Answer
The mass of puck 2 is $~~1.0~kg$
Work Step by Step
Since the vertical drop is the same for both pucks, both pucks took the same time to fall to the lower level. Note that puck 1 traveled in the air a horizontal distance of $2d$ while puck 2 traveled through the air a horizontal distance of $d$. Therefore, after the collision, the speed of puck 1 is twice the speed of puck 2.
Let $v_{1i}$ be the initial velocity of puck 1.
Note that $v_{1f} = -2~v_{2f}$
We can use Equation (9-67) to write an equation for $v_{2f}$:
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}~v_{1i}$
$-2~v_{2f} = \frac{m_1-m_2}{m_1+m_2}~v_{1i}$
$v_{2f} = (-\frac{1}{2})~\frac{m_1-m_2}{m_1+m_2}~v_{1i}$
We can use Equation (9-68) to write an equation for $v_{2f}$:
$v_{2f} = \frac{2~m_1}{m_1+m_2}~v_{1i}$
We can equate the two expressions for $v_{2f}$ to find $m_2$:
$(-\frac{1}{2})~\frac{m_1-m_2}{m_1+m_2}~v_{1i} = \frac{2~m_1}{m_1+m_2}~v_{1i}$
$m_2-m_1 = 4~m_1$
$m_2 = 5~m_1$
$m_2 = (5)(0.20~kg)$
$m_2 = 1.0~kg$
The mass of puck 2 is $~~1.0~kg$
