p201f04-homework08-sol

p201f04-homework08-sol - Version One Homework 8 Heinz 81204...

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Unformatted text preview: Version One Homework 8 Heinz 81204 Oct 14, 2004 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Billiard Collision 02 09:04, trigonometry, multiple choice, > 1 min, normal. 001 (part 1 of 1) 10 points Assume an elastic collision (ignoring friction and rotational motion). A queue ball initially moving at 8 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue balls final speed is 4 . 6 m / s . 8 m / s 4 . 6 m / s Before After Find the queue balls angle with respect to its original line of motion. Correct answer: 54 . 9004 . Explanation: Let : v q i = 8 m / s and v q f = 4 . 6 m / s . Given m q = m e = m , ~p e i = 0, ~p m~v , and ~p ~p p 2 . Conservation of energy (and multiplying by 2 m ) gives p 2 q f + p 2 e f = p 2 q i . (1) Conservation of momentum (specifically, ~p q f + ~p e f = ~p q i , and squaring) gives ( ~p q f + ~p e f ) ( ~p q f + ~p e f ) = ~p q i ~p q i . Carring out the scalar multiplication term by term gives ~p q f ~p q f + ~p e f ~p e f + 2 ~p q f ~p e f = ~p q i ~p q i . Rewriting in a simplified form p 2 q f + p 2 e f + 2 ~p q f ~p e f = p 2 q i . (2) Subtracting the equation (1) for the conser- vation of energy we have 2 ~p q f ~p e f = 0 . (3) Dividing equation (3) by 2 m ~v q f ~v e f = 0 , (4) yields three possibilities 1) ~v e f = 0, where m q misses m e . 2) ~v q f = 0, where a head-on collision results. 3) ~v e f and ~v e f are + = 90 . Most pool play- ers know that the queue ball and the target ball scatter at 90 to one-another after a two-body collision (to a close approxima- tion). Use this third possibility. 8 m / s 4 . 6 m / s 6 . 5 5 m / s 54 . 9 90 Before After y The impact parameter is y . Dividing by the mass, Eq. 1 gives us v 2 q i = v 2 q f + v 2 e f , rewriting v 2 e f = v 2 q i- v 2 q f , then v e f = q v 2 q i- v 2 q f = q (8 m / s) 2- (4 . 6 m / s) 2 = 6 . 54523 m / s , and = arctan v e f v q f = arctan 6 . 54523 m / s 4 . 6 m / s = 54 . 9004 , also v e f = v q i sin = (8 m / s)sin(54 . 9004 ) = 6 . 54523 m / s , and v q f = v q i cos Version One Homework 8 Heinz 81204 Oct 14, 2004 2 = (8 m / s)cos(54 . 9004 ) = 4 . 6 m / s , finally = 90 - = 35 . 0996 y R = 2 R cos R = 2 cos(54 . 9004 ) = 1 . 15 . Car Collision 09:05, trigonometry, numeric, > 1 min, nor- mal. 002 (part 1 of 1) 10 points Two cars, one of mass 1000 kg, and the second of mass 2000 kg, are moving at right angles to each other when they collide and stick to- gether. The initial velocity of the first car is 10 m / s in the positive x direction and that of the second car is 16 m / s in the positive y direction....
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This note was uploaded on 07/02/2011 for the course PHYS 201 taught by Professor Woodahl during the Fall '09 term at Indiana.

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p201f04-homework08-sol - Version One Homework 8 Heinz 81204...

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