Answer
Block 2 comes to a stop a distance of $~~2.22~m~~$ into the region with friction.
Work Step by Step
We can use conservation of energy to find $v_{1i}$, the velocity of block 1 just before the collision:
$\frac{1}{2}m_1v_{1i}^2 = m_1~gh$
$v_{1i}^2 = 2gh$
$v_{1i} = \sqrt{2gh}$
$v_{1i} = \sqrt{(2)(9.8~m/s^2)(2.50~m)}$
$v_{1i} = 7.0~m/s$
We can use Equation (9-68) to find $v_{2f}$, the velocity of block 2 just after the collision:
$v_{2f} = \frac{2m_1}{m_1+2.00~m_1}~v_{1i}$
$v_{2f} = (\frac{2}{3})~(7.0~m/s)$
$v_{2f} = 4.67~m/s$
We can find the magnitude of deceleration in the region with friction:
$F = m~a$
$mg~\mu_k = m~a$
$a = g~\mu_k$
$a = (9.8~m/s^2)(0.500)$
$a = 4.9~m/s^2$
In the next part of the solution, we can let $~v_i = 4.67~m/s~$ and $~v_f = 0$
We can find the distance $d$ that block 2 slides:
$v_f^2 = v_i^2+2ad$
$d = \frac{v_f^2 - v_i^2}{2a}$
$d = \frac{0 - (4.67~m/s)^2}{(2)(-4.9~m/s^2)}$
$d = 2.22~m$
Block 2 comes to a stop a distance of $~~2.22~m~~$ into the region with friction.
