Dept. of Physics & Astronomy

Office: Watanabe Hall, Rm. 426, (808) 956-2922

E-mail: nassir @ hawaii.edu

*Return to PHYS 151 Old Exams & Answers page*

**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* = *m*v*r* 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 = (2*L*)*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
= 2*L*)

**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 / (2*L*), so if *L* decreases,
*f* rises)

**4f(ii).** B (f = v / (2*L*), so diameter has no effect)

**4f(iii).** C (f = v / (2*L*), 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)