# PHYSICS 151

##### University of Hawaii, Manoa Dept. of Physics & Astronomy

Instructor: Michael Nassir
Office: Watanabe Hall, Rm. 426, (808) 956-2922
E-mail: nassir @ hawaii.edu

## ANSWERS to FINAL EXAM from SPRING 2008

Entirely Multiple-Choice (on bubble sheet):

unit conversions:
1.
D  6.6x10^-4 GW
2. C  0.72 cm^3
3. E  3.0x10^18 mm^3
4. B  5.5x10^4 um/ns

5. A  "a" only  (graph of constant v = horizontal graph, slope = 0)
6. E  "a, b, & f" (constant accel -> graph with constant slope = straight line)
7. C  "e & f" (negative accel -> graph with negative slope at all times)
8. D  "e & f" (slowing down -> graph getting closer to zero at all times)
9. C  17.1 m/s (v_0 = sqrt(2*g*h_peak) )
10. A  1.7 s (t = v_0 / g)
11. D  momentum of cannonballer (p = m*v; since v decreases over time, p also decreases)

12. C  700 N (F = (coeff of friction)*m*g)
13. B  0.35 (coeff of friction = F / (m*g) )
14. B  1.9 kJ (Work = F * d)
15. D  803 N (F_drag = weight = m*g)
16. D  constant momentum (p = m*v, and v = constant)
17. C  2.0 m (want center-of-mass position at point where suspension cable attaches = 1.5 m)

18. D  point D (E_tot is conserved, so K is largest when U=m*g*y is smallest)
19. B  equal to v_A (E_tot is conserved; since U_C=U_A, we see that K_C=K_A, so v_C=v_A)
20. A  20. m (E_B = E_D, so (K + U)_B = (K + U)_D, and use U_B=m*g*y to find y_B)
21. B  equal to (E_tot is conserved)
22. E  cannot determine using only the information given (conserv of E gives us v^2_A - v^2_E, but NOT v_A - v_E!)
23. C  greater than (friction causes (K + U) to decrease over time)

24. E  2.8 m/s (impulse = delta(p) = p_f - p_i; use p_f = m*v_f to find v_f)
25. B  75 N (F = impulse / t)
26. B  equal in strength (Newton's 3rd Law: force of paddle on puck = equal & opposite to force of puck on paddle)
27. B  1/3  (T = 2*pi / omega)
28. B  1/3 (L is conserved, so I_init * omega_init = I_final * omega_final )
29. C  (L is conserved/constant, but K_rot = (1/2)*I*omega^2 becomes 3x larger)
30. D  1.7 rad/s^2  (alpha = delta(omega) / t)
31. B  13 rev (use rotational kinematics to find: delta(theta) = 84.0 rad, then convert rad -> revolutions)

32. A  190 kg/m^3 (density = m/V, where m = 14*75kg, and V = 5500 L = 5.5 m^3 )
33. A  yes (object floats if: density(object) < density(fluid); here: 190 kg/m^3 < 1000 kg/m^3)
34. D 97 atm (P = F/A, where F = m*g, and convert final answer from [Pa] to [atm])
35. A  roof will bulge upward/outward (Bernoulli's Principle: as v increases, P decreases)
36. E  20. kcal (Q = m*c*delta(T), where delta(T) = 37degC - 10degC = 27degC = 27K)
37. B  heat of vaporization ("vaporization"=boiling -- NOT needed here; "fusion"=melting)
38. A  entropy increases (delta(S) = Q/T, where Q = heat of fusion and T=273K, so delta(S) is positive)

39. D  30. m (wavelength = 2*L for fundamental mode)
40. A  wire A has higher-frequency fundamental (wire A has lower (m/L), so wire A has faster v, and both wires have same wavelength)
41. B transverse waves (displacement of wire is perpendicular to direction of wave propagation)
42. E  60. m/s (v = wavelength * frequency)
43. D  10. Hz (frequency of nth overtone = n*f_1 = 2 Hz, 4 Hz, 6 Hz, 8 Hz, 10 Hz, ...)
44. E 7.0 m (period of ideal pendulum: T = 2*pi*sqrt(L/g) )
45. B  period is unchanged (formula for T does NOT depend on mass of pendulum bob)

46. C  1  (period: T = 2*pi*sqrt(m/k) -- does NOT depend on amplitude of oscillations)
47. B  1/4  (E_tot = (1/2)*k*A^2, so when A=d, energy is 1/4 as large as when A=2d )
48. A  1/2 (v_max = A*sqrt(k/m), so when A=d, v_max is 1/2 as large as when A=2d )
49. D  7.2 cm (Hooke's Law: F_spring = k*x)
50. A 0.92 Hz (resonance occurs when f_driving = f_natural = sqrt(k/m) / (2*pi) )
51. C  amplitude grows larger (description of resonant growth)

52. D  isothermal (constant temperature)