Answer
Block 1 slides a distance of $~~30~cm~~$ into the region with friction.
Work Step by Step
We can use Equation (9-67) to find $v_{1f}$, the velocity of the ball just after the collision:
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}~v_{1i}$
$v_{1f} = \frac{m_1-0.40~m_1}{m_1+0.40~m_1}~v_{1i}$
$v_{1f} = \frac{0.60~m_1}{1.40~m_1}~v_{1i}$
$v_{1f} = \frac{0.60~m_1}{1.40~m_1}~v_{1i}$
$v_{1f} = (\frac{3}{7})~(4.0~m/s)$
$v_{1f} = 1.714~m/s$
We can find the magnitude of deceleration in the region with friction:
$F = m~a$
$mg~\mu_k = m~a$
$a = g~\mu_k$
$a = (9.8~m/s^2)(0.50)$
$a = 4.9~m/s^2$
In the next part of the solution, we can let $~v_i = 1.714~m/s~$ and $~v_f = 0$
We can find the distance $x$ that block 1 slides:
$v_f^2 = v_i^2+2ax$
$x = \frac{v_f^2 - v_i^2}{2a}$
$x = \frac{0 - (1.714~m/s)^2}{(2)(-4.9~m/s^2)}$
$x = 0.30~m$
$x = 30~cm$
Block 1 slides a distance of $~~30~cm~~$ into the region with friction.
