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

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ANSWERS to MIDTERM #1 from SPRING 2006

Part A

1a. 6.3 x 10^3 angstroms or 6300 angstroms
1b. 1.3 x 10^2 micrograms or 130 micrograms
1c. 0.22 m^2
1d. 79 km/y
2a. D (all digits following the "8" are significant)
2b. 8.0900 x 10^-2 kg
3a. C (must add two velocity vectors, then find magnitude and direction of sum)
3b. A
4. B
5a. segment A: 2.0 m/s^2
segment B: zero
segment C: 4.0 m/s^2
5b. B (graphs shows constant positive velocity)
5c. D (either find cumulative area under velocity graph; or use kinematics equations twice - separately over segment A and segment B)
6a. A
6b. C

Part B

1a. free-body diagram should have 4 force vectors:  F_A to left, F_B to right, normal force upward, weight downward
1b. -0.47 m/s^2, to the left  (sum of horizontal force vectors: F_net = m * a)
1c. 834 N, upward  (here, normal force = weight = m * g)
1d. E  (if speed is constant, acceleration is zero; a=0 when net horizontal force is also zero; therefore F_B = F_A)

2a. v_0x = v_0 * cos(theta) = 20.5 m/s
v_0y = v_0 * sin(theta) = 14.3 m/s
2b. 2.92 s  (use kinematics eqns: either find total time of flight from y_0 = 0m to y_f = 0m; or else find time to top of arc, then double it [using symmetry of path])
2c. 10.5 m  (using kinematics eqns)
BONUS: 18.7 m AND 41.2 m  (using kinematics eqns)