{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam2-F03-ans

# exam2-F03-ans - J 2000 W v m W K 2 2 1 = = → = ∆ s m...

This preview shows pages 1–2. Sign up to view the full content.

Physics I – Exam 2 – Fall 2003 Answer Key Part A – 1. C, 2. B, 3. B, 4. C 4 points each, no partial credit. Part B 16 Points 1. Knowing to use (m g h) for the potential energy at the top. 2. Knowing to use ( K + U = –Fd) for the kinetic energy at the bottom. 3. Straight line for PE from whatever at d = 0 to PE = 0 at d = 100 cm. 4. Straight line for KE from KE = 0 at d = 0 to whatever at d = 100 cm. 5. PE = (0.5) (9.8) (0.05) = 0.245 J at d = 0 cm. 6. KE = 0.245 – (0.1) (1) = 0.145 J at d = 100 cm. Part C C-1 16 points s / rad 377 60 2 3600 0 = π × = ϖ 0 f = ϖ 2 0 f s / rad 02618 . 0 3600 4 377 t - = × - = ϖ - ϖ = α m N 00 . 1 02618 . 0 2 . 38 I - = - × = α = τ (The minus sign is dropped because the problem asks for magnitude, but OK if there.) For # of revolutions, there are three methods: A: rad 10 71 . 2 2 14400 02618 . 0 14400 377 ) t ( ) t ( 6 2 2 2 1 0 × = ÷ × - × = α + ϖ = θ B: rad 10 71 . 2 2 14400 02618 . 0 0 ) t ( ) t ( 6 2 2 2 1 f × = ÷ × + = α - ϖ = θ C: rev 432000 2 ) 60 4 )( 0 3600 ( ) t ( ) ( f 0 2 1 = ÷ × + = ϖ + ϖ = θ In method C, we can use units of rev and min.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
C-2 16 points 1D work for variable force: J 2000 8000 5 . 0 Area dx F W 2 1 = × × = = = Work-Kinetic Energy Theorem:
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: J 2000 W v m W K 2 2 1 = = → = ∆ s / m 1000 004 . 2000 2 m W 2 v = × = = No points for trying to solve this with “Favg = m a” even if getting right answer! C-3 8 points Conservation of angular momentum: → ϖ = ϖ 2 2 1 1 I I s / rad 20 75 . 12 25 . 1 I I 2 1 1 2 = × = ϖ = ϖ C-4 8 points Work-Kinetic Energy Theorem: 2 1 1 2 1 2 2 2 2 1 I I W W K ϖ-ϖ = → = ∆ J 60 90 150 12 25 . 1 5 . 20 75 . 5 . W 2 2 =-= × ×-× × = C-5 12 points 1D elastic collision equations (for second object initially at rest): s / m 2 4 5 . 1 5 . 5 . 1 5 . v m m m m v i 1 2 1 2 1 f 1-= × +-= +-= s / m 2 4 5 . 1 5 . 5 . 2 v m m m 2 v i 1 2 1 1 f 2 + = × + × = + = C-6 8 points Conservation of mechanical energy: 2 2 1 2 2 1 x k v m U K = → = ∆ + ∆ cm 00 . 4 m 0400 . 3750 5 . 1 2 k m v x = = × = =...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

exam2-F03-ans - J 2000 W v m W K 2 2 1 = = → = ∆ s m...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online