Midterm1Solution

Midterm1Solution - PHYSICS 9A MIDTERM 1 October 27 , 2011 I...

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PHYSICS 9A MIDTERM 1 October 27 , 2011 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) . If the x component of a vector A in the x-y plane is half as large as the magnitude of the vector, its y component is A) A/2 B) 2A C) 3A/4 D) 3 2 A E) 5 2 A Solution: Denote x component of vector A as A x , y component of vector A as A y , and its length as |A|. Then we have A x 2 +A y 2 =|A| 2 (|A|/2) 2 +A y 2 =|A| 2 it then follows that A y = 3 2 A Answer D.
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Problem 2. (two points) A missile has to be launched from some point on the Earth. Assuming that it travels over flat area and its initial velocity is always the same, what would be the optimal angle to fire this missile so that it covers the largest possible distance? (neglect air resistance) A) 15 degrees B) 30 degrees C) 45 degrees D) 60 degrees E) 75 degrees Solution: Along horizontal axis x(t)=v*cos( φ )*t , where v is the initial velocity, and φ is the firing angle. Along vertical axis y(t)=v*sin( )*t-gt 2 /2 . The time during which the missile is in the air can be estimated from y(t)=0 which gives t 0 =2*v*sin( )/g , and the distance along horizontal axis x(t 0 )=v*cos( )*2*v*sin( )/g=v*sin(2 )/g . It is then clear that the distance is at maximum when sin(2 )=1 , or =45 degrees. Answer C.
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Problem 3 (two points). SOLUTION.
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Midterm1Solution - PHYSICS 9A MIDTERM 1 October 27 , 2011 I...

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