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oldmidterm3 #1

# oldmidterm3 #1 - oldmidterm3 01 Turner(58185 This print-out...

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This print-out should have 25 questions. Multiple-choice 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 ball’s final speed is 1 . 2 m / s at an angle of θ with respect to its original line of motion. 3 . 1 m / s 1 . 2 m / s θ φ Before After Find the eight ball’s 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 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). 3 . 1 m / s 1 . 2 m / s 2 . 86 m / s 67 . 2 22 . 8 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 (3 . 1 m / s) 2 (1 . 2 m / s) 2 = 2 . 85832 m / s , and θ = 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 θ = (3 . 1 m / s) sin(67 . 226 ) = 2 . 85832 m / s , and v q f = v q i cos θ = (3 . 1 m / s) cos(67 . 226 ) = 1 . 2 m / s , finally oldmidterm3 01 – Turner – (58185) 1

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hinojosa (jlh3938) – oldmidterm3 01 – Turner – (58185) 2 φ = 90 θ = 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. F ext Find the frictional force exerted by the belt on the sand. Correct answer: 3 . 219 N. Explanation: Let : ˙ m = 2 . 9 kg / s and v = 1 . 11 m / s .
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