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Unformatted text preview: Overton, Mays – Homework 19 – Due: Mar 9 2005, 4:00 am – Inst: Turner 1 This printout should have 14 questions. Multiplechoice 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 Assume an elastic collision (ignoring friction and rotational motion). A queue ball initially moving at 2 . 7 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s final speed is 1 m / s . 2 . 7 m / s 1 m / s θ Before After Find the queue ball’s angle θ with respect to its original line of motion. Correct answer: 68 . 2615 ◦ . Explanation: Let : v q i = 2 . 7 m / s and v q f = 1 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 θ + φ = 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. 2 . 7 m / s 1 m / s 2 . 5 1 m / s 68 . 3 ◦ φ 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 (2 . 7 m / s) 2 (1 m / s) 2 = 2 . 50799 m / s , and θ = arctan µ v e f v q f ¶ = arctan µ 2 . 50799 m / s 1 m / s ¶ = 68 . 2615 ◦ , also v e f = v q i sin θ = (2 . 7 m / s)sin(68 . 2615 ◦ ) = 2 . 50799 m / s , and v q f = v q i cos θ = (2 . 7 m / s)cos(68 . 2615 ◦ ) = 1 m / s , finally Overton, Mays – Homework 19 – Due: Mar 9 2005, 4:00 am – Inst: Turner 2 φ = 90 ◦ θ = 21 . 7385 ◦ y R = 2 R cos θ R = 2 cos(68 . 2615 ◦ ) = 0 . 740741 . 002 (part 1 of 2) 10 points A 0 . 0805 kg block is released from rest from the top of a 61 . 7 ◦ frictionless incline. When it has fallen a vertical distance of 2 . 22 m, a . 0149 kg bullet is fired into the block along a path parallel to the slope of the incline, and momentarily brings the block to rest, stopping in the block....
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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