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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. Express your answer numerically in meters per second.

Ask by Turnbull Ross. in the United States
Mar 06,2025

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Tutor-Verified Answer

Answer

After grabbing the box, the girl and the box move upward together with a velocity of approximately 3.98 meters per second.

Solution

Below is a step‐by‐step solution in markdown format. --- ### Step 1. Determine the Girl’s Speed at \( h = 2.00 \) m The girl is launched with an initial upward velocity of \[ v_i = 8.00 \text{ m/s}. \] As she rises, gravity decelerates her motion. Using energy conservation (or kinematics), we have: \[ v^2 = v_i^2 - 2gh, \] where - \( g = 9.80 \text{ m/s}^2 \), - \( h = 2.00 \text{ m} \). Plug in the values: \[ v^2 = (8.00)^2 - 2(9.80)(2.00) = 64.0 - 39.2 = 24.8. \] Taking the square root gives the upward speed at \( h = 2.00 \) m: \[ v = \sqrt{24.8} \approx 4.98 \text{ m/s}. \] --- ### Step 2. Apply Conservation of Momentum During the Collision At \( h = 2.00 \) m, the girl (mass \( m_1 = 60.0 \) kg) grabs a stationary box (mass \( m_2 = 15.0 \) kg). Since the collision is perfectly inelastic (they stick together), linear momentum is conserved. Before the collision, the total momentum is: \[ p_{\text{initial}} = m_1 v \quad \text{(since the box is at rest)}. \] After the collision, the total mass is \[ m_1 + m_2, \] and they move together with velocity \( v_f \). Setting the momenta equal: \[ m_1 v = (m_1 + m_2) v_f. \] Solving for \( v_f \): \[ v_f = \frac{m_1 v}{m_1 + m_2}. \] Substitute the known values: \[ v_f = \frac{60.0 \text{ kg} \times 4.98 \text{ m/s}}{60.0 \text{ kg} + 15.0 \text{ kg}} = \frac{298.8 \text{ kg·m/s}}{75.0 \text{ kg}} \approx 3.98 \text{ m/s}. \] --- ### Final Answer The combined system (girl plus box) moves upward with a velocity of approximately: \[ \boxed{3.98 \text{ m/s}}. \]

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The Deep Dive

To find the speed of the girl and the box just after she grabs the box at height \( h = 2.00 \) meters, we can use the principle of conservation of momentum and the kinematic equations. First, calculate the speed of the girl just before she grabs the box. We use the kinematic equation: \[ v^2 = v_i^2 + 2gh \] Given: - \( v_i = 8.00 \) m/s (initial velocity) - \( g = -9.80 \) m/s² (negative because it acts downward) - \( h = 2.00 \) m Calculate the final velocity \( v \): \[ v^2 = (8.00)^2 + 2(-9.80)(-2.00) \] \[ v^2 = 64 + 39.2 = 103.2 \] \[ v = \sqrt{103.2} \approx 10.16 \, \text{m/s} \] Now, using conservation of momentum: Let \( v_f \) be the final velocity after she grabs the box. The initial momentum before she grabs the box is: \[ p_i = m_1 \cdot v + m_2 \cdot 0 = 60.0 \cdot 10.16 + 15.0 \cdot 0 \] \[ p_i = 609.6 \, \text{kg m/s} \] After grabbing the box, the total mass is \( m_{total} = m_1 + m_2 = 60.0 + 15.0 = 75.0 \) kg. The final momentum \( p_f \) is given by: \[ p_f = m_{total} \cdot v_f \] Setting the initial momentum equal to the final momentum: \[ 609.6 = 75.0 \cdot v_f \] Solve for \( v_f \): \[ v_f = \frac{609.6}{75.0} \approx 8.13 \, \text{m/s} \] So, the speed of the girl and the box just after she grabs the box is approximately \( 8.13 \) meters per second.

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