"CSULA- PHYSICS 211"
CSULA- PHYSICS 211 Summer’12 Solution: Material to Study Chapter 3 OBJECTIVE QUESTIONS OQ-3 (7e- Q3) The vector 2D1 will be twice as long as D1 and in the opposite direction, namely northeast. Adding D 2 , which is about equally long and southwest, we get a sum that is still longer and due east, choice (a). OQ-8 (7e-None) 8m/s, choice (c). OQ-10 (7e-Q8) Take the difference of the coordinates of the ends of the vector. Final first means head end first: 1 (2) = 3 cm, answer (a). CONCEPTUAL QUESTIONS CQ-2 (7e-Q10) Vectors A and B are perpendicular to each other. CQ-3 (7e-Q12) Addition of a vector to a scalar is not defined. Think of numbers of apples and of clouds. CQ-4 (7e-Q11) No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude. PROBLEMS P6 (7e-P5) y We have r x 2 y 2 and tan 1 . x y (a) The radius for this new point is: x 2 y 2 x 2 y 2 r and its angle is tan 1 180 . x (b) (2x)2 (2y)2 2r This point is in the third quadrant if x , y is in the first quadrant or in the fourth quadrant if x , y is in the second quadrant. It is at an angle of 180 . (c) (3x)2 (3y)2 3r This point is in the fourth quadrant if x , y is in the first quadrant or in the third quadrant if x , y is in the second quadrant. It is at an angle of . P7 (7e-P7) x tan 35.0 100 m x 100 m tan 35.0 70.0 m P15 (7e-P15) A x 25.0 Ay A y 40.0 We observe that: tan . Ax A Ax Ay 2 2 25.02 40.02 47.2 units Ay 40.0 So: tan 1 tan tan 1 1.60 58.0 . Ax 25.0 The diagram shows that the angle from the +x axis can be found by subtracting from 180°: 180 58 122 . P23 (7e-P25) (a) A B 3ˆ 2ˆj ˆ 4ˆj i i 2ˆ 6ˆ i j (b) A B 3ˆ 2ˆj iˆ 4ˆj i ˆ 4i 2ˆ j (c) A B 22 62 6.32 (d) A B 42 22 4.47 A B tan 1 71.6 288 A B tan 1 26.6 6 2 (e) AND 2 4 P29 (7e-P31) (a) F F1 F2 ˆ ˆ F 120cos 60.0 i 120sin 60.0 ˆ 80.0cos 75.0 i 80.0sin 75.0 ˆ j j F 60.0i 104ˆ 20.7ˆ 77.3ˆ 39.3i 181ˆ N ˆ j i j ˆ j (b) F3 F 39.3iˆ 181ˆj N F 39.32 1812 185 N tan 1 181 77.8 39.3 P45 (7e-P40) (a) ˆ E 17.0 cm cos27.0i 17.0 cm sin 27.0ˆ j E 15.1iˆ 7.72ˆj cm (b) ˆ F 17.0 cm sin 27.0i 17.0 cm cos27.0ˆ j F 7.72ˆi 15.1ˆj cm Note that we do not need to explicitly identify the angle with the positive x axis. (c) ˆ G 17.0 cm sin 27.0i 17.0 cm cos27.0ˆ j G 7.72iˆ 15.1ˆj cm y 27.0° 27.0° F G E 27.0° x P57 (7e-P47) Let represent the angle between the directions of A and B . Since A and B have the same magnitudes, A , B , and R A B form an isosceles triangle in which the angles are 180 , , and . The magnitude of R is then R 2A cos . This can be seen 2 2 2 from applying the law of cosines to the isosceles triangle and using the fact that B A . Again, A , – B , and D A B form an isosceles triangle with apex angle . Applying the law of cosines and the identity 1 cos 2sin 2 gives the magnitude of D as D 2A sin . The problem requires that R 100D . 2 2 Thus, 2A cos 200A sin . This gives tan 0.010 and 1.15 . R B 2 2 2 /2 A A D –B