Physics: Principles with Applications (7th Edition)

Let $m_A = 0.440~kg$ and let $m_B = 0.220~kg$. We can use conservation of momentum to set up an equation. $m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$ Since the collision is elastic, we can use Equation 7-7 to set up another equation. $v_A - 0 = v_B' - v_A'$ $v_A' = v_B' - v_A$ We can use this expression for $v_A'$ in the first equation. $m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$ $v_B' = \frac{2m_A~v_A}{m_a+m_B} = \frac{(2)(0.440~kg)(3.80~m/s)}{(0.440~kg)+(0.220~kg)}$ $v_B' = 5.07~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = 5.07~m/s - 3.80~m/s = 1.27~m/s$