Answer
The magnitude of the 0.150-kg glider's final velocity is 3.20 m/s, and it is moving to the left.
The magnitude of the 0.300-kg glider's final velocity is 0.20 m/s, and it is moving to the left.
Work Step by Step
Let $v_A'$ be the final velocity of the 0.150-kg glider.
Let $v_B'$ be the final velocity of the 0.300-kg glider.
Let $m_A = 0.150~kg$ and let $m_B = 0.300~kg$.
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 use Equation (8-27) to 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.
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_B’ = \frac{(2)(0.150~kg)(0.80~m/s)+(0.300~kg)(-2.20~m/s)- (0.150~kg)(-2.20~m/s)}{(0.150~kg)+(0.300~kg)}$
$v_B' = -0.20~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = -0.20~m/s - 0.80~m/s + (-2.20~m/s)$
$v_A' = -3.20~m/s$
The magnitude of the 0.150-kg glider's final velocity is 3.20 m/s, and it is moving to the left.
The magnitude of the 0.300-kg glider's final velocity is 0.20 m/s, and it is moving to the left.
