G.C. Sloan: Physics 2305, Exam 1

Physics 2305: Exam 1 Results

Remember that grades assigned to exams are for comparative purposes only. Final course grades will be based on numerical results.

Answer keys are below

Students: 254

Mean: 67.2

Std Dev: 19.0

A 32

B 43

C 107

D 43

F 29

Answers to multiple-choice questions

	Form A	Form B	Form C	Form D	Form E
1.	B	B	D	C	C
2.	D	C	C	D	A
3.	D	B	C	B	C
4.	A	C	B	A	B
5.	C	A	C	B	C
6.	C	D	D	C	C
7.	B	B	B	C	D
8.	C	A	C	A	B
9.	B	D	B	A	B
10.	D	B	A	D	D

Answers to free-form questions

Derive the kinematic equations using calculus: See Sec. 2-8.

Harry's potted plant
a) v₁ = (h₂/t) - (gt/2) = 7.60 m/s (Forms A,C) or 5.65 m/s (B,D)
b) h₁ = v₁²/(2g) = 2.95 m (A,C) or 1.63 m (B,D)

Mischevious Molly
a) v = (2gr)^1/2 = 2.6 m/s (Form A) or 3.0 m/s (B)
b) v = (T_maxr/m)^1/2 = 6.5 m/s (A) or 8.4 m/s (B)

Paquette
a) v = (gr)^1/2 = 2.4 m/s (Forms C,E) or 2.0 m/s (D)
b) v = [(T/m - g) r]^1/2 = 3.4 m/s (C,E) or 2.7 m/s (D)

The ballista
a) v_o = (gr)^1/2 = 66.5 m/s
b) a = v²/2d = gR/2d = 920 m/s²
c) The initial angle = (1/2) sin^-1(gR/v_o^s) which gives either 21 or 69 degrees.

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