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276_Physics ProblemsTechnical Physics

276_Physics ProblemsTechnical Physics - 278*P9.63 Linear...

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278 Linear Momentum and Collisions *P9.63 (a) Each object swings down according to mgR mv = 1 2 1 2 MgR Mv = 1 2 1 2 v gR 1 2 = The collision: + = + + mv Mv m M v 1 1 2 a f v M m M m v 2 1 = + Swinging up: 1 2 1 35 2 2 M m v M m gR + = + ° a f a f a f cos v gR 2 2 1 35 = ° cos a f 2 1 35 2 0 425 0 425 1 425 0 575 0 403 gR M m M m gR M m M m m M m M ° + = + = = = cos . . . . . a fa f a f (b) No change is required if the force is different. The nature of the forces within the system of colliding objects does not affect the total momentum of the system. With strong magnetic attraction, the heavier object will be moving somewhat faster and the lighter object faster still. Their extra kinetic energy will all be immediately converted into extra internal energy when the objects latch together. Momentum conservation guarantees that none of the extra kinetic energy remains after the objects join to make them swing higher. P9.64 (a) Use conservation of the horizontal component of momentum for the system of the shell, the cannon, and the carriage, from just before to just after the cannon firing. p p xf xi = : m v m v shell shell cannon recoil cos . 45 0 0 °+ = 200 125 45 0 5 000 0 a fa f b g cos . °+ = v recoil or v recoil m s =
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