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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")