Answer
Block 2 comes to a stop a distance of $~~0.556~m~~$ 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:
$\frac{1}{2}m_1v_{1i}^2 = m_1~gh$
$v_{1i}^2 = 2gh$
$v_{1i} = \sqrt{2gh}$
$v_{1i} = \sqrt{(2)(9.8~m/s^2)(2.50~m)}$
$v_{1i} = 7.0~m/s$
Since the collision is completely inelastic, the two blocks stick together and move with the same velocity after the collision.
We can use conservation of momentum to find the velocity of the blocks just after the collision:
$p_f = p_i$
$(m_1+2.00~m_1)~v_{2f} = m_1~(7.0~m/s)$
$v_{2f} = \frac{7.0~m/s}{3.00}$
$v_{2f} = 2.33~m/s$
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 $~v_i = 2.33~m/s~$ and $~v_f = 0$
We can find the distance $d$ that block 2 slides:
$v_f^2 = v_i^2+2ad$
$d = \frac{v_f^2 - v_i^2}{2a}$
$d = \frac{0 - (2.33~m/s)^2}{(2)(-4.9~m/s^2)}$
$d = 0.556~m$
Block 2 comes to a stop a distance of $~~0.556~m~~$ into the region with friction.
