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Unformatted text preview: This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Assume an elastic collision (ignoring friction and rotational motion). A queue ball initially moving at 3 . 1 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue balls final speed is 1 . 2 m / s at an angle of with respect to its original line of motion. 3 . 1 m / s 1 . 2 m / s Before After Find the eight balls speed after the colli sion. Correct answer: 2 . 85832 m / s. Explanation: Let : v q i = 3 . 1 m / s and v q f = 1 . 2 m / s . Given m q = m e = m , vectorp e i = 0, vectorp mvectorv , and vectorp vectorp 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, vectorp q f + vectorp e f = vectorp q i , and squaring) gives ( vectorp q f + vectorp e f ) ( vectorp q f + vectorp e f ) = vectorp q i vectorp q i . Carrying out the scalar multiplication term by term gives vectorp q f vectorp q f + vectorp e f vectorp e f + 2 vectorp q f vectorp e f = vectorp q i vectorp q i . Rewriting in a simplified form p 2 q f + p 2 e f + 2 vectorp q f vectorp e f = p 2 q i . (2) Subtracting the Eq. 1 for conservation of en ergy we have 2 vectorp q f vectorp e f = 0 . (3) Dividing Eq. 3 by 2 m vectorv q f vectorv e f = 0 , (4) yields three possibilities 1) vectorv e f = 0, where m q misses m e . 2) vectorv q f = 0, where a headon collision results. 3) vectorv e f vectorv e f ; i.e. , + = 90 . This third possibility agrees with the condi tions shown in the figure. Note: Most pool players already know that the queue ball and the target ball scatter at 90 to oneanother after a twobody collision (to a close approximation). 3 . 1 m / s 1 . 2 m / s 2 . 8 6 m / s 67 . 2 22 . 8 90 Before After Equation 1 gives us 1 2 mv 2 q i = 1 2 mv 2 q f + 1 2 mv 2 e f , rewriting v 2 e f = v 2 q i v 2 q f , then v e f = radicalBig v 2 q i v 2 q f = radicalBig (3 . 1 m / s) 2 (1 . 2 m / s) 2 = 2 . 85832 m / s , and = arctan parenleftbigg v e f v q f parenrightbigg = arctan parenleftbigg 2 . 85832 m / s 1 . 2 m / s parenrightbigg = 67 . 226 , also v e f = v q i sin = (3 . 1 m / s) sin(67 . 226 ) = 2 . 85832 m / s , and v q f = v q i cos = (3 . 1 m / s) cos(67 . 226 ) = 1 . 2 m / s , finally oldmidterm3 01 Turner (58185) 1 hinojosa (jlh3938) oldmidterm3 01 Turner (58185) 2 = 90 = 22 . 774 002 10.0 points Sand from a stationary hopper falls on a mov ing conveyor belt at the rate of 2 . 9 kg / s, as shown in the figure. The belt is supported by frictionless rollers and moves at 1 . 11 m / s un der the action of a horizontal external force supplied by the motor that drives the belt....
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Friction

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