\begin{tabular}{l} A girl of mass \( m_{1}=60.0 \) kilograms springs from a trampoline with an initial \\ upward velocity of \( v_{i}=8.00 \) meters per second. At height \( h=2.00 \) meters \\ above the trampoline, the girl grabs a box of mass \( m_{2}=15.0 \) kilograms. ( \\ Figure 1) \\ For this problem, use \( g=9.80 \) meters per second per second for the \\ magnitude of the acceleration due to gravity. \\ \begin{tabular}{l} Part A \\ \hline\end{tabular} \\ \hline\end{tabular}

Below is a step‐by‐step solution in Markdown with LaTeX formatting. --- ### Problem Recap - A girl of mass \( m_{1}=60.0\,\text{kg} \) is launched upward from a trampoline with an initial speed of \( v_{i}=8.00\,\text{m/s} \). - At a height of \( h=2.00\,\text{m} \) above the trampoline, she catches (grabs) a box of mass \( m_{2}=15.0\,\text{kg} \). - Use \( g=9.80\,\text{m/s}^2 \). **Objective (Part A):** Determine the speed of the system after the girl catches the box. --- ### Step 1. Find the Girl’s Speed at \( h = 2.00\,\text{m} \) Before the girl grabs the box, we need to know her speed when she reaches \( h = 2.00\,\text{m} \). We can use conservation of mechanical energy (neglecting air resistance): 1. **At the trampoline** (initial point): - Kinetic Energy: \( \frac{1}{2} m_{1} v_{i}^{2} \) - Potential Energy: \( 0 \) (we take the trampoline as our zero potential reference) 2. **At \( h=2.00\,\text{m} \)**: - Kinetic Energy: \( \frac{1}{2} m_{1} v^{2} \) - Potential Energy: \( m_{1}gh \) Equate the total energies: \[ \frac{1}{2} m_{1} v_{i}^{2} = \frac{1}{2} m_{1} v^{2} + m_{1}gh \] Since \( m_{1} \) appears in every term, it cancels out: \[ \frac{1}{2} v_{i}^{2} = \frac{1}{2} v^{2} + gh \] Now, solve for \( v \): \[ \frac{1}{2} v^{2} = \frac{1}{2} v_{i}^{2} - gh \] Multiply both sides by 2: \[ v^{2} = v_{i}^{2} - 2gh \] Plug in the given values: \[ v^{2} = (8.00\,\text{m/s})^{2} - 2(9.80\,\text{m/s}^{2})(2.00\,\text{m}) \] \[ v^{2} = 64.0\,\text{m}^{2}\!/\text{s}^{2} - 39.2\,\text{m}^{2}\!/\text{s}^{2} = 24.8\,\text{m}^{2}\!/\text{s}^{2} \] Taking the square root: \[ v = \sqrt{24.8}\,\text{m/s} \approx 4.98\,\text{m/s} \] So, the girl’s speed at \( h = 2.00\,\text{m} \) is approximately \( 4.98\,\text{m/s} \) upward. --- ### Step 2. Use Conservation of Momentum When the Box is Grabbed At the moment the girl catches the box, the box (of mass \( m_{2} \)) is initially at rest (unless stated otherwise). We assume the collision is quick enough that external forces (gravity) have a negligible effect during the impact, so we use conservation of linear momentum. Let: - \( v \) be the girl’s speed just before grabbing the box (\( \approx 4.98\,\text{m/s} \)). - \( v_{\text{final}} \) be the speed of the combined mass immediately after she grabs the box. Conservation of momentum gives: \[ m_{1}v = (m_{1}+m_{2})v_{\text{final}} \] Solve for \( v_{\text{final}} \): \[ v_{\text{final}} = \frac{m_{1}v}{m_{1}+m_{2}} \] Plug in the numbers: \[ v_{\text{final}} = \frac{60.0\,\text{kg}\times 4.98\,\text{m/s}}{60.0\,\text{kg}+15.0\,\text{kg}} = \frac{298.8\,\text{kg}\cdot\text{m/s}}{75.0\,\text{kg}} \] \[ v_{\text{final}} \approx 3.98\,\text{m/s} \] Thus, after the girl grabs the box, the combined system moves upward at about \( 3.98\,\text{m/s} \). --- ### Final Answer After catching the box, the speed of the combined system is approximately: \[ v_{\text{final}} \approx 3.98\,\text{m/s} \]