ps01sol

# ps01sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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PS01-1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2010 Problem Set 1 Solutions Problem 1: Vectors (10 points) Consider the two vectors shown in the figure below. The magnitude of 2.88 = A r and the vector A r makes an angle 33.7 o with the positive x -axis. The magnitude of 3.44 = B r and the vector B r makes an angle 35.5 o with the positive x -axis pointing down to the right as shown in the figure below. Find the x and y components of a) A r and B r ; b) + AB r r ; c) r r ; d) a unit vector pointing in the direction of A r ; e) a unit vector pointing in the direction of B r . Solution: We need to use 33.7 A θ = o in order to determine the x and y components of the vector A r : cos (2.88)(cos(33.7 ) 2.40 xA A == = A o r , sin (2.88)(sin(33.7 ) 1.60 yA A = A o r . Thus

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PS01-2 ˆˆ 2.40 1.60 =+ Ai j r . We need to use 35.5 B θ =− o in order to determine the x and y components of the vector B r : cos (3.44)(cos( 35.5 ) 2.80 xB B == = B o r , cos (3.44)(sin( 35.5 ) 2.00 yB B = B o r . Thus 2.80 2.00 Bi j r . b) The vector sum is then (2.40 1.60 ) (2.80 2.00 ) (5.20) ( .40) += + + AB i j i j ij r r c) The vector difference is (2.40 1.60 ) (2.80 2.00 ) ( .40) (3.60) −= + + i j i j r r d) The unit vector pointing in the direction of A r is given by 2.40 1.60 ˆ 0.83 0.69 2.88 = AA i j j A rr r e) The unit vector pointing in the direction of B r is given by 2.80 2.00 ˆ 0.81 0.58 3.44 = j j B r r Problem 2 Vectors (10 points) Consider two points located at 1 r r and 2 r r , separated by distance 12 1 2 r . Find a vector A r from the origin to the point on the line between 1 r r and 2 r r at a distance x from the point at 1 r r , where x is some number. Express your answer in terms of 1 r r , 2 r r , 12 r , and x . Show your work.
PS01-3 Solution: Consider the unit vector pointing from 1 r r and 2 r r given by 1 2 1212 12 1 2 ˆ // r =− − =− rr r r r r rrr rr . The vector α r in the figure connects A r to the point at 1 r r , therefore we can write () 12 1 2 12 ˆ x x r = α r r . The vector 1 = + rA α r r r . Therefore 111 2 1 2 12 12 12 1 x xx r ⎛⎞ =−=− − = − + ⎜⎟ ⎝⎠ Ar α r r r r rrr rr r r .

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PS01-4 Problem 3 Concept Questions (10 points) (a) (5 points) Two objects with charges q and 3 q + are placed on a line as shown in the figure below. Besides an infinite distance away from the charges, where else can the electric field possibly be zero? 1. Between the two charges. 2. To the right of the charge on the right. 3. To the left of the charge on the left. 4. The electric field is only zero an infinite distance away from the charges.
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## ps01sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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