problem08_73

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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8.73: Let v be the speed of the mass released at the rim just before it strikes the second mass. Let each object have mass m . Conservation of energy says gR v mgR mv 2 ; 2 2 1 = = This is speed 1 v for the collision. Let 2 v be the speed of the combined object just after the collision. Conservation of momentum applied to the collision gives 2 2 so 2 1 2 2 1 gR v v mv mv = = = Apply conservation of energy to the motion of the combined object after the collision.
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Unformatted text preview: Let 3 y be the final height above the bottom of the bowl. ( 29 ( 29 3 2 2 2 1 2 2 gy m v m = 4 / 2 2 1 2 2 2 3 R gR g g v y = = = Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the initial gravitational potential energy....
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