Homework

1. Kinematics Part II (Chapter 3, Projectile Motion)

100 points

3.1 Components of Motion

p. 91-92 : 1-6, 9, 12, 13 17

Helpful Tips:

3. Try b and c with a mallet and a basketball. Roll the ball and then strike it with the mallet at an angle ^ to its motion.

12. You are solving for the q and hypotenuse of this triangle formed by y and x

13. V_y 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

Helpful Tips:

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 0^o, the angle between w and F_^ approaches 0^o. 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

Helpful Tips:

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:

V_bt = V_{b(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 V_ba = V_b-V_a = V_b + (-V_a). To do this with the vectors, just FLIP the direction of V_a 180^o and ADD it head to tail with V_b.

3.4 Projectile Motion

p. 95-96: 64, 66-68, 70, 72, 76, 79, 81, 83, 90

Helpful Tips:

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 V_oy component and then use y = V_oyt – 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 (V_ox) which is 8.485 m/s (12 m/s at an angle of 45^o gives this V_ox result). So, once you know t from the quadratic equation, multiply it by V_ox to find the x range.

For part b: V_y = V_oy-gt where V_y = the y component velocity of the rock hitting the water and V_oy = 8.435 m/s. Realize that the rock hits the water with a V_y AND a V_x. Therefore, you need to find a resultant vector for V_y and V_x. Of course V_x remains constant throughout the entire flight, so V_x = V_ox.

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.

Even Answers:

2) c

4) 28 m/s, 21 m/s

6) ± 6.3 m/s

8) 1.1 x 102m, 27^o North of East

10) x = -10.8cm; y = -6.25 cm

12) 2.5 m at 53^o above the +x axis

14) a. 75.2 m b. 99.8 m

16) (10m, -6.0m)

18) c

20) d

26) a. 35mx b. 5mx c. –5mx

30) b) 145N at 50.1^o 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, 4.25^o upstream b. 44.6s

64) b

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. 51.8^o 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

84) 1.4^o