Question

Clear Answer and steps plz, I solved it many times but never worked out

(13%) Problem 6: Suppose that a 110 kg football player running at 7.5 m/s catches a 0.49 kg ball moving at a speed of 25 m/s with his feet off the ground, while both of them are moving horizontally 2500 Part (a) Calculate the final speed of the player in meters per second, if the ball and player are initially moving in the same direction. * 25% Part(b) Calculate the change in kinetic energy of the system, in Joules, after the player catches the ball AKE1 Grade Summary Deductions Potential cancel cosO cotan) cosh cotanh(0) -cosO -cotanO -cosh -cotanh0 Submissions Attempts remainin % per attempt) detailed view cos co 0 END Degrees Radians DELİ CLEAR Submit Hints: 1 for a 0% deduction. Hints remaining:- Feedback: 0% deduction per feedback. Use the result of part (a). Change is final value minus initial value 쏘 2500 Part (c) Calculate the final speed of the player in meters per second if the ball and player are initially moving in opposite directions Δ 2596 Part (d) Calculate the change in kinetic energy of the system, in joules, in this case

Answer 1

Given data The mass are, m_1 = 110 kg m_2 = 0.49 kg The speed is, v_1 = 7.5 m/s v_2 = 25 m/s (a) Using conservation of energy m_1 v_1 + m_1 v_2 = (m_1 + m_2)v v = m_1 v_1 + m_2 v_2/m_1 + m_2 = (110)(7.5) + (0.49)(25)/110 + 0.49 = 7.58 m/s (b) The change in Kinetic energy is delta KE = -(1/2 m_1 v_1^2 + 1/2 m_2 v_2^2) + 1/2 (m_1 + m_2)v^2 = -(1/2 (110)(7.5)^2 + 1/2 (0.49)(25.0)^2) + 1/2 (110 + 0.49)(7.58)^2 = -72.70 J (c) Using conservation of energy m_1 v_1 - m_2 v_2 = (m_1 + m_2)v v = m_1 v_1 - m_2 v_2/m_1 + m_2 = (110)(7.5) - (0.49)(25)/110 + 0.49 = 7.36 m/s (d) The change in kinetic energy is delta KE = -1/2(1/2 m_1 v_1^2 + 1/2 m_2 v_2^2) + 1/2(m_1 + m_2)v^2 = -(1/2(110)(7.5)^2 - 1/2(0.49)(25)^2) + 1/2(110 + 0.419)(7.36)^2 = 50.1 J