Find the kinetic energy loss

You're on a roll — Step 1: Calculate the initial kinetic energy. The initial velocities are $u_A = 4u$ and $u_B = -u$. The masses are $m_A = m$ and $m_B = 3m$. $$KE_{initial} = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2$$ $$KE_{initial} = \frac{1}{2}m(4u)^2 + \frac{1}{2}(3m)(-u)^2$$ $$KE_{initial} = \frac{1}{2}m(16u^2) + \frac{1}{2}(3m)(u^2)$$ $$KE_{initial} = 8mu^2 + \frac{3}{2}mu^2$$ $$KE_{initial} = \frac{16mu^2 + 3mu^2}{2}$$ $$KE_{initial} = \frac{19}{2}mu^2$$ Step 2: Calculate the final velocities using $e = \frac{3}{5}$. From the previous calculation, the final velocities are: $$v_A = \frac{u(1 - 15e)}{4}$$ $$v_B = \frac{u(1 + 5e)}{4}$$ Substitute $e = \frac{3}{5}$: $$v_A = \frac{u(1 - 15(\frac{3}{5}))}{4} = \frac{u(1 - 9)}{4} = \frac{-8u}{4} = -2u$$ $$v_B = \frac{u(1 + 5(\frac{3}{5}))}{4} = \frac{u(1 + 3)}{4} = \frac{4u}{4} = u$$ Step 3: Calculate the final kinetic energy. $$KE_{final} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2$$ $$KE_{final} = \frac{1}{2}m(-2u)^2 + \frac{1}{2}(3m)(u)^2$$ $$KE_{final} = \frac{1}{2}m(4u^2) + \frac{1}{2}(3m)(u^2)$$ $$KE_{final} = 2mu^2 + \frac{3}{2}mu^2$$ $$KE_{final} = \frac{4mu^2 + 3mu^2}{2}$$ $$KE_{final} = \frac{7}{2}mu^2$$ Step 4: Calculate the kinetic energy loss. Kinetic energy loss is the initial kinetic energy minus the final kinetic energy. $$\Delta KE = KE_{initial} - KE_{final}$$ $$\Delta KE = \frac{19}{2}mu^2 - \frac{7}{2}mu^2$$ $$\Delta KE = \frac{19 - 7}{2}mu^2$$ $$\Delta KE = \frac{12}{2}mu^2$$ $$\Delta KE = \boxed{6mu^2}$$ Got more? Send 'em!