#### Answer

After the collision, the 100-g ball is moving to the left at a speed of 0.8 m/s
After the collision, the 400-g ball is moving to the right at a speed of 2.2 m/s

#### Work Step by Step

Let $m_A = 0.10~kg$ and let $m_B = 0.40~kg$
Let $v_A$ be the initial velocity of the 100-g ball.
Let $v_B$ be the initial velocity of the 400-g ball.
Let $v_A'$ be the final velocity of the 100-g ball.
Let $v_B'$ be the final velocity of the 400-g ball.
We first 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 then set up another equation.
$v_A - v_B = v_B' - v_A'$
$v_A' = v_B' - v_A + v_B$
Next, we use this expression for $v_A'$ in the first equation.
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_Bâ€™ = \frac{(2)(0.10~kg)(4.0~m/s)+(0.40~kg)(1.0~m/s)- (0.10~kg)(1.0~m/s)}{(0.10~kg)+(0.40~kg)}$
$v_B' = 2.2~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = 2.2~m/s - 4.0~m/s + 1.0~m/s$
$v_A' = -0.8~m/s$
After the collision, the 100-g ball is moving to the left at a speed of 0.8 m/s.
After the collision, the 400-g ball is moving to the right at a speed of 2.2 m/s.