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Online Homework Answers 14 (Dec. 14, 2006)

# Online Homework Answers 14 (Dec. 14, 2006) - Hood Charles...

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Hood, Charles – Homework 14 – Due: Dec 14 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 8 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 2) 10 points A 36 . 9 kg girl is standing on a 235 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless sup- porting surface. The girl begins to walk along the plank at a constant speed of 1 . 72 m / s to the right relative to the plank. What is her velocity relative to the ice sur- face? Correct answer: 1 . 48658 m / s. Explanation: v g = velocity of girl relative to ice v p = velocity of plank relative to ice v gp = relative velocity of girl. Let : m g = 36 . 9 kg , m p = 235 kg , and v gp = 1 . 72 m / s . By conservation of momentum 0 = m g v g + m p v p , v p = - m g v g m p The relative velocity is v gp = v g - v p (1) = v g - - m g v g m p = v g m p + m g v g m p = v g ( m p + m g ) m p . Thus v g = m p v gp m g + m p = (235 kg) (1 . 72 m / s) 235 kg + 36 . 9 kg = 1 . 48658 m / s . 002 (part 2 of 2) 10 points What is the velocity of the plank relative to the ice surface? Correct answer: - 0 . 233424 m / s. Explanation: Using Eq. (1), v p = v g - v gp = 1 . 48658 m / s - 1 . 72 m / s = - 0 . 233424 m / s , directed opposite to the girl’s direction.

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Online Homework Answers 14 (Dec. 14, 2006) - Hood Charles...

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