Answer
The speed of block 3 is $1.78~m/s$
Work Step by Step
We can use Equation (9-68) to find $v_{2f}$, the velocity of block 2 after the collision:
$v_{2f} = \frac{2~m_1}{m_1+m_2}~v_{1i}$
$v_{2f} = \frac{2~m_1}{m_1+2.00~m_1}~v_{1i}$
$v_{2f} = \frac{2~m_1}{3.00~m_1}~v_{1i}$
$v_{2f} = \frac{2}{3.00}~(4.00~m/s)$
$v_{2f} = 2.67~m/s$
In the next part of the solution, we can let $v_{2i} = 2.67~m/s$
We can use Equation (9-68) to find $v_{3f}$, the velocity of block 3 after the collision:
$v_{3f} = \frac{2~m_2}{m_2+m_3}~v_{2i}$
$v_{3f} = \frac{2~m_2}{m_2+2.00~m_2}~v_{2i}$
$v_{3f} = \frac{2~m_2}{3.00~m_2}~v_{2i}$
$v_{3f} = \frac{2}{3.00}~(2.67~m/s)$
$v_{3f} = 1.78~m/s$
The speed of block 3 is $1.78~m/s$
