Question: 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?

\(\begin{aligned}{}{\rm{mass}}\;{\rm{of}}\;{\rm{ball}} &= 5\;{\rm{kg (drpped}}\;{\rm{from}}\;{\rm{a height 12}}\;{\rm{m)}}\\{\rm{mass}}\;{\rm{of}}\;{\rm{bar}} &= 8\;{\rm{kg}}\\{\rm{length (Diameter)}} &= 4\;{\rm{m}}\\{\rm{radius, r = 2}}\;{\rm{m}}\end{aligned}\)

The trading height for velocity is equal to the velocity divided by the gravitational constant.

Velocity,

\(\begin{aligned}{}v &= \sqrt {2gh} \\ &= \sqrt {2 \times 9.8 \times 12} \\ & = 15.34\;{\rm{m/s}}\end{aligned}\)

Angular momentum,

\(\begin{aligned}{}Ib & = mvr\\ & = 5 \times 15.34 \times 2\\ & = 153.44\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

\(\begin{aligned}{}la & = \left( {\frac{{m{l^2}}}{{12}} + {m_1}{r^2} + {m_2}{r^2}} \right)\\ & = \left( {\frac{{8 \times {4^2}}}{{12}} + 5 \times {2^2} + 5 \times {2^2}} \right)\\ & = 50.67\;{\rm{kg}}{{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

Angular velocity,

\(\begin{aligned}{}\omega & = \frac{{153.44}}{{50.67}}\\ & = 3.02\;{\rm{rad/s}}\end{aligned}\)

Linear Velocity,

\(\begin{aligned}{}v & = r\omega \\ & = 2 \times 3.02\\ & = 6.04\;{\rm{m/s}}\end{aligned}\)

\(\begin{aligned}{}h & = \frac{{{v^2}}}{{2g}}\\ & = \frac{{{{6.04}^2}}}{{2 \times 9.81}}\\ & = 1.86\;{\rm{m}}\end{aligned}\)

Hence, the ball will go \(1.86\;{\rm{m}}\)high after the collision.