Answer
Block 2 slides a distance of $~~3.3~m~~$ into the region with friction.
Work Step by Step
We can use Equation (9-68) to find $v_{2f}$, the velocity of block 2 just after the collision:
$v_{2f} = \frac{2~m_1}{m_1+m_2}~v_{1i}$
$v_{2f} = \frac{2~m_1}{m_1+0.4~m_1}~v_{1i}$
$v_{2f} = \frac{2~m_1}{1.4~m_1}~v_{1i}$
$v_{2f} = \frac{10}{7}~v_{1i}$
$v_{2f} = \frac{10}{7}~(4.0~m/s)$
$v_{2f} = 5.714~m/s$
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 $~v_i = 5.714~m/s~$ and $~v_f = 0$
We can find the distance $x$ that block 2 slides:
$v_f^2 = v_i^2+2ax$
$x = \frac{v_f^2 - v_i^2}{2a}$
$x = \frac{0 - (5.714~m/s)^2}{(2)(-4.9~m/s^2)}$
$x = 3.3~m$
Block 2 slides a distance of $~~3.3~m~~$ into the region with friction.
