2005MechSolutions

# 2005MechSolutions - 1 2 1 6 1 3 1 2 1 2 1 2 1 2 1 2 2 2 2 1...

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

2005 AP Physics Mechanics Exam 1) a) v a v m k g a ma kv mg F + = = + = Since acceleration is proportional to speed, and the speed decreases as the ball ascends, the acceleration decreases as the ball ascends. b) v m k g dt dv - - = c) at terminal speed, k mg v kv mg F t = = - = 0 d) maximum height 2 2 1 t a h avg = so time avg a h t 2 = ascending: v m k g a + = descending: v m k g a - = so a up > a down and t up < t down The time required to fall is larger than the time required to ascend. e) 2) a) 2 R GmM F s = where m = the mass of a moon b) 2 2 3 2 2 2 2 2 2 2 4 4 2 T GM R T R T R R GM v R mv R GmM F s s s π π π = = = = = = c) Graph T 2 vs. R 3 d) T (s) R (m) T 2 (s 2 ) R 3 (m 3 ) 8.14 x 10 4 1.85 x 10 8 6.63 x 10 9 6.33 x 10 24 1.18 x 10 5 2.38 x 10 8 1.39 x 10 10 1.35 x 10 25 1.63 x 10 5 2.95 x 10 8 2.66 x 10 10 2.57 x 10 25 2.37 x 10 5 3.77 x 10 8 5.62 x 10 10 5.36 x 10 25 e) Best fit line: 23 2 14 3 10 95 . 1 10 52 . 9 × + × = T R f) ( 29 kg 10 63 . 5 10 67 . 6 10 52 . 9 4 slope 4 slope 4 26 11 14 2 2 2 3 2 × = × × = = = = - π π π G M T R GM s s kv mg kv mg ascending descending v o v t speed, m/s time, s t up t down v o > v t t down – t up > t up

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

View Full Document
3) a) ϖ ϖ 2 1 3 1 d M I L = = b) ( 29 2 1 2 2 1 2 3 3 1 M d M v vd M d M L L particle vd M L f o ϖ ϖ = = = = c) 2 2 2 2 1 2 2 1 2 2 2 1 2 2 18 3 2 1 3 1 2 1 2 1 2 1 M d M M d M M d M mv I K K f o ϖ ϖ ϖ ϖ = =
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 2 1 6 1 3 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 = = = = = = = M M M M M d M d M M d M M d M M mv d M d M I d) using the results of part (b), with M 1 = M 2 and M 2 lever arm r = x : ( 29 x d x M d M x M d M v vx M d M vx M particle vr M L 3 3 3 3 1 2 1 2 1 2 2 1 2 2 1 2 2 = = = = = = using the results of part c) 3 3 18 6 18 6 1 18 6 1 18 3 2 1 2 1 6 1 2 1 2 1 2 2 2 2 2 2 1 2 2 4 2 2 2 1 2 2 4 2 2 2 2 2 2 2 1 2 d x d M d M x x d M M x d M d M x d M x d M mv d M I = = = = = = = =...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern