OldMidterm3 - oldmidterm 03 PAPAGEORGE MATT Due 4:00 am...

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oldmidterm 03 – PAPAGEORGE, MATT – Due: Mar 31 2008, 4:00 am 1 Question 1, chap 10, sect 99. part 1 of 1 10 points Assume an elastic collision (ignoring fric- tion and rotational motion). A queue ball initially moving at 2 . 5 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s final speed is 1 . 1 m / s at an angle of θ with respect to its original line of motion. 2 . 5 m / s 1 . 1 m / s θ φ Before After Find the eight ball’s speed after the colli- sion. Correct answer: 2 . 24499 m / s (tolerance ± 1 %). Explanation: Let : v q i = 2 . 5 m / s and v q f = 1 . 1 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 the conservation of energy 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 head-on collision results. 3) vectorv e f vectorv e f ; i.e. , θ + φ = 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 one-another after a two-body collision (to a close approximation). 2 . 5 m / s 1 . 1 m / s 2 . 24 m / s 63 . 9 26 . 1 90 Before After Equation 1 gives us 1 2 m v 2 q i = 1 2 m v 2 q f + 1 2 m v 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 (2 . 5 m / s) 2 (1 . 1 m / s) 2 = 2 . 24499 m / s , and θ = arctan parenleftbigg v e f v q f parenrightbigg = arctan parenleftbigg 2 . 24499 m / s 1 . 1 m / s parenrightbigg = 63 . 8961 , also v e f = v q i sin θ = (2 . 5 m / s) sin(63 . 8961 ) = 2 . 24499 m / s , and v q f = v q i cos θ = (2 . 5 m / s) cos(63 . 8961 ) = 1 . 1 m / s , finally φ = 90 θ = 26 . 1039
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oldmidterm 03 – PAPAGEORGE, MATT – Due: Mar 31 2008, 4:00 am 2 Question 2, chap 10, sect 1. part 1 of 1 10 points Sand from a stationary hopper falls on a moving conveyor belt at the rate of 4 . 36 kg / s, as shown in the figure. The belt is supported by frictionless rollers and moves at 0 . 47 m / s under the action of a horizontal external force supplied by the motor that drives the belt. F ext Find the frictional force exerted by the belt on the sand. Correct answer: 2 . 0492 N (tolerance ± 1 %). Explanation: Let : ˙ m = 4 . 36 kg / s and v = 0 . 47 m / s . The momentum of the sand on the belt is p x = m v .
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