Physics 1303 - Summer 1998

Homework Assignment #7

Due: 3 August 1998

READING

Chapters 13 and 14.

QUESTIONS

Chapter 13 - 1, 4, 10, 11, 17, 19.
Chapter 14 - 1, 3, 5, 8, 9, 12, 13, 14, 19.

PROBLEMS

Chapter 13 - 1, 5, 9, 10, 21, 25, 60.
Chapter 14 - Review Problem, 1, 5, 6, 11, 12, 15, 18, 19, 30, 33, 53, 55.

Bonus:
Visit one of the Web sites at the bottom of the Physics 1303 Web Page (different from the site you visited in Assignment #1) and write a 200 word description/critique of it.

ANSWERS

These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to copy these and turn them in as homework. You must show your work.

Q 13-1) 4A

Q 13-4)
  1. yes
  2. yes
  3. no
Q 13-10) If L is doubled, then T increases by sqrt(2).
If M is doubled, then T remains unchanged.
Q 13-11)
  1. period decreases
  2. period increases
  3. period remains unchanged
Q 13-17) to avoid destroying the bridge by resonance.
Q 13-19) shorten the length of the pendulum.
Q 14-1) order of magnitude: 10-7 newtons
Q 14-3) 10
Q 14-5) No.
Q 14-8) U = -2 K for a circular orbit. For a general elliptical orbit, the absolute value of U is always greater than K, and U is always negative. The total energy E=U+K is therefore always negative.
Q 14-9) F is always perpendicular to displacement in a circular orbit. In an elliptical orbit, sometimes positive work is done (and the particle speeds up), sometimes negative work is done (the particle slows down), but the NET work done in one full revolution is zero.
Q 14-12) Speed is a maximum when the planet is closest to the Sun. Speed is a minimum when the planet is farthest from the Sun.
Q 14-13) gX = G MX/RX2
Q 14-14) The gravitational force on m1 at the center of the Earth is zero. Equation 14-1 is correct if m2 is the mass "under" m1. That is, one discounts all the mass in the Earth further from the center than m2's.
Q 14-19) The centripetal acceleration at the equator is about 0.003 of a "g". (0.034 m/s2)
P 13-1)
  1. f=1.50 Hz, T=0.667 s
  2. 4.00 m
  3. Pi rad = 3.1416 rad
  4. 2.83 m
P 13-5)
  1. v=13.9 cm/s, a=16.0 cm/s2
  2. vmax = 16.0 cm/s at t=0.262 s
  3. amax = 32 cm/s2 at t=1.05 s
P 13-9)
  1. 0.542 kg
  2. 1.81 s
  3. 1.20 m/s2
P 13-10) 0.627 s
P 13-21)
  1. E increases by 4 times
  2. vmax is doubled
  3. amax is doubled
  4. T is unchanged
P 13-25)
  1. 1.55 m
  2. 6.06 s
P 13-60)
  1. 0.50 m/s
  2. 0.0856 m
P 14-Review)
  1. sqrt(GM/R)
  2. 2 Pi sqrt(R3 / GM)
  3. GmM / 2R
  4. - GmM / R
  5. sqrt(GM / g)
  6. 2 Pi (GM / g3 )1/4
  7. m/2 sqrt(gGM)
  8. - m sqrt(gGM)
  9. sqrt(2GM / R) this is sqrt(2) times faster than its orbital speed.
P 14-1)
  1. 3.46 x 108 m
  2. 3.34 x 10-3 m/s2
P 14-5) g = Gm / L2 [sqrt(2) + 1/2 ] toward the opposite corner
P 14-6) g/16 = 0.613 m/s2
P 14-11) M = 1.26 x 1032 kg = 63 solar masses for each star
P 14-12)
  1. 1024 m/s
  2. 0.0014 m = 1.4 mm = 1/20 inch
P 14-15) 1.9 x 1027 kg ( = 316 Earth masses)
P 14-18) 226 N
P 14-19) 89,800 km above THE SURFACE
P 14-30) 469 megajoules
P 14-33) Proof required (see review problem)
P 14-53) 313 rad/s = 3000 rpm (revolutions per minute)
P 14-55)
  1. v1 = m2 sqrt[2G / d(m1+m2)] and v2 = m1 sqrt[2G / d(m1+m2)]
    relative velocity = sqrt[2G(m1+m2) / d]
  2. K1 = 1.07 x 1032 J and K2 = 2.67 x 1031 J

Please report any corrections to Professor Scalise.

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