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# PHYSICS 151

Instructor: Michael Nassir

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

E-mail: nassir @ hawaii.edu
*Return to PHYS 151 Old Exams & Answers page*

## ANSWERS to FINAL
EXAM from SPRING 2003

**Part A**

**1.** C

**2.** D

**3.** Note: both masses are launched with *same initial
speed*

3a. B

**3b.** A

**3c.** C

**4a.** T

**4b.** F

**4c.** T

**4d.** F

**5a.** A (Law of Universal Gravitation)

**5b.** B (*F* = *mmg*)

**5c.** C (*F* = *mmg*,
and *P* = *F***v*)

**6a.** F (total momentum before collision = total momentum
after collision)

**6b.** B (inelastic, since some KE is lost)

**7.** F (*P* = *F*÷*A*, where *F*
= total weight, and *A* = total area (both heels))

**8.** C (object floats if density of object < density of
water)

**Part B**

**9.** A

**10.** C (weight = restoring force, so *mg* = *kx*)

**11.** B

**12a.** C

**12b.** E

**12c.** G (initial angle or amplitude does NOT affect period)

**13a.** draw vibration with 4 equally-spaced nodes (including
both ends) and 3 interspersed antinodes

**13b.** F (3rd harmonic: *f* = *v* ÷ (2*L/3*))

**13c.** A and E

**14.** E

**15a.** C (*P_*1/*T_*1 = *P_*2/*T_*2)

**15b.** D and F

**16.** E (*e* = (*T*_H - *T*_L) / *T*_H)

**17.** Diagram missing -- please ignore.

**Part C**

**1a.** 690 N (*m* = density * *V*; weight = *mg*)

**1b.** 0.89 N = 0.13% of weight (buoyant force = density of
air * *g* * *V*)

**2a.** 1.0 x 10^3 N

**2b.** horizontal line at 5.0 m/s^2 (*t* = 0s to 6s),
followed by horizontal line at -9.8 m/s^2 (*t* = 6s to 14.33s)

**2c.** -52 m/s (first must calculate: *v* = +30.m/s at
*t*=6s; then use that as new initial velocity to find *v*
at 14.33s)

**2d.** straight line from (0,0) to (6,30), then straight line
from (6,30) to (14.33,-52)

**2e.** upward-facing parabola from *t*=0s to 6s; connected
(smoothly) to downward-facing parabola from *t*=6s to 14.33s
(peaking at 9.1s)

**Part D**

**3a.** 8.9 m/s (using either KE_init = PE_final, OR kinematics:
*v*^2 = *v*0^2 + 2 *a Dy*)

**3b.** 5.5 J lost

**3c.** B (inelastic, since some KE is lost in the bounce)

**3d.** 0.041 degree C (*Q* = *m c DT*)

**4a.** 430 N*m (total torque = 2 * one child's torque =
2 * (*R * F*_p))

**4b.** 1.3 s (first: a = total
torque ÷ *I* = 1.35 rad/s^2; then: w
= a *t* = 4.73 rad/s; then: *T*
= 2 p ÷ w)

**4c.** 3600 J (total work = increase in rotational KE = (1/2)
*I* w^2)

**4d.** B (via "right-hand rule")