Chapter 8
 2. 15.7 kg m/s down
 6. (a) proof
    (b) r/R=M/m
        (R is the radius of M's orbit)
24. (Ki-Kf)/Ki=m/(m+m_0)
26. 10.8 m/s up
30. 192.5 m/s
46. (a) -1 m/s
    (b) 60 J
    (c) 30 J
49. the initial speed is 83.5 mph
76. 1.19 kg

Chapter 11
10. 0.48 m
42. F_j=(8/29)Mg N;
    F_l=(12.5/29)Mg N; 
    F_r=(8.5/29)Mg N
47. tan(theta/2) or sin(theta)/(1+cos(theta))
Chapter 12
12. 0.10058 m/s²
13. ANSWER IN BACK IS WRONG.  Correct answer is
    38200 km.
55. new minimum ground speed for orbit = 1.76 m/s
Serway Problems
12:17.  16.6 km/s
12:19.  (a) 188 GJ
        (b) 103.5 kW
12:20.  0.47 GJ
12:41.  (a) v_relative = sqrt(2G(m_1+m_2)/d)
            (the question should ask for relative speed
            rather than relative velocity)
        (b) K_1 = 1.07 x 10^32 J; K_2 = 2.67 x 10^31 J
12:48.  (a) 0.85 GJ
        (b) 2.7 GJ
12:49.  proof
Chapter 13
 2. T = 0.8 s; omega = 7.85 rad/s
    x(t) = (0.15 m)sin((7.85 rad/s)t+1.63 rad) or
    x(t) = (0.15 m)sin((7.85 rad/s)t+6.228 rad)
    (since v(.1 s) is not given you can't choose
     between these two possible values of the phase)
    (also 6.228 rad may instead be written as -0.056 rad)
 6. 14.2 cm
 7. THE ANSWER IN THE BACK IS INCOMPLETE.  Since no
    information is given about the velocity at time
    zero, the answer could be -38 mm or +38 mm.
10. f=4.98 Hz
16. (a) x(t)=(0.05m)sin((7.1rad/s)t+(pi/2) rad)
    (b) -0.05 m
    (c) 2.5 m/s²
    (d) 0.354 m/s
22. (a) (15.24 cm) sin(3.49 s-1 t)
    (b) A=15.24 cm; omega=3.49 s-1; v_max=53.2 cm/s
30. (a) 7.06 N/m  (b) 6.26 rad/s  (c) 5.09 cm above table
40. proof
72. 4.98 taps/second
78. (a) delta=3.343 rad
    (b) f=0.325 Hz
    (c) a_0=0.417 m/s² (+x direction)
    (d) k=40 N/m
    (e) m=9.6 kg
88. (a) zero
    (b) opposite to the displacement
    (c) netF = 16GMmy/L³ for y<<L;
    (d) SHM with omega = 4sqrt(GM/L³)