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ParthKohli

  • one year ago

cool question

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  1. ParthKohli
    • one year ago
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    |dw:1441641575975:dw|

  2. ParthKohli
    • one year ago
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    The 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 \(\alpha, \theta\).

  3. ParthKohli
    • one year ago
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    \[\rm m \sqrt{2gl} = (m)(-v_{bob}) + (2m)(v_{box})\tag{conservation of momentum}\]\[\rm \Rightarrow \sqrt{2gl} = 2v_{box} - v_{bob} \]\[\rm 2v_{approach} = 3 v _{separation} \Rightarrow 2 \sqrt{2gl} = 3 (v_{box} + v_{bob})\]\[\rm v_{box}^2 = 2\cdot g\alpha \cdot 3l/2 = 3gl\alpha \Rightarrow v_{box} = \sqrt{3gl\alpha}\]Now,\[\rm 4v_{box} - 2 v_{bob} = 3 v_{box} + 3 v_{bob}\]\[\boxed{\rm v_{box} = 5 v_{bob}}\]

  4. ParthKohli
    • one year ago
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    Hey Ganeshie...

  5. ParthKohli
    • one year ago
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    So from conservation of momentum I have\[\rm \sqrt{2gl} = 9 v_{bob} = 9/5 v_{box} = 9/5 \sqrt{3gl\alpha } = \sqrt{243gl\alpha/25}\]\[\Rightarrow \alpha = 50/243 \]

  6. ParthKohli
    • one year ago
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    Yeeeeeeeee!

  7. ParthKohli
    • one year ago
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    \[\rm v_{bob} = \sqrt{2gl/81}\Rightarrow 1/2 m v_{bob}^2 = mgl/81 = mgh \Rightarrow h = l/81 \]

  8. ParthKohli
    • one year ago
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    \[l\cos \alpha = 80l/81 \Rightarrow \alpha = \cos^{-1}(80/81)\]

  9. ParthKohli
    • one year ago
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    I GOT IT!!! WOW!!! Didn't expect to get this right after this long break.

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