Answer
$m_2 = 2.2~kg$
Work Step by Step
Note that initially, block 1 is at rest while block 2 is moving.
We can use Equation (9-68) to find an expression for the velocity of block 1 just after the collision:
$v_{1f} = \frac{2m_2}{m_1+m_2}~v_{2i}$
We can use Equation (9-67) to find an expression for the velocity of block 2 just after the collision:
$v_{2f} = \frac{m_2-m_1}{m_1+m_2}~v_{2i}$
Note that the condition required just after the collision is $~~v_{1f} = -v_{2f}$
We can find $m_2$:
$~~v_{1f} = -v_{2f}$
$\frac{2m_2}{m_1+m_2}~v_{2i} = -\frac{m_2-m_1}{m_1+m_2}~v_{2i}$
$2m_2 = -(m_2-m_1)$
$3m_2 = m_1$
$m_2 = \frac{m_1}{3}$
$m_2 = \frac{6.6~kg}{3}$
$m_2 = 2.2~kg$
