Question

10) A 5.00-kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 8.00 kg and is 4.00 m in length. At the other end of the bar sits another 5.00-kg 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

Answer 1

Answer:

$\Delta h = 10.547\,m$

Explanation:

The velocity of the ball just before the collision with one end of the bar is:

$v = -\sqrt{(0\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (12\,m)}$

$v = 15.342\,\frac{m}{s}$

As the bar is pivoted at its center and collision is entirely inelastic, final velocity is determined by the Principle of Angular Momentum Conservation:

$(5\,kg)\cdot (-15.342\,\frac{m}{s} ) = \left\{-\left[\frac{1}{12}\cdot (8\,kg)\cdot (4\,m)^{2}\cdot (\frac{1}{2\,m} )+(5\,kg)\right] + (5\,kg)\right\}\cdot v$

The final velocity of the another ball is:

$v = 14.383\,\frac{m}{s}$

The maximum height of the other ball is:

$\Delta h = \frac{v^{2}-v_{o}^{2}}{2\cdot g}$

$\Delta h = \frac{(0\,\frac{m}{s} )^{2}-(14.383\,\frac{m}{s} )^{2}}{2\cdot (-9.807\,\frac{m}{s^{2}} )}$

$\Delta h = 10.547\,m$