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A 4.80kg ball is dropped from a height of 10.0m above one end of a uniform bar that pivots at its center. The bar has mass 6.50kg and is 7.20m in length. At the other end of the bar sits another 5.60kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision.

How high will the other ball go after the collision?
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angular momentum conservation

mvl/2 = [I+m(l/2)^2]w

v=sqrt(2gh) =sqrt(2*9.8*10)=14 m/sec

4.8*14*7.2/2 = [6.5*7.2*7.2/12 +4.8*3.6*3.6]w

Velocity at other end=wl/2 =2.68*3.6=9.65 m/sec

height reached by 2nd ball=V^2/2g =9.65*9.65/(2*9.8) =4.75 m (ans)
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