Answer
$x_{max}=0.839m$
Work Step by Step
We can determine the required maximum distance as follows:
We know that:
$\frac{1}{2}m_Av_{A_\circ}^2-Fs=\frac{1}{2}m_Av_{A_1}^2$
We plug in the known values to obtain:
$\frac{1}{2}(15)(10)^2_(0.3)(10)(9.81)(4)=\frac{1}{2}v_{A_2}^2$
This simplifies to:
$v_{A_1}=8.744m/s\leftarrow$
According to the conservation of linear momentum
$m_Av_{A_1}+m_Bv_{B_1}=m_Av_{A_2}+m_Bv_{B_2}$
We plug in the known values to obtain:
$(15)(8.744)+(10)(0)=15v_{A_2}+10v_{B_2}$
This simplifies to:
$3v_{A_2}+2v_{B_2}=26.232~~~$[eq(1)]
We know that:
$e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$
$\implies 0.6=\frac{v_{B_2}-v_{A_2}}{8.744-0}$
$\implies 5.246=v_{B_2}-v_{A_2}~~~$[eq(2)]
After solving eq(1) and eq(2), we obtain:
$v_{A_2}=3.148m/s\leftarrow$
and $v_{B_2}=8.394m/s\leftarrow$
Now we apply the conservation of energy for block $B$
$\frac{1}{2}m_Bv_B^2+\frac{1}{2}ks_1^2=\frac{1}{2}m_Bv^2_{{B^{\prime}_2}}+\frac{1}{2}kx_{max}^2$
We plug in the known values to obtain:
$\frac{1}{2}(10)(8.394)^2+\frac{1}{2}(1000)(0)^2=\frac{1}{2}(10)(0)^2+\frac{1}{2}(1000)x_{max}^2$
This simplifies to:
$x_{max}=0.839m$
