## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let $m_A = 0.10~kg$ and let $m_B = 0.30~kg$. Let $v_A$ and $v_B$ be the initial velocity of each ball. Let $v_A'$ and $v_B'$ be the final velocity of each ball. (a) Let's assume that the collision is perfectly elastic. 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 set up another equation. $v_A – v_B = v_B' - v_A'$ $v_A - 0 = v_B' - v_A'$ $v_A' = v_B' - v_A$ We can use this expression for $v_A'$ in the first equation. $m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$ $v_B' = \frac{2m_A~v_A}{m_A+m_B}$ $v_B’ = \frac{(2)(0.10~kg)(10~m/s)}{(0.10~kg)+(0.30~kg)}$ $v_B' = 5.0~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A$ $v_A' = 5.0~m/s - 10~m/s$ $v_A’ = -5.0~m/s$ The final velocity of ball 1 is -5.0 m/s The final velocity of ball 2 is 5.0 m/s (b) Let's assume that the collision is perfectly inelastic. We can set up an equation. $m_A~v_A + 0 = (m_A+m_B)~v_f$ $v_f = \frac{m_A~v_A}{m_A+m_B}$ $v_f = \frac{(0.10~kg)(10~m/s)}{0.10~kg+0.30~kg}$ $v_f = 2.5~m/s$ The final velocity of both balls is 2.5 m/s