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MASTERING PHYSICS CHAPTER 3

MASTERING PHYSICS CHAPTER 3 - 3.1 Solve(a If one component...

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3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if A x = 0 m and A y = 5 m, then the magnitude of the vector is A = + = ( ) ( ) 0 5 5 2 2 m m m (b) A zero magnitude says that the length of the vector is zero, thus each component must be zero.

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3.2. Visualize: Solve: (a) C A B = + only if r A and r B are in the same direction. Size does not matter. (b) C A B = if r A and r B are in the opposite direction to each other. Size matters only in that A > B because C as a magnitude can only be positive.
3.3. Visualize: Solve: (a) C = A B if r B is directed in the same direction as r A . Size does not matter, except that A > B because C as a magnitude cannot be negative. (b) C = A + B if r B is directed opposite to the direction of r A . Size does not matter.

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3.4. Visualize: Solve: (a) To find r r A B + , we place the tail of vector r B on the tip of vector r A and connect the tail of vector r A with the tip of vector r B . (b) Since r r r r A B A B = + − ( ) , we place the tail of the vector ( ) r B on the tip of vector r A and then connect the tail of vector r A with the tip of vector ( ) r B .
3.5. Visualize: Solve: (a) To find r r A B + , we place the tail of vector r B on the tip of vector r A and then connect vector r A ’s tail with vector r B ’s tip. (b) To find r r A B , we note that r r r r A B A B = + − ( ) . We place the tail of vector r B on the tip of vector r A and then connect vector r A ’s tail with the tip of vector r B .

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3.6. Visualize: The position vector r r whose magnitude r is 10 m has an x -component of 6 m. It makes an angle θ with the + x -axis in the first quadrant. Solve: Using trigonometry, r r x = cos , θ or 6 10 m m = ( )cos . θ This gives θ = ° 53 1 . . Thus the y -component of the position vector r r is r r y = = ° = sin ( )sin . θ 10 53 1 8 m m. Assess: The y -component is positive since the position vector is in the first quadrant.
3.7. Visualize: The figure shows the components v x and v y , and the angle θ . Solve: We have, v v y = − ° sin , 40 or = − ° 10 40 m/s v sin , or v = 15 56 . m/s. Thus the x -component is v v x = ° = ° = cos ( . ) cos . 40 15 56 40 11 9 m/s m/s. Assess: Note that we had to insert the minus sign manually with v y since the vector is in the fourth quadrant.

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3.8. Visualize: Solve: Vector r E points to the left and up, so the components E x and E y are negative and positive, respectively, according to the Tactics Box 3.1. (a) E E x = − cos θ and E E y = sin . θ (b) E E x = − sin φ and E E y = cos . φ Assess: Note that the role of sine and cosine are reversed because we are using a different angle. θ and φ are complementary angles.
3.9. Visualize: We will follow rules in the Tactics Box 3.1. Solve: (a) Vector r r points to the right and down, so the components r x and r y are positive and negative, respectively: r r x = = ° = cos ( )cos . θ 100 45 70 7 m m r r y = − = − ° = − sin ( )sin . θ 100 45 70 7 m m (b) Vector r v points to the right and up, so the components v x and v y are both positive: v v x = = ° = cos ( ) cos θ 300 20 282 m/s m/s v v y = = ° = sin ( )sin θ 300 20 103 m/s

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