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Unformatted text preview: Version One Homework 8 Heinz 81204 Oct 14, 2004 1 This printout should have 16 questions. Multiplechoice 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 headon 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 oneanother after a twobody 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.
 Fall '09
 WOODAHL
 Physics, Work

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