University of Hawaii, Manoa
Dept. of Physics & Astronomy
Instructor: Michael Nassir
Office: Watanabe Hall, Rm. 426, (808) 956-2922
E-mail: nassir @

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Part A

1. D (there are 1,000,000,000 m^3 in every 1 km^3, and 1000 L in every 1 m^3)
2a. B, C, and D (whenever velocity graph is negative)
2b. A and B (whenever velocity graph has negative slope)
2c. +2.0 m/s^2 (slope of velocity graph over 30s to 35s interval)
2d. D (cumulative area of velocity graph is zero at 20s)
3a. D (since it is thrown exactly vertically, it is momentarily at rest at top of arc)
3b. B, C, and D
4a. F (f = v / (2 p r))
4b. F (F_cp = m v^2 / r)
4c. A (F_cp and momentary displacement are at 90 degrees to each other, so zero work is done)
4d. B (w remains constant, so a is zero; a_cp = (v^2 / r) is non-zero for uniform circular motion)
4e. E (L = mvr for a point mass)
4f. A (using "right-hand rule")

Part B

5. B
6. D
7. D (calculate position of center of mass for both masses OR find position of pivot at which sum of torques is zero)
8. B
9a. B (period = 2 p sqrt(L / g))
9b. C (the other choices have NO effect on the period -- see formula for #9a)
10a. E (in this case, v is same for waves on both strings: for short string, v = (2L)f = 1065 m/s; then use this to find L for long string)
10b. C (speed of waves on string = sqrt(tension force / (m / L) ), and want to decrease speed to decrease f)
10c. C (f_beat = difference of the two frequencies)
11. in this order: B, D, J, A, G, C

Part C

1a. 0.19 N (F_buoyant = density of air * g * volume of air displaced by balloon)
1b. 12 g (total downward weight of balloon + card should equal upward buoyant force from part (a))

2a. 0.88 N (applied force must be as great as spring's restoring force = kx)
2b. 1.5 s (period = 2 p sqrt(m / k))
2c. 0.42 m/s (from conservation of energy, can show that: max speed = amplitude * sqrt(k / m))
2d. displacement graph: cosine curve with amplitude 10 cm, completing one cycle every T
velocity graph: negative sine curve with amplitude 0.42 m/s, completing one cycle every T
2e. A, D, and E (as energy is gradually lost to friction: max PE diminishes, thereby decreasing amplitude; and max KE diminishes, thereby decreasing v_max)

Part D

3a. 12 m/s (total momentum before collision equals total momentum afterward)
3b. B = inelastic (kinetic energy before collision is 7.2 J, but total kinetic energy afterward is only 4.2 J, so KE is lost in collision)
3c. 2.5 m (puck takes 0.495 s to fall)
3d. B = no change (time of fall does not depend on mass of puck)
3e. B = same period of time (time of fall does not depend on initial horizontal velocity of puck)

4a. 1st harmonic: one node at center, and one antinode at each end
2nd harmonic: antinodes at both ends and at center; nodes at (1/4)L and (3/4)L
4b. F (from inspection of sketch in part (a), wavelength = 2L)
4c. D (from inspection of sketch in part (a), wavelength = L)
4d. 1st harmonic: f = 61.1 Hz (f = v ÷ wavelength from part (b))
2nd harmonic: f = 122 Hz (f = v ÷ wavelength from part (c), OR: f_2 = 2 * f_1)
4e. A (human hearing covers approx. 20 Hz to 20,000 Hz)
4f(i). A (f = v / (2L), so if L decreases, f rises)
4f(ii). B (f = v / (2L), so diameter has no effect)
4f(iii). C (f = v / (2L), so if v decreases, f also decreases)
4g. B (compressional waves in air are sound waves, and are longitudinal: the wave displacement is in the same direction as the wave propagation)