3.1 Components of Motion
p. 91-92 : 1-6, 9, 12, 13 17
12. You are solving for the q and hypotenuse of this triangle formed by y and x
13. Vy is unaffected by an acceleration in the x direction. Component vector addition will give you the q. You can assume that the x direction acceleration is constant over the time interval.
3.2 Vector addition and Subtraction
p. 92-94: 18-20, 29,30,33,41,43
18. Think about the “extreme” cases. Added together or subtracted
33. notice that in the x direction the forces are equal but opposite and thus cancel.
41. as q approaches 0o, the angle between w and F^ approaches 0o. The angle between w and F^ vary directly – meaning that they each increase or decrease by the same amount.
3.3 Relative Velocity
p. 94-95: 48,50,52,53,56,59,60,63
52. the velocity of the ball to the stationary observer must be the SUM of all of the velocities acting on the ball. Realize that some of these velocities are positive and some are negative.
In the case of the moving truck:
Vbt = Vb(sum of velocities) – Vt(which is 90 km/h)
56. There will be two possible answers since the direction of the passenger train isn’t given. Be sure to convert km/h into m/s
59. Head to tail addition of vectors will work well here
63. Recall that Vba = Vb-Va = Vb + (-Va). To do this with the vectors, just FLIP the direction of Va 180o and ADD it head to tail with Vb.
3.4 Projectile Motion
p. 95-96: 64, 66-68, 70, 72, 76, 79, 81, 83, 90
68. Hit the cup lab
76. convert km/h to m/sec and km to m. Use y = 1/2gt2 to determine t. multiply t by the horizontal velocity of the plane (in m/s) to tell you the horizontal distance (x) traveled by the package during time (t). Find q now that you know distances for y (500m) and x (393m).
79. Don’t try this at home!
83. For part a: Find the Voy component and then use y = Voyt – gt2 where y = -20m (the height above the river). Use quadratic equation to solve and disregard the negative answer.
While it is true that the rock first goes up before falling (and therefore will fall a greater distance than 20 m), for purposes of range, all you need to know is the horizontal velocity component (Vox) which is 8.485 m/s (12 m/s at an angle of 45o gives this Vox result). So, once you know t from the quadratic equation, multiply it by Vox to find the x range.
For part b: Vy = Voy-gt where Vy = the y component velocity of the rock hitting the water and Voy = 8.435 m/s. Realize that the rock hits the water with a Vy AND a Vx. Therefore, you need to find a resultant vector for Vy and Vx. Of course Vx remains constant throughout the entire flight, so Vx = Vox.
90. You should make up some “fake” numbers and that way you can see if it works or not. This is a classic physics demo.
4) 28 m/s, 21 m/s
6) ± 6.3 m/s
8) 1.1 x 102m, 27o North of East
10) x = -10.8cm; y = -6.25 cm
12) 2.5 m at 53o above the +x axis
14) a. 75.2 m b. 99.8 m
16) (10m, -6.0m)
26) a. 35mx b. 5mx c. –5mx
30) b) 145N at 50.1o North of East
32) 2.0 mx
34) a. 19.2 Nx b. –19.2Nx c. yes
42) 83.2 m at 2.0o above horizontal
46) 23 Km at 27o North of West
48) a. zero b. 4.0 m/s
50) a. –30 Km/h b. +30 Km/h
52) a. +55 Km/h b. –35 Km/h
54) 60 m
56) +70m/s or –20m/s
60) a. 0.075 m/s b. 30o relative to a line across the river
62) a. same both ways, 4.25o upstream b. 44.6s
66) Horizontal motion does not affect vertical motion
68) a. 0.64s b. 0.96m
70) 17 m
72) 7.6 m/s
74) a. 0.123s b. 1.10m
76) a. 51.8o b. directly over impact point
78) a. 1.37 m b. 20.4 m c. kick harder and/or increase q
80) 0.68 s
82) 3.8 x 102 m