Answer:
[tex]\Delta h = 10.547\,m[/tex]
Explanation:
The velocity of the ball just before the collision with one end of the bar is:
[tex]v = -\sqrt{(0\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (12\,m)}[/tex]
[tex]v = 15.342\,\frac{m}{s}[/tex]
As the bar is pivoted at its center and collision is entirely inelastic, final velocity is determined by the Principle of Angular Momentum Conservation:
[tex](5\,kg)\cdot (-15.342\,\frac{m}{s} ) = \left\{-\left[\frac{1}{12}\cdot (8\,kg)\cdot (4\,m)^{2}\cdot (\frac{1}{2\,m} )+(5\,kg)\right] + (5\,kg)\right\}\cdot v[/tex]
The final velocity of the another ball is:
[tex]v = 14.383\,\frac{m}{s}[/tex]
The maximum height of the other ball is:
[tex]\Delta h = \frac{v^{2}-v_{o}^{2}}{2\cdot g}[/tex]
[tex]\Delta h = \frac{(0\,\frac{m}{s} )^{2}-(14.383\,\frac{m}{s} )^{2}}{2\cdot (-9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]\Delta h = 10.547\,m[/tex]
