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Answer :
After the tackle, the wide receiver's velocity is 5.06 m/s east, and the linebacker's velocity is 6.62 m/s at 111.8° west of north.
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. Initially, the momentum of the system is the sum of the momenta of the wide receiver and the linebacker, and finally, it remains the same after the tackle.
1. First, we calculate the initial momentum of the system:
[tex]\(p_{\text{initial}} = m_1 \cdot v_{1x} + m_2 \cdot v_{2x}\),[/tex] where m represents mass and [tex]\(v_x\)[/tex] represents velocity in the x-direction.
[tex]\(p_{\text{initial}} = (90.0 \, \text{kg} \cdot 15.0 \, \text{m/s}) + (115 \, \text{kg} \cdot 8.0 \, \text{m/s})\) \(p_{\text{initial}} = 1350 \, \text{kg} \cdot \text{m/s} + 920 \, \text{kg} \cdot \text{m/s} = 2270 \, \text{kg} \cdot \text{m/s}\).[/tex]
2. Next, we apply conservation of momentum to find the final velocities of the wide receiver and the linebacker. Since momentum is a vector, we need to consider both magnitude and direction. We'll break down the velocities into their x and y components.
[tex]\(p_{\text{final}} = m_1 \cdot v_{1x}' + m_2 \cdot v_{2x}'\), where \(v_x'\)[/tex]represents final velocity in the x-direction after the tackle.
3. Solving for the final velocities, we find:
[tex]\(p_{\text{final}} = (90.0 \, \text{kg} \cdot v_{1x}') + (115 \, \text{kg} \cdot v_{2x}')\) \(2270 \, \text{kg} \cdot \text{m/s} = (90.0 \, \text{kg} \cdot v_{1x}') + (115 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(300^\circ))\)[/tex]
4. We can now solve for the final velocities:
[tex]\(90.0 \, \text{kg} \cdot v_{1x}' = 2270 \, \text{kg} \cdot \text{m/s} - 115 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(300^\circ)\) \(v_{1x}' = (2270 \, \text{kg} \cdot \text{m/s} - 115 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(300^\circ)) / 90.0 \, \text{kg}\) \(v_{1x}' ≈ 5.06 \, \text{m/s}\)[/tex]
5. For the linebacker's final velocity:
[tex]\(v_{2x}' = (2270 \, \text{kg} \cdot \text{m/s} - 90.0 \, \text{kg} \cdot 5.06 \, \text{m/s}) / 115 \, \text{kg}\) \(v_{2x}' ≈ 6.62 \, \text{m/s}\)[/tex]
6. Finally, to determine the direction of the linebacker's velocity, we use trigonometry:
[tex]\(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\) \(\theta = \arctan(\frac{v_{2y}'}{v_{2x}'})\) \(\theta ≈ \arctan(\frac{115 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \sin(300^\circ)}{2270 \, \text{kg} \cdot \text{m/s} - 115 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(300^\circ)})\) \(\theta ≈ 111.8^\circ\)[/tex]
Therefore, the wide receiver's final velocity is approximately[tex]\(5.06 \, \text{m/s}\) east, and the linebacker's final velocity is approximately \(6.62 \, \text{m/s}\) at \(111.8^\circ\) west of north.[/tex]
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