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HW24 - Pickering Tracey Homework 24 Due noon Inst Drummond...

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Pickering, Tracey – Homework 24 – Due: Mar 31 2006, noon – Inst: Drummond 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider the two vectors ~ M = ( a, b ) = a ˆ ı + b ˆ and ~ N = ( c, d ) = c ˆ ı + d ˆ where, a and c represent the x -displacement and b and d represent the y -displacement in a Cartesian xy co-ordinate system. Note: ˆ ı and ˆ represent unit vectors ( i.e. vectors of length 1) in the x and y directions, respectively. What is the magnitude of the vector prod- uct k ~ M × ~ M k ? 1. k ~ M × ~ M k = a 2 - 2 a b + b 2 2. k ~ M × ~ M k = 2 a b 3. k ~ M × ~ M k = a 2 + b 2 4. k ~ M × ~ M k = a 2 - b 2 5. k ~ M × ~ M k = a b 6. k ~ M × ~ M k = p a 2 + b 2 7. k ~ M × ~ M k = a 2 + 2 a b + b 2 8. k ~ M × ~ M k = a + b 9. k ~ M × ~ M k = 0 correct 10. k ~ M × ~ M k = - a b Explanation: The magnitude of the vector product of two vectors k ~ A × ~ B k = A B sin θ , where θ is the angle between the two vectors placed with their tails at the same point. When ~ A and ~ B are the same vector, θ = 0 . Since sin 0 = 0, the vector product of a vector with itself is zero. 002 (part 2 of 3) 10 points What is the magnitude of the vector product k ~ M × ~ N k ? 1. k ~ M × ~ N k = a c + b d 2. k ~ M × ~ N k = a 2 + b 2 + c 2 + d 2 3. k ~ M × ~ N k = a b c d 4. k ~ M × ~ N k = a d - b c correct 5. k ~ M × ~ N k = a b - c d 6. k ~ M × ~ N k = 0 7. k ~ M × ~ N k = a d + b c 8. k ~ M × ~ N k = a b + c d 9. k ~ M × ~ N k = a - b 10. k ~ M × ~ N k = a c - b d Explanation: Take the vector products of the x - and y - displacement of ~ M and ~ N individually ~ M × ~ N = ( a ˆ ı + b ˆ ) × ( c ˆ ı + d ˆ ) = a c ı × ˆ ı ) + b c × ˆ ı ) + a d ı × ˆ ) + b d × ˆ ) = a d - b c . since ˆ ı ˆ , we have ˆ ı × ˆ ı = 0, ˆ × ˆ ı = +1, ˆ ı × ˆ = - 1, and ˆ × ˆ = 0. Note: sin 0 = 0 and sin 90 = 1. The magnitude of k ˆ ı × ˆ k = (1)(1) sin 90 = 1 , and k ˆ × ˆ ı k = (1)(1) sin 90 = 1 . The vector product of two vectors, when non-zero, is a vector. These two vector prod- ucts have direction given by ˆ ı × ˆ = ˆ k and ˆ × ˆ ı = - ˆ k , where ˆ k is a unit vector pointing along the positive z -axis. Thus the vectors ˆ ı × ˆ and ˆ × ˆ ı point in opposite directions. The result is ~ M × ~ N = ( a d - b c ) ˆ k .

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Pickering, Tracey – Homework 24 – Due: Mar 31 2006, noon – Inst: Drummond 2 003 (part 3 of 3) 10 points What is the direction of the vector product ~ M × ~ N ? 1. along the x -axis 2. in the yz plane, but not along y or z 3. in the xy plane, but not along x or y 4. none of these since its direction cannot be determined 5. in the xz plane, but not along x or z 6. along the y -axis 7. along the z -axis correct Explanation: ~ M × ~ N points along the z -axis. It only has a displacement in the ± ˆ k direction. 004 (part 1 of 1) 10 points A satellite of mass 3 m moves on a plane in a circular orbit of radius r 3 m i about a spherical planet of mass M and radius R . The magni- tude of the velocity is v 3 m i . M is much larger than 3 m , so the center of mass of the system can be regarded as being located at the center of M
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HW24 - Pickering Tracey Homework 24 Due noon Inst Drummond...

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