quiz6sol - October24,2007 PHY2053 Quiz6(Chapter6) Name:UFID:

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October 24, 2007 PHY2053  Quiz 6 (Chapter 6) Name:                                   UFID: 1. (5pts) A 5.00 kg ball strikes a wall at 8.00 m/s at an angle of 60.0º with the plane of  the wall. It bounces off the wall at 4.00 m/s at an angle of 45.0º with the plane of the  wall. If the ball is in contact with the wall for 0.200 s, what is the magnitude of the  average force exerted by the wall on the ball? We express the initial and final velocity in components, and then calculate the change in   momentum. Vi = (8sin60º, 8cos60º) = (6.93 m/s, 4 m/s), Vf = (-4sin45º, 4cos45º) = (-2.83, 2.83) ΔV = Vf-Vi = (-2.83-6.93, 2.83-4) = (-9.76, -1.17) ΔP = mΔV = (5 × -9.76, -5 × 1.17) = (-48.8 kg m/s, -5.85 kg m/s) Since the change in momentum of the ball equals the average force times the time   interval, we get F = ΔP/Δt = (-48.8/0.2, -5.85/0.2) = (-244 N, -29.3 N)
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quiz6sol - October24,2007 PHY2053 Quiz6(Chapter6) Name:UFID:

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