# 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 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 ÷ (2L/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 = v0^2 + 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")