hw11_solns

hw11_solns - Answer Key Homework 11 David McIntyre 45123...

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Answer, Key – Homework 11 – David McIntyre – 45123 – May 10, 2004 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore making your selection. The due time is Central time. Chapter 9 problems. 001 (part 1 oF 2) 0 points Two blocks oF masses M and 3 M are placed on a horizontal, Frictionless surFace. A light spring is attached to one oF them, and the blocks are pushed together with the spring between them. M 3M M 3M Before After (a) (b) v A cord holding them together is burned, aFter which the block oF mass 3 M moves to the right with a speed oF 2 . 83 m / s. What is the speed oF the block oF mass M? Correct answer: 8 . 49 m / s. Explanation: Let : v = 2 . 83 m / s . ±rom conservation oF momentum Δ p = 0, in our case we obtain 0 = Mv M - 3 Mv ThereFore v M = 3 v = (3) (2 . 83 m / s) = 8 . 49 m / s . 002 (part 2 oF 2) 0 points IF M is 5 kg and the spring between the masses has a spring constant oF 6300 N / m, how much was the spring originally compressed From its equilibrium length? Correct answer: 0 . 27618 m. Explanation: Let : M = 5 kg and k = 6300 N / m . 1 2 k L 2 = 1 2 M v 2 m + 1 2 (3 M ) v 2 L = s M v 2 M + 3 M v 2 k = s (5 kg) (8 . 49 m / s) 2 + 3 (5 kg) (2 . 83 m / s) 2 6300 N / m = 0 . 27618 m . 003 (part 1 oF 2) 0 points A 1070 kg car skidding due north on a level Frictionless icy road at 278 . 08 km / h collides with a 2107 . 9 kg car skidding due east at 176 km / h in such a way that the two cars stick together. 2107 . 9 kg 176 km / h 278 . 08 km / h 1070 kg v f θ N At what angle ( - 180 θ +180 ) East oF North do the two coupled cars skid o² at? Correct answer: 51 . 2693 . Explanation: Let : m 1 = 1070 kg , v 1 = 278 . 08 km / h , m 2 = 2107 . 9 kg , and v 2 = 176 km / h .
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Answer, Key – Homework 11 – David McIntyre – 45123 – May 10, 2004 2 m 2 v 2 p 2 p 1 m 1 v 1 p f = m f v f θ N Basic Concepts: Momentum Conserva- tion, K = 1 2 mv 2 , ~p = m~v . During the collision, the total momentum of the two car system will be conserved ~p f = ~p 1 + ~p 2 = m 1 ~v 1 + m 2 ~v 2 p fx ˆ ı + p fy ˆ = m 1 v 1 ˆ + m 2 v 2 ˆ ı. Looking at the x and y components of mo- mentum, p fx = m 2 v 2 = (2107 . 9 kg) (176 km / h) × µ 10 3 m km ¶µ 1 h 3600 s = 103053 kg m / s , and p fy = m 1 v 1 = (1070 kg) (278 . 08 km / h) × µ 10 3 m km ¶µ 1 h 3600 s = 82651 .
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This note was uploaded on 09/15/2009 for the course PHY 557 taught by Professor Rijssenbeek during the Spring '08 term at Adelphi.

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hw11_solns - Answer Key Homework 11 David McIntyre 45123...

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