) 2.1 m s 52. 14 m s 54. See Solution Section 56. 29.4 m s 58. 10.7 m s 60. (a) , (b) , 132 cm at 69.6m=°dJG146 cm at 69.6m′=°dJG111 cm at 70.0f=°dJG132 cm at 70.0f′=°dJG cm at 65.8° m14.0f′′ ′∆= − =dd dJGJGJG64. (a) 1.20 m s , 0 (b) 0.960 m (c) 0.500 m s 66. 3.96 m s 68. 26 knots at 50° south of east; 20 knots due south 70. (a) 26.6° (b) 0.950 72. 4.12 mVectors and Two-Dimensional Motion 59 Problem Solutions 3.1 Your sketch should be drawn to scale, and be