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Homework 9 - Solutions

Homework 9 - Solutions - homework 09 FRENNEA KYLE Due 4:00...

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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 Δ p = 0 m A v A ˆ ı + m B v B ˆ = ( 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 α β 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 α = 67 with a velocity of v 3 = 6 . 4 m / s. What is the angle of the can’s 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 α + m 2 v 4 cos β , so that m 2 v 4 cos β = m 1 v 1 - m 1 v 3 cos α (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 α + m 2 v 4 sin β , so that m 2 v 4 sin β = m 2 v 2 - m 1 v 3 sin α (2) = (1 . 6 kg) (5 . 6 m / s) - (0 . 6 kg) (6 . 4 m / s) sin67 = 5 . 42526 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (5 . 42526 kg m / s) (3 . 41959 kg m / s) = 1 . 58652 , and β = 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 α m 2 cos β = (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

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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 α m 2 sin β = (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 θ N At what angle ( - 180 θ +180 ) East of North do the two coupled cars skid off at? Correct answer: 55 . 7231 (tolerance ± 1 %). Explanation: Let : m 1 = 2040 kg , v 1 = 219 . 6 km / h , m 2 = 3651 . 6 kg , and v 2 = 180 km / h . 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 , vectorp = mvectorv . During the collision, the total momentum of the two car system will be conserved vectorp f = vectorp 1 + vectorp 2 = m 1 vectorv 1 + m 2 vectorv 2 p fx ˆ ı + p fy ˆ = m 1 v 1 ˆ + m 2 v 2 ˆ ı .
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