2D Collision and Center of mass-solutions

# 2D Collision and Center of mass-solutions - le (txl65) 2D...

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le (txl65) – 2D Collision and Center of mass – mohanty – (81440) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points Consider an elastic collision (ignoring fric- tion and rotational motion). A queue ball initially moving at 3 . 7 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s Fnal speed is 1 . 7 m / s . 3 . 7 m / s 1 . 7 m / s θ φ Before After ±ind the queue ball’s angle θ with respect to its original line of motion. Correct answer: 62 . 6478 . Explanation: Let : v = 3 . 7 m / s and v 1 = 1 . 7 m / s . KE is conserved, so 1 2 m v 2 = 1 2 m v 2 1 + 1 2 m v 2 2 v 2 = v 2 1 + v 2 2 v v 1 v 2 θ This also means that Vv forms the hy- potenuse of a right triangle, so θ + φ = 90 and cos θ = v 1 v θ = arccos p v 1 v P = arccos ± 1 . 7 m / s 3 . 7 m / s ² = 62 . 6478 . 002 (part 1 of 2) 10.0 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 . 4 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 . 5 kg rock traveling at v 1 = 8 . 8 m / s. The rock ricochets o² at an angle of α = 52 with a velocity of v 3 = 6 . 1 m / s. What is the angle of the can’s motion after the collision? Correct answer: 66 . 7 . 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 . 5 kg) (8 . 8 m / s) - (0 . 5 kg) (6 . 1 m / s) cos 52 = 2 . 52223 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 . 4 kg) (5 . 9 m / s) - (0 . 5 kg) (6 . 1 m / s) sin52 = 5 . 85657 kg m / s .

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le (txl65) – 2D Collision and Center of mass – mohanty – (81440) 2 Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (5 . 85657 kg m / s) (2 . 52223 kg m / s) = 2 . 32198 , and β = arctan(2 . 32198) = 66 . 7 . 003 (part 2 of 2) 10.0 points With what speed does the can move immedi- ately after the collision? Correct answer: 4 . 55472 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1 - m 1 v 3 cos α m 2 cos β = (0 . 5 kg) (8 . 8 m / s) (1 . 4 kg) cos66 . 7 - (0 . 5 kg) (6 . 1 m / s) cos(52 ) (1 . 4 kg) cos(66 . 7 ) = 4 . 55472 m / s 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(66 . 7 ) - (0 . 5 kg) (6 . 1 m / s) sin(52 ) (1 . 4 kg) sin(66 . 7 ) = 4 . 55472 m / s . 004 (part 1 of 3) 10.0 points A 1000 kg car moving South at speed v passes an intersection and hits a 2000 kg car moving East at speed v/ 2 .
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## This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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2D Collision and Center of mass-solutions - le (txl65) 2D...

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