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ParthKohli
 one year ago
cool question
ParthKohli
 one year ago
cool question

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441641575975:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2The bob of mass \(m\) hangs from its pivot and I set it free from \(A\) when it collides with this box of mass \(2m\) at point \(O\) on the ground. The coefficient of restitution \(e = 2/3\) and now the bob goes to point \(C\) after the collision. Meanwhile the box travels a distance \(3l/2\) before stopping (it stops because of friction, the coefficient of friction being \(\alpha\)) So we have to find \(\alpha, \theta\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[\rm m \sqrt{2gl} = (m)(v_{bob}) + (2m)(v_{box})\tag{conservation of momentum}\]\[\rm \Rightarrow \sqrt{2gl} = 2v_{box}  v_{bob} \]\[\rm 2v_{approach} = 3 v _{separation} \Rightarrow 2 \sqrt{2gl} = 3 (v_{box} + v_{bob})\]\[\rm v_{box}^2 = 2\cdot g\alpha \cdot 3l/2 = 3gl\alpha \Rightarrow v_{box} = \sqrt{3gl\alpha}\]Now,\[\rm 4v_{box}  2 v_{bob} = 3 v_{box} + 3 v_{bob}\]\[\boxed{\rm v_{box} = 5 v_{bob}}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So from conservation of momentum I have\[\rm \sqrt{2gl} = 9 v_{bob} = 9/5 v_{box} = 9/5 \sqrt{3gl\alpha } = \sqrt{243gl\alpha/25}\]\[\Rightarrow \alpha = 50/243 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[\rm v_{bob} = \sqrt{2gl/81}\Rightarrow 1/2 m v_{bob}^2 = mgl/81 = mgh \Rightarrow h = l/81 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[l\cos \alpha = 80l/81 \Rightarrow \alpha = \cos^{1}(80/81)\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I GOT IT!!! WOW!!! Didn't expect to get this right after this long break.
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