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# FinalExamSolutions - PHYSICS 9A FINAL EXAM I certify by my...

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PHYSICS 9A FINAL EXAM December 12, 2007 I certify by my signature below that I will abide by the UC Davis Code of Academic Conduct. This includes not copying from anyone else’s exam not letting any other student copy from my exam not discussing this exam with any student who has not yet taken it, nor providing any information, written or oral, that might get to a student who has not yet taken it. Anyone suspected of cheating will be automatically reported to Student Judicial Affairs. LAST NAME: FIRST NAME: STUDENT ID: (P R I N T) (P R I N T) (LAST 4 DIGITS) Signature: PROBLEM 1 (two points). Two balls start at the same time moving to the right with the same velocity. On their way one passes a circular well while another one passes a circular bump of exactly the same radius. Which ball reaches the wall on the right side first? A. Top ball comes first B. Bottom ball comes first C. Both balls come together.

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SOLUTION Consider times for the first ball to reach the bottom (t 1left ) and to climb back (t 1right ) and, similarly, for the second ball to roll up (t 2left ) and to roll down (t 2right ). We see that the first ball is climbing back faster than the second ball (t 1right <t 2left ) is rolling up because the speed of the first ball at the bottom is larger of its initial speed while for the second ball its speed at the bottom is equal to its original speed. Also, we see that the time for the first ball to reach the bottom (t 1left ) is smaller than the time for the second ball to roll down (t 2right ), so we get t 1left < t 2right . This is because the first ball starts with original speed v while the second ball lost some of its speed since it had to climb up. Therefore: t 1right <t 2left t 1left < t 2right Summing up the two inequalities we obtain t 1right + t 1left <t 2left +t 2right which assumes that the first ball passes its hill faster than the second ball passes its well, although at the end they will have the same speed. Hence the first ball comes first! Answer A. v v v 1 >v v 2 <v v v t 1right t 2left t 2right t 1left
PROBLEM 2 (two points). SOLUTION. The bullet deviates from the straight trajectory due to gravity, therefore the question is whether the vertical value of this deviation is exactly equal to the displacement of the criminal or not. The vertical position of the bullet at each time moment t: h(t)=v0*sin( α )*t - gt 2 /2 where angle is measured with respect to horizontal. A the absence of gravity, the same distance would just be h’(t)=v0*sin( α )*t as the bullet would travel along straight line. It is then clear that the vertical deviation from the straight line for the bullet at each time moment is the difference h’(t)-h(t)= gt 2 /2. Now the criminal falls off so its distance with respect to the gutter increases simply as gt

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## This note was uploaded on 11/11/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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FinalExamSolutions - PHYSICS 9A FINAL EXAM I certify by my...

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