91Ó°ÊÓ

First, calculate the gravitational potential energy (PE) of the 5.00-kg ball before it is dropped using the formula: \[ PE = mgh \] where \( m = 5.00 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 12.0 \, \text{m} \). Substitute the values: \[ PE = 5.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 12.0 \, \text{m} = 588.6 \, \text{J} \] This is the initial potential energy of the dropped ball.

Calculate the moment of inertia (I) for the system immediately before collision, which includes contribution from the bar and the two balls: For the bar (which pivots at its center), use the formula: \[ I_{\text{bar}} = \frac{1}{12} M L^2 \] For the ball that will stick, use: \[ I_{\text{sticking ball}} = m r^2 \] For the other ball at the end of the bar, also use: \[ I_{\text{other ball}} = m r^2 \] Substitute the values: \[ M = 8.00 \, \text{kg}, \; m = 5.00 \, \text{kg}, \; L = 4.00 \, \text{m}, \; r = \frac{L}{2} = 2.00 \, \text{m} \] \[ I_{\text{bar}} = \frac{1}{12} \times 8.00 \, \text{kg} \times (4.00 \, \text{m})^2 = 10.67 \, \text{kg} \cdot \text{m}^2 \] \[ I_{\text{sticking ball}} = 5.00 \, \text{kg} \times (2.00 \, \text{m})^2 = 20.00 \, \text{kg} \cdot \text{m}^2 \] \[ I_{\text{other ball}} = 5.00 \, \text{kg} \times (2.00 \, \text{m})^2 = 20.00 \, \text{kg} \cdot \text{m}^2 \] Thus, the total moment of inertia: \[ I_{\text{total}} = 10.67 + 20.00 + 20.00 = 50.67 \, \text{kg} \cdot \text{m}^2 \]

Since the collision is inelastic, conservation of angular momentum applies: Initial angular momentum = Final angular momentum For the initial angular momentum, only the falling ball contributes: \[ L_i = mvr = m \sqrt{2gh} \cdot r \] For the final angular momentum we have: \[ L_f = I_{\text{total}} \cdot \omega \] Set them equal and substitute the known quantities, solving for \( \omega \):\[ m \sqrt{2gh} \cdot r = I_{\text{total}} \cdot \omega \] \[ 5.00 \, \text{kg} \cdot \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 12.0 \, \text{m}} \cdot 2.00 \, \text{m} = 50.67 \, \text{kg} \cdot \text{m}^2 \cdot \omega \] \[ 5.00 \cdot 15.34 \cdot 2.00 = 50.67 \cdot \omega \] \[ \omega = \frac{153.4}{50.67} \approx 3.03 \, \text{rad/s} \]

The other ball gains kinetic energy that is converted to potential energy at its highest point. Use: \[ mgh = \frac{1}{2} m v^2 \] The linear velocity \( v \) at the pivot is \( v = \omega r \), where \( r = 2.00 \, \text{m} \): \[ v = 3.03 \, \text{rad/s} \times 2.00 \, \text{m} = 6.06 \, \text{m/s} \] Substitute \( v \) in the energy equation to find \( h \): \[ 5.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h = \frac{1}{2} \times 5.00 \, \text{kg} \times (6.06 \, \text{m/s})^2 \] \[ 49.05 \times h = 91.83 \] \[ h = \frac{91.83}{49.05} \approx 1.87 \, \text{m} \] The other ball will rise approximately 1.87 meters.