## ParthKohli one year ago cool question

1. ParthKohli

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2. ParthKohli

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

$\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

Hey Ganeshie...

5. ParthKohli

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

Yeeeeeeeee!

7. ParthKohli

$\rm v_{bob} = \sqrt{2gl/81}\Rightarrow 1/2 m v_{bob}^2 = mgl/81 = mgh \Rightarrow h = l/81$

8. ParthKohli

$l\cos \alpha = 80l/81 \Rightarrow \alpha = \cos^{-1}(80/81)$

9. ParthKohli

I GOT IT!!! WOW!!! Didn't expect to get this right after this long break.