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

`\Delta h = 10.547\,m`

Step-by-step explanation:

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

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

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

The final velocity of the another ball is:

`v = 14.383\,(m)/(s)`

The maximum height of the other ball is:

`\Delta h = (v^(2)-v_(o)^(2))/(2\cdot g)`

`\Delta h = ((0\,(m)/(s) )^(2)-(14.383\,(m)/(s) )^(2))/(2\cdot (-9.807\,(m)/(s^(2)) ))`

`\Delta h = 10.547\,m`