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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 μ C and 2 μ 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 μ C = − 4 × 10 − 6 C , q 2 = 2 μ 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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This note was uploaded on 09/19/2011 for the course PHY 317 taught by Professor Gositz during the Fall '08 term at University of Texas.
 Fall '08
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