solutions 13

# solutions 13 - Patel Kinal – Homework 13 – Due 7:00 pm...

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Unformatted text preview: Patel, Kinal – Homework 13 – Due: Oct 19 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 12 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 1) 10 points Two ice skaters approach each other at right angles. Skater A has a mass of 38 . 8 kg and travels in the + x direction at 2 . 05 m / s. Skater B has a mass of 79 . 4 kg and is moving in the + y direction at 2 . 19 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 1 . 61772 m / s. 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 = p ( m A v A ) 2 + ( m B v B ) 2 m A + m B = p (79 . 54 kg m / s) 2 + (173 . 886 kg m / s) 2 38 . 8 kg + 79 . 4 kg = 1 . 61772 m / s keywords: 002 (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 . 5 kg can of soup is thrown upward with a velocity of v 2 = 5 . 9 m / s. It is immediately struck from the side by an m 1 = 0 . 43 kg rock traveling at v 1 = 8 . 6 m / s. The rock ricochets off at an angle of α = 40 ◦ with a velocity of v 3 = 6 m / s. What is the angle of the can’s motion after the collision? Correct answer: 76 . 5373 ◦ . 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 . 43 kg)(8 . 6 m / s)- (0 . 43 kg)(6 m / s)cos40 ◦ = 1 . 72161 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 . 5 kg)(5 . 9 m / s)- (0 . 43 kg)(6 m / s)sin40 ◦ = 7 . 19161 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (7 . 19161 kg m / s) (1 . 72161 kg m / s) = 4 . 17727 , and β = arctan(4 . 17727) = 76 . 5373 ◦ . 003 (part 2 of 2) 10 points With what speed does the can move immedi- ately after the collision? Correct answer: 4 . 92989 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1- m 1 v 3 cos α m 2 cos β = (0 . 43 kg)(8 . 6 m / s) (1 . 5 kg)cos76 . 5373 ◦- (0 . 43 kg)(6 m / s)cos(40 ◦ ) (1 . 5 kg)cos(76 . 5373 ◦ ) = 4 . 92989 m / s Patel, Kinal – Homework 13 – Due: Oct 19 2007, 7:00 pm – Inst: D Weathers 2 or using equation (2) above, v 4 = m 2 v 2- m 1 v 3 sin α m 2 sin β = (5 . 9 m / s) sin(76 . 5373 ◦ )- (0 . 43 kg)(6 m / s)sin(40 ◦ ) (1 . 5 kg)sin(76 . 5373 ◦ ) = 4 . 92989 m / s . keywords: 004 (part 1 of 1) 10 points Assume an elastic collision (ignoring friction and rotational motion)....
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solutions 13 - Patel Kinal – Homework 13 – Due 7:00 pm...

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