{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2007MechSolutions

# 2007MechSolutions - 2006 AP Physics Mechanics Exam FN 1 a...

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

2006 AP Physics Mechanics Exam 1) a) b) vertical forces: θ θ sin sin 1 1 F mg F mg F F N N - = = + c) horizontal forces: θ θ θ μ μ θ sin cos cos cos 1 1 1 1 F mg ma F F ma F ma F F N N - - = - = = - d) constant acceleration v-t graph: linear, positive slope through origin x-t graph: 2 nd order polynomial through origin e) Lose contact with ground: F N = 0: θ θ sin 0 sin 1 1 mg F F mg F N = = - = ( 29 θ μ θ μ θ μ θ sin cos cos cos 1 1 1 1 F mg F F F ma ma F F N N - - = - = = - θ θ μ μ θ θ θ μ μ θ μ θ sin sin cos sin sin cos cos 1 1 1 + - = + - = - = mg mg mg F mg F F F ma N θ θ θ θ μ μ θ θ cot cot sin cos sin cos g a mg mg mg mg mg ma = = = + - = F N F 1 μ F N mg

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

View Full Document
2) a) ( 29 ( 29 m vT R T R v 6 3 3 10 83 . 3 2 10 08 . 7 10 4 . 3 2 2 × = × × = = = π π π b) ( 29 ( 29 kg G Rv M R GmM R mv 23 11 2 3 6 2 2 2 10 64 . 6 10 67 . 6 10 4 . 3 10 83 . 3 × = × × × = = = - c) R GmM mv R GmM mv R GmM R mv 2 2 1 2 2 2 2 = = = Total Energy: R GmM R GmM R GmM R GmM mv U K E 2 2 2 1 2 - = - = - = + = ( 29 ( 29 ( 29 ( 29 J R GmM E 9 6 23 11 10 4 . 5 10 83 . 3 2 10 64 . 6 930 10 67 . 6 2 × - = × × × - = - = - d) GM R T T R T R R GM v R GmM R mv 3 2 2 2 2 2 2 2 2 2 4 4 2 π π π = = = = = Decreasing the orbital radius decreases the orbital period The new orbital period would be LESS THAN the given period. e) R 1 = 4.36 x 10 5 + 3.43 x 10 6 = 3.866 x 10 6 m R 2 = 3.71 x 10 5 + 3.43 x 10 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

2007MechSolutions - 2006 AP Physics Mechanics Exam FN 1 a...

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

View Full Document
Ask a homework question - tutors are online