Question

Can someone help solve the second question and explain it?

< Chapter 8 Exercise 8.21 く) 5of17 Part A On a kg ) is moving toward puck B (with mass 0.355 kg ), that is initially at rest. After the collision, puck A has a velocity of 0.121 m/s to the left, and puck B has velocity 0.660 m/s to the right. horizontal air table, puck A (with mass 0.245 What was the speed of puck A before the collision? = 0.835 m/s Correct ▼ Part B Calculate the change in the total kinetic eergy of the system that occurs during the collision Submit Incorrect; Try Again: 3 attempts remaining Provide Feedbachk Ne

Answer 1

m_A = 0.245 kg m_B = 0.355 kg V_2 A = 0.121 m/s V_2 B = 0.660 m/s V_1 A = ? V_1 B = 0 V_ Collision = m_a V_a + m_b V_b/m_a = (0.245) (-121) + (0.355) ()/m_a = -0.029645 + 0.2343/0.245 V A_1 = 0.835 m/s K_i = 1/2 m v_i^2 = 1/2 (0.245) (0.835)^2 = 0.0854 J \r\n Kf = 1/2 m_A V_A^2 + 1/2 m_B V_B^2 = 1/2 (0.245) (0.121)^2 + 1/2 (0.355) (0.6601)^2 = 0.00179 + 0.07731 = 0.079109 J Delta K = Kf = Ki = -0.006291 J The total kinetic energy decrease.