Answer
Particle 2 collides with particle 1 at the position $~~x = -28~cm$
Work Step by Step
We can use Equation (9-67) to find $v_{1f}$, the velocity of particle 1 just after the collision:
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}~v_{1i}$
$v_{1f} = \left(\frac{0.30~kg-0.40~kg}{0.30~kg+0.40~kg}~\right)~(2.0~m/s)$
$v_{1f} = \left(-\frac{1}{7}~\right)~(2.0~m/s)$
$v_{1f} = -0.2857~m/s$
After the initial collision, particle 1 moves toward the left with a speed of $0.2857~m/s$
We can use Equation (9-68) to find $v_{2f}$, the velocity of particle 2 just after the collision:
$v_{2f} = \frac{2~m_1}{m_1+m_2}~v_{1i}$
$v_{2f} = \left[\frac{(2)~(0.30~kg)}{0.30~kg+0.4~kg}\right]~(2.0~m/s)$
$v_{2f} = (\frac{6}{7})~(2.0~m/s)$
$v_{2f} = 1.7143~m/s$
Suppose that particle 1 moves toward the left a distance $d$
The time it takes is $t = \frac{d}{0.2857}$
In order to collide with particle 1, particle 2 must travel a distance of $d+2~x_w$ which is $d+1.4$
The time it takes is $t = \frac{d+1.4}{1.7143}$
We can equate the expressions for $t$ to find $d$:
$\frac{d}{0.2857} = \frac{d+1.4}{1.7143}$
$1.7143~d = 0.2857~d+0.4$
$1.4286~d = 0.4$
$d = \frac{0.4}{1.4286}$
$d = 0.28~m$
$d = 28~cm$
Since this distance is to the left of $x = 0$, particle 2 collides with particle 1 at the position $~~x = -28~cm$
