Answer
Afterwards, the speed of block 2 is $~~6.0~m/s$
Work Step by Step
We can use the speed of the center of mass to find $v_{1i}$, the initial velocity of block 1:
$v_{com} = \frac{m_1~v_{1i}+m_2~v_{2i}}{m_1+m_2}$
$v_{com} = \frac{m_1~v_{1i}+m_2~(0)}{m_1+3m_1}$
$v_{com} = \frac{m_1~v_{1i}}{4m_1}$
$v_{1i} = 4~v_{com}$
$v_{1i} = (4)(3.00~m/s)$
$v_{1i} = 12.0~m/s$
We can use equation (9-68) to find $v_{2f}$, the velocity of block 2 after the collision:
$v_{2f} = \frac{2~m_1}{m_1+m_2}~v_{1i}$
$v_{2f} = \frac{2~m_1}{m_1+3m_1}~v_{1i}$
$v_{2f} = \frac{2~m_1}{4~m_1}~v_{1i}$
$v_{2f} = \frac{1}{2}~v_{1i}$
$v_{2f} = \frac{1}{2}~(12.0~m/s)$
$v_{2f} = 6.0~m/s$
Afterwards, the speed of block 2 is $~~6.0~m/s$
