Answer
After the collision, the speed of the 6.0-kg object is $5.0~m/s$
Work Step by Step
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 $5.0~m/s$
