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# hw06b - aljabr(faa335 Hw06 Ross(89251 This print-out should...

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aljabr (faa335) – Hw06 – Ross – (89251) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two insulating spheres having radii 0 . 25 cm and 0 . 45 cm, masses 0 . 1 kg and 0 . 4 kg, and charges - 2 μ C and 3 μ C are released from rest when their centers are separated by 1 . 1 m . How fast is the smaller sphere moving when they collide? Correct answer: 11 . 0683 m / s. Explanation: Let : k e = 8 . 99 × 10 9 N · m 2 / C 2 , r 1 = 0 . 25 cm = 0 . 0025 m , r 2 = 0 . 45 cm = 0 . 0045 m , m 1 = 0 . 1 kg , m 2 = 0 . 4 kg , q 1 = - 2 μ C = - 2 × 10 6 C , q 2 = 3 μ C = 3 × 10 6 C , and d = 1 . 1 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 . 4 kg) ( 8 . 99 × 10 9 N · m 2 / C 2 ) (0 . 1 kg) (0 . 1 kg + 0 . 4 kg) × ( - 2 × 10 6 C) (3 × 10 6 C) × parenleftbigg 1 0 . 0025 m + 0 . 0045 m - 1 1 . 1 m parenrightbigg = 122 . 507 m 2 / s 2 . and v 1 = radicalBig 122 . 507 m 2 / s 2 = 11 . 0683 m / s . 002 10.0 points Note: The force of gravity does not enter into this problem. 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 below is viewed from above the table). The particle is initially released from rest when the string makes an angle 59 with a uniform electric field in the horizontal plane (shown in the figure).

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hw06b - aljabr(faa335 Hw06 Ross(89251 This print-out should...

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