## College Physics (4th Edition)

After the collision, the speed of the 6.0-kg object is $5.0~m/s$
Let $m_A = 2.0~kg$ and let $m_B = 6.0~kg$ Let $v_A$ be the initial velocity of the 2.0-kg object. Let $v_B$ be the initial velocity of the 6.0-kg object. Let $v_A'$ be the final velocity of the 2.0-kg object. Let $v_B'$ be the final velocity of the 6.0-kg object. We can use conservation of momentum to set up an equation. $m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$ Since the collision is elastic, we can set up another equation. $v_A - v_B = v_B' - v_A'$ $v_A' = v_B' - v_A + v_B$ We can use this expression for $v_A'$ in the first equation. $m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$ $v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$ $v_B’ = \frac{(2)(2.0~kg)(10~m/s)+(6.0~kg)(0)- (2.0~kg)(0)}{(2.0~kg)+(6.0~kg)}$ $v_B' = 5.0~m/s$ After the collision, the speed of the 6.0-kg object is $5.0~m/s$