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# hw05a - aldawsari(ma28683 – HW05 – Betancourt –(16856...

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Unformatted text preview: aldawsari (ma28683) – HW05 – Betancourt – (16856) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 23.2-23.5 001 10.0 points Two insulating spheres having radii 0 . 19 cm and 0 . 39 cm, masses 0 . 16 kg and 0 . 56 kg, and charges- 4 μ C and 4 μ C are released from rest when their centers are separated by 1 . 6 m . How fast is the smaller sphere moving when they collide? Correct answer: 15 . 4996 m / s. Explanation: Let : k e = 8 . 99 × 10 9 N · m 2 / C 2 , r 1 = 0 . 19 cm = 0 . 0019 m , r 2 = 0 . 39 cm = 0 . 0039 m , m 1 = 0 . 16 kg , m 2 = 0 . 56 kg , q 1 =- 4 μ C =- 4 × 10 − 6 C , q 2 = 4 μ C = 4 × 10 − 6 C , and d = 1 . 6 m . 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 . 56 kg) ( 8 . 99 × 10 9 N · m 2 / C 2 ) (0 . 16 kg) (0 . 16 kg + 0 . 56 kg) × (- 4 × 10 − 6 C) (4 × 10 − 6 C) × parenleftbigg 1 . 0019 m + 0 . 0039 m- 1 1 . 6 m parenrightbigg = 240 . 237 m 2 / s 2 . and v 1 = radicalBig 240 . 237 m 2 / s 2 = 15 . 4996 m / s . 002 10.0 points A charged particle is connected to a string that is is tied to the pivot point P . The particle, string, and pivot point all lie on a horizontal table (consequently the figure...
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hw05a - aldawsari(ma28683 – HW05 – Betancourt –(16856...

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