Answer
The final speed of the 0.450-kg puck will be 1.93 m/s toward the west.
The final speed of the 0.900-kg puck will be 3.87 m/s toward the east.
Work Step by Step
Let $m_A = 0.450~kg$ and let $m_B = 0.900~kg$.
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 use Equation 7-7 to set up another equation:
$v_A - 0 = v_B' - v_A'$
$v_A' = v_B' - v_A$
We can use this expression for $v_A'$ in the first equation:
$m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$
$v_B' = \frac{2m_A~v_A}{m_a+m_B} = \frac{(2)(0.450~kg)(5.80~m/s)}{(0.450~kg)+(0.900~kg)}$
$v_B' = 3.87~m/s$
We can then use $v_B'$ to find $v_A'$:
$v_A' = 3.87~m/s - 5.80~m/s = -1.93~m/s$
