Exam1 06 Solutions - Physics 150 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Answer = B Answer = C Answer = D Answer = D Answer = E Answer =

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11. Answer = E 12. Answer = D 13. Answer = E 14. Answer = E 15. Answer = A 16. Answer = C 17. Answer = D 18. Answer = E 19. Answer = A 20. Answer = B Physics 150 Exam #1 Solution 10/9/06 1. Answer = B 2. Answer = C 3. Answer = D 4. Answer = D 5. Answer = E 6. Answer = E 7. Answer = E 8. Answer = C 9. Answer = D 10. Answer = B
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y 61.224 m = yy o v o sin θ () t + 1 2 gt 2 := t 3.535sec = t v o sin θ g := At the top v y = 0. B) θ f 12.841 deg = θ f atan v y v x := v f 20.513 msec -1 = v f v x 2 v y 2 + := v x 20msec -1 = v x v o cos θ := v y 4.559 msec -1 = v y v o sin θ := y 60.164 m = o v o sin θ t + 1 2 2 := A) y o 0m := t 4 sec := θ 60 deg := v o 40 m sec := Problem #1: Physics 150 Exam #1 Solution 10/9/06
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a m C g m A m B + m C + := a 3.267msec -2 = Tm A m B + () a := T 32.667 N = B) Necessary force of static friction on block A to accelerate it. F netyA F NA m A g := F NA m A g m A a y := a y = 0 for mass A. F NA m A g := a x = a for mass A. F netxA f s := f s m A a := f s 6.533N = f smax µ s F NA := f smax 3.92 N = Hence f s > f smax so block A will not remain at rest relative to block B. Physics 159 Exam #1 Solution 10/9/06 Problem #2: m A 2kg := m B 8kg := m C 5kg := µ s 0.20 := A) F netyAB F NAB m A m B + g := F NAB m a m B + g m A m B + a y := a y = 0 for masses A and B. F NAB m A m B + g := F netyC C g := T C g m C a y := a y = -a for mass C.
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This note was uploaded on 04/10/2008 for the course PHYSICS 150 taught by Professor Naik during the Winter '08 term at University of Michigan.

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Exam1 06 Solutions - Physics 150 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Answer = B Answer = C Answer = D Answer = D Answer = E Answer =

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