POSSIBLE POINTS: 8.3 During a goal-line stand a75-kg fullback moving eastward with a speed of 10m/s collides head-on with a 100-kg lineman moving westward with a speed of a m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the the post;collision velocity of the two players. 2m/sEast 4m/s East 2m/sWest 4m/sWest

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POSSIBLE POINTS: 8.3 During a goal-line stand a75-kg fullback moving eastward with a speed of 10m/s collides head-on with a 100-kg lineman moving westward with a speed of a m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the the post;collision velocity of the two players. 2m/sEast 4m/s East 2m/sWest 4m/sWest

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To determine the post-collision velocity of the two players, we can use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if no external forces are acting on it. In this case, the system is the two players, and we'll assume no external forces are acting on them during the collision.The formula for momentum (\(p\)) is:\[p = m \times v\]where \(m\) is the mass and \(v\) is the velocity.Before the collision, the fullback has a momentum \(p_{\text{fullback}}\) and the lineman has a momentum \(p_{\text{lineman}}\). We'll take eastward as the positive direction and westward as the negative direction.For the fullback:\[p_{\text{fullback}} = m_{\text{fullback}} \times v_{\text{fullback}}\]\[p_{\text{fullback}} = 75 \, \text{kg} \times 10 \, \text{m/s}\]\[p_{\text{fullback}} = 750 \, \text{kg} \cdot \text{m/s}\]For the lineman:\[p_{\text{lineman}} = m_{\text{lineman}} \times v_{\text{lineman}}\]\[p_{\text{lineman}} = 100 \, \text{kg} \times (-4) \, \text{m/s}\]\[p_{\text{lineman}} = -400 \, \text{kg} \cdot \text{m/s}\]The negative sign indicates that the lineman is moving in the opposite direction to the fullback.The total momentum before the collision (\(p_{\text{total}}\)) is the sum of the momenta of the fullback and the lineman:\[p_{\text{total}} = p_{\text{fullback}} + p_{\text{lineman}}\]\[p_{\text{total}} = 750 \, \text{kg} \cdot \text{m/s} - 400 \, \text{kg} \cdot \text{m/s}\]\[p_{\text{total}} = 350 \, \text{kg} \cdot \text{m/s}\]After the collision, the two players stick together, so their combined mass (\(m_{\text{combined}}\)) is the sum of their individual masses:\[m_{\text{combined}} = m_{\text{fullback}} + m_{\text{lineman}}\]\[m_{\text{combined}} = 75 \, \text{kg} + 100 \, \text{kg}\]\[m_{\text{combined}} = 175 \, \text{kg}\]Let \(v_{\text{combined}}\) be the post-collision velocity of the two players together. Since momentum is conserved:\[p_{\text{total}} = m_{\text{combined}} \times v_{\text{combined}}\]\[350 \, \text{kg} \cdot \text{m/s} = 175 \, \text{kg} \times v_{\text{combined}}\]Now we can solve for \(v_{\text{combined}}\):\[v_{\text{combined}} = \frac{350 \, \text{kg} \cdot \text{m/s}}{175 \, \text{kg}}\]\[v_{\text{combined}} = 2 \, \text{m/s}\]Since the result is positive, the direction is eastward.The post-collision velocity of the two players is \(2 \, \text{m/s}\) East.Final Answer: \(2 \, \text{m/s}\) East

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