qz6sm08sol - July11,2008 PHY2053Discussion Quiz6(Chapter6)...

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July 11, 2008 PHY2053 Discussion Quiz 6 (Chapter 6) Name:                                   UFID: *1. (5pts) A 0.800-kg steel ball strikes a massive wall at 15.0 m/s at an angle of 60.0˚  with the plane of the wall. It bounces off the wall with the same speed and angle. If the  ball is in contact with the wall for 0.0500 s, what is the average force exerted by the  wall on the ball? Taking the +x direction perpendicular to and into the wall and +y along the plane of the   wall, the initial and final velocity of the ball are v i  = (15cos30˚, 15sin30˚) = (13.0 m/s, 7.5 m/s) v f  = (-15cos30˚, 15sin30˚) = (-13.0 m/s, 7.5 m/s) Thus the change in momentum is given by Δp = m(v f -v i ) = 0.8 × (-13-13, 7.5-7.5) = (-20.8 kgm/s, 0 kgm/s) According to the impulse-momentum theorem, this equals impulse I = FΔt, thus F = I/Δt = (-20.8, 0)/0.05 = (-416 N, 0N)
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This note was uploaded on 12/13/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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qz6sm08sol - July11,2008 PHY2053Discussion Quiz6(Chapter6)...

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