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Unformatted text preview: nanni (arn437) HW#3 Erskine (56905) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Four charges are fixed at the corners of a square centered at the origin as follows: q at ( a, + a ); 2 q at (+ a, + a ); 3 q at (+ a, a ); and 6 q at ( a, a ). A fifth charge + q with mass m is placed at the origin and released from rest. Find the speed when it is a great distance from the origin, where the potential energy of the fifth charge due to the four point charges is negligible. 1. bardbl vectorv bardbl = q radicalBigg 2 5 k ma 2. bardbl vectorv bardbl = q radicalBigg 3 5 k ma 3. bardbl vectorv bardbl = q radicalBigg 6 6 k ma 4. bardbl vectorv bardbl = q radicalBigg 2 2 k ma 5. bardbl vectorv bardbl = q radicalBigg 3 2 k ma 6. bardbl vectorv bardbl = q radicalBigg 6 2 k ma correct 7. bardbl vectorv bardbl = q radicalBigg 6 5 k ma 8. bardbl vectorv bardbl = q radicalBigg 6 3 k ma 9. bardbl vectorv bardbl = q radicalBigg 3 3 k ma 10. bardbl vectorv bardbl = q radicalBigg 3 6 k ma Explanation: x a +6 q a 3 q +2 q + q + q, m 2 a The initial energy of the charge is E i = K i + U i = U i = q parenleftbigg k q 2 a + 2 k q 2 a + ( 3 q ) k 2 a + 6 k q 2 a parenrightbigg = 6 k q 2 2 a . The final energy is E f = 1 2 mv 2 . From energy conversation, we have E i = E f 6 k q 2 2 a = 1 2 mv 2 v = q radicalBigg 6 2 k ma . 002 10.0 points Two insulating spheres having radii 0 . 2 cm and 0 . 4 cm, masses 0 . 14 kg and 0 . 55 kg, and charges 4 C and 2 C are released from rest when their centers are separated by 0 . 9 m . How fast is the smaller sphere moving when they collide? Correct answer: 11 . 6441 m / s. Explanation: Let : k e = 8 . 99 10 9 N m 2 / C 2 , r 1 = 0 . 2 cm = 0 . 002 m , r 2 = 0 . 4 cm = 0 . 004 m , m 1 = 0 . 14 kg , m 2 = 0 . 55 kg , q 1 = 4 C = 4 10 6 C , q 2 = 2 C = 2 10 6 C , and d = 0 . 9 m . nanni (arn437) HW#3 Erskine (56905) 2 By conservation of momentum 0 = m 1 v 1 m 2 v 2 v 2 = m 1 v 1 m 2 By conservation of energy ( K + U ) i = ( K + U ) f 0 + k e q 1 q 2 d = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 + k e q 1 q 2 r 1 + r 2 k e q 1 q 2 d = 1 2 m 1 v 2 1 + 1 2 m 2 parenleftbigg m 1 v 1 m 2 parenrightbigg 2 + k e q 1 q 2 r 1 + r 2 k e q 1 q 2 parenleftbigg 1 r 1 + r 2 1 d parenrightbigg = m 1 v 2 1 2 m 2 ( m 1 + m 2 ) . So v 2 1 = 2 m 2 k e q 1 q 2 m 1 ( m 1 + m 2 ) parenleftbigg 1 r 1 + r 2 1 d parenrightbigg = 2 (0 . 55 kg) ( 8 . 99 10 9 N m 2 / C 2 ) (0 . 14 kg) (0 . 14 kg + 0 . 55 kg) ( 4 10 6 C) (2 10 6 C) parenleftbigg 1 . 002 m + 0 . 004 m 1 . 9 m parenrightbigg = 135 . 584 m 2 / s 2 ....
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 Fall '08
 gositz

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