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Unformatted text preview: homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Two ice skaters approach each other at right angles. Skater A has a mass of 72 . 5 kg and travels in the + x direction at 2 . 88 m / s. Skater B has a mass of 62 . 5 kg and is moving in the + y direction at 0 . 982 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 1 . 6121 m / s (tolerance 1 %). Explanation: From conservation of momentum p = 0 m A v A + m B v B = ( m A + m B ) v f Therefore v f = radicalbig ( m A v A ) 2 + ( m B v B ) 2 m A + m B = radicalbig (208 . 8 kg m / s) 2 + (61 . 375 kg m / s) 2 72 . 5 kg + 62 . 5 kg = 1 . 6121 m / s Question 2 part 1 of 2 10 points m 1 1 m 2 before after v v v 2 v m 1 m 2 3 4 An m 2 = 1 . 6 kg can of soup is thrown upward with a velocity of v 2 = 5 . 6 m / s. It is immediately struck from the side by an m 1 = 0 . 6 kg rock traveling at v 1 = 8 . 2 m / s. The rock ricochets off at an angle of = 67 with a velocity of v 3 = 6 . 4 m / s. What is the angle of the cans motion after the collision? Correct answer: 57 . 7764 (tolerance 1 %). Explanation: Basic Concepts: Conservation of Mo mentum p before = p after . Solution: Horizontally m 1 v 1 + m 2 (0) = m 1 v 3 cos + m 2 v 4 cos , so that m 2 v 4 cos = m 1 v 1 m 1 v 3 cos (1) = (0 . 6 kg) (8 . 2 m / s) (0 . 6 kg) (6 . 4 m / s) cos 67 = 3 . 41959 kg m / s . Vertically m 1 (0) + m 2 v 2 = m 1 v 3 sin + m 2 v 4 sin , so that m 2 v 4 sin = m 2 v 2 m 1 v 3 sin (2) = (1 . 6 kg) (5 . 6 m / s) (0 . 6 kg) (6 . 4 m / s) sin67 = 5 . 42526 kg m / s . Thus tan = m 2 v 4 sin m 2 v 4 cos = (5 . 42526 kg m / s) (3 . 41959 kg m / s) = 1 . 58652 , and = arctan(1 . 58652) = 57 . 7764 . Question 3 part 2 of 2 10 points With what speed does the can move imme diately after the collision? Correct answer: 4 . 00815 m / s (tolerance 1 %). Explanation: Using equation (1) above, v 4 = m 1 v 1 m 1 v 3 cos m 2 cos = (0 . 6 kg) (8 . 2 m / s) (1 . 6 kg) cos57 . 7764  (0 . 6 kg) (6 . 4 m / s) cos(67 ) (1 . 6 kg) cos(57 . 7764 ) = 4 . 00815 m / s homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am 2 or using equation (2) above, v 4 = m 2 v 2 m 1 v 3 sin m 2 sin = (5 . 6 m / s) sin(57 . 7764 ) (0 . 6 kg) (6 . 4 m / s) sin(67 ) (1 . 6 kg) sin(57 . 7764 ) = 4 . 00815 m / s . Question 4 part 1 of 2 10 points A 2040 kg car skidding due north on a level frictionless icy road at 219 . 6 km / h collides with a 3651 . 6 kg car skidding due east at 180 km / h in such a way that the two cars stick together. 3651 . 6 kg 180 km / h 219 . 6 km / h 2040 kg v f N At what angle ( 180 +180 ) East of North do the two coupled cars skid off at?...
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 Spring '08
 Turner
 Mass, Work

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