oldmidterm3 - oldmidterm 03 VALENCIA, DANIEL Due: Mar 31...

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oldmidterm 03 – VALENCIA, DANIEL – 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 . 9 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s Fnal speed is 0 . 94 m / s at an angle of θ with respect to its original line of motion. 2 . 9 m / s 0 . 94 m / s θ φ Before After ±ind the eight ball’s speed after the colli- sion. Correct answer: 2 . 74343 m / s (tolerance ± 1 %). Explanation: Let : v q i = 2 . 9 m / s and v q f = 0 . 94 m / s . Given m q = m e = m , Vp e i = 0, Vp mVv , and Vp · Vp 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 (speciFcally, Vp q f + Vp e f = Vp q i , and squaring) gives ( Vp q f + Vp e f ) · ( Vp q f + Vp e f ) = Vp q i · Vp q i . Carrying out the scalar multiplication term by term gives Vp q f · Vp q f + Vp e f · Vp e f + 2 Vp q f · Vp e f = Vp q i · Vp q i . Rewriting in a simpliFed form p 2 q f + p 2 e f + 2 Vp q f · Vp e f = p 2 q i . (2) Subtracting the Eq. 1 for the conservation of energy we have 2 Vp q f · Vp e f = 0 . (3) Dividing Eq. 3 by 2 m Vv q f · Vv e f = 0 , (4) yields three possibilities 1) Vv e f = 0, where m q misses m e . 2) Vv q f = 0, where a head-on collision results. 3) Vv e f Vv e f ; i.e. , θ + φ = 90 . This third possibility agrees with the condi- tions shown in the Fgure. 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 . 9 m / s 2 74 m 71 . 1 18 . 9 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 = r v 2 q i v 2 q f = r (2 . 9 m / s) 2 (0 . 94 m / s) 2 = 2 . 74343 m / s , and θ = arctan p v e f v q f P = arctan p 2 . 74343 m / s 0 . 94 m / s P = 71 . 0866 , also v e f = v q i sin θ = (2 . 9 m / s) sin(71 . 0866 ) = 2 . 74343 m / s , and v q f = v q i cos θ = (2 . 9 m / s) cos(71 . 0866 ) = 0 . 94 m / s , Fnally φ = 90 θ = 18 . 9134
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oldmidterm 03 – VALENCIA, DANIEL – 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 6 . 84 kg / s, as shown in the Fgure. The belt is supported by frictionless rollers and moves at 0 . 615 m / s under the action of a horizontal external force supplied by the motor that drives the belt. F ext ±ind the frictional force exerted by the belt on the sand.
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This note was uploaded on 03/26/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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oldmidterm3 - oldmidterm 03 VALENCIA, DANIEL Due: Mar 31...

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