91Ó°ÊÓ

Before the collision, we have only one ice chunk moving with a velocity of 12.0 m/s and the other chunk at rest. Using conservation of momentum, the total momentum before collision is equal to the total momentum after collision.The momentum before the collision is given by:\[ p_i = m_1 v_1 = 5.00 \, \text{kg} \times 12.0 \, \text{m/s} = 60.0 \, \text{kg m/s} \]Both chunks stick together, so the combined mass after the collision is:\[ m_f = m_1 + m_2 = 5.00 \, \text{kg} + 5.00 \, \text{kg} = 10.00 \, \text{kg} \]The initial velocity of the system after the collision \( v_f \) can be found using:\[ p_f = m_f v_f \Rightarrow 60.0 \, \text{kg m/s} = 10.00 \, \text{kg} \times v_f \]Solving for \( v_f \):\[ v_f = \frac{60.0}{10.00} = 6.0 \, \text{m/s} \]

The kinetic energy of the system after the collision will be converted to gravitational potential energy at the highest point when the chunks stop moving upward. The expression for kinetic energy \( KE \) after the collision is:\[ KE = \frac{1}{2} m_f v_f^2 = \frac{1}{2} \times 10.00 \, \text{kg} \times (6.0 \, \text{m/s})^2 = 180 \, \text{J} \]At the maximum height, all kinetic energy is converted into potential energy \( PE \):\[ PE = m_f g h \]Set \( KE = PE \):\[ 180 \, \text{J} = 10.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h \]Solving for \( h \):\[ h = \frac{180}{10.00 \times 9.81} = 1.83 \, \text{m} \]