PH2100 Fall 2000 - Final Exam Answers

Answers (explanations are below):

Multiple Choice:
1. C
2. C
3. B
4. B
5. A
6. F
7. E
8. D
9. D
10. C
11. D
12. C
13. F


Worked Problems:
14. A) 48.6 m/s     B) 16.2 m
15. A) 26.8 N     B) 0.555
16. A) 67.0 J     B) 1.67 m/s2
17. A) y = 0.133 m     B) 37.2 kg.m2     C) -i
18. (-17.2 i + 5.70 j ) N or 18.1 N at 162o CCW from x-axis
19. A) 7.50 Hz     B) 158 N
20. 3.84 m/s


Explanations:

1. A+B = (5.67 + 2.31) i + (-3.12 + 1.76) j = 7.98 i - 1.36 j
       Magnitude = (7.982 + (-1.36)2)1/2 = 8.09

2. A.B = (5.30)(-2.28) + (-2.23)(-1.78) = -8.11

3. AXB = ( (6.52)(3.37) - (3.76)(0) ) k = 22.0 k
       Magnitude = 22.0
or use |A||B|sin(60o)

4. Vertical component is (v sin(theta)).

5. Initial horizontal component is (v cos(theta)), and since there is no acceleration in the horizontal direction, it is constant. At the peak, the vertical component is zero, so the speed is just (v cos(theta)).

6. The frequency for a mass on a spring is proportional to 1/(M)1/2
Hence, if f2/f1 = 3, then M2/M1 = 1/9

7. Since m1 is not equal to m2, then v1 is not equal to v2. The kinetic energy has no direction associated with it, so you don't know what direction the particles are traveling. Hence, the only answer which must always be true is "None of the above."

8. radius = diameter/2 = 0.099 m. For rolling motion omega = v/r = 2.18/0.099 s-1 = 22.0 rad/s.

9. T2 is proportional to R3 (Kepler's 3rd Law)
       so (T/1 year)2 = (5RE/RE)3
       or T = 11 year.

10. v = omega/k = 36/12 = 3.0 m/s.

11. Wave speed is proportional to the square root of the tension, T. Hence if T is doubled, the wave speed increases by 21/2 = 1.41 times.

12. Intensity is proportional to 1/r2, so with r doubled, the intensity drops by a factor of 4. Hence, the reduction in dB is 10log10(4) = 6 dB. So the final intensity is (60-6) dB = 54 dB. Or simply remember that a factor of two is 3 dB. Here we have 2 factors of two, or a reduction of 2 times 3 dB.

13. To travel toward +x, need a minus sign in the arguement. To have wavelength lambda, have k = 2 pi/lambda, omega = 2 pi f . Hence answer is F.

14. A) Simplest to use conservation of energy. Ui = 0, Ki = 1/2 m vi2
       Uf = m g (8m), Kf = 1/2 m vf2
       or vf2 = (50.2 m/s)2 - 2 g (8m)
       so vf = 48.6 m/s

    B) vxi = 138m/3s = 46 m/s, so vyi2 = (50.2 m/s)2 - (46 m/s)2
       or vyi = 20.1 m/s
       Now y = y0 + vyi t - 1/2 g t2
       with t = 3 s, y0 = 0, and vyi from above, get y = 16.2 m

15. A) For m2 (defining positive as down) have m2g - T = m2 a2
       for m1 (defining positive to the right) have T - mu m1 g = m1 a1
       for the directions chosen, a1 = a2 = a, and adding the two equations gives
       a = (m2 - mu m1)/(m1 + m2) = 2.02 m/s2
       substituting into the first equation gives T = 26.8 N.

    B) The equations are the same as in (A), but a = 0 and mu is unknown. Must have (m2 - mu m1) = 0,
       so mu = m2/m1 = 0.555 is the smallest mu can be.

16. A) Work = change in kinetic energy = 1/2 (5.36 kg) (5.00 m/s)2 = 67.0 N .

    B) ave a = (vf - vi)/t = 5/3 m/s2 .

17. A) xcm = 0 by symmetry. ycm = 1/6 ((3)(2.5) + (-1)(0.7) + (-2)(3)) = 0.133 m

    B) I = (3kg)(2.5m)2 + (1kg)(-0.7m)2 + (2kg)(3m)2 = 37.2 kg.m2

    C) Use right hand rule -> get -i direction.

18. Fx = -dU/dx = -8.6 x, Fy = -dU/dy = 5.7 N
       so F at x = 2m, y = 3m is (-17.2 i + 5.7 j ) N

19. A) for a string fixed at both ends, get frequencies fn = n f1 , n = 1, 2, 3, ...
       The fundamental is f1 = 3.75 Hz, the next higher freq. is 2f1 = 7.50 Hz

    B) from v = (T/mu)1/2, T = mu v2.
   Now v = lambda f, and for the fundamental given, lambda = 2L = 40 m, and f = 3.75 Hz, so v = 150 m/s.
   Hence, T = (0.007 kg/m)(150 m/s)2 = 158 N.

20. Since the 2nd train is approaching, its frequency is heard higher. f' = (220 + 2.5) Hz = 222.5 Hz.
       but also f' = 220 Hz (341/(341 - vs)). Solve for vs = 3.84 m/s.

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