#### Answer

$t=0.226s$

#### Work Step by Step

We can determine the required time as follows:
According to the conservation of linear momentum
$m_A(v_A)1+m_B(v_B)1=m_A(v_A)2+m_B(v_B)2$
This simplifies to:
$(\frac{1}{32.2})(20)+(\frac{10}{32.2})(0)=\frac{1}{32.2}(v_A)2+(\frac{10}{32.2})(v_B)2$
This simplifies to:
$(v_A)2=20-10(v_B)2~~~$[eq(1)]
We know that:
$e=\frac{(v_B)2-(v_A)2}{20-0}$
This simplifies to:
$(v_A)2=(v_B)2-12~~~$[eq(2)]
After solving eq(1) and (2), we obtain:
$(v_B)2=2.909ft/s\rightarrow$
and $(v_A)2=9.09ft/s\leftarrow$
Now we apply the impulse momentum principle
$m_B(v_B)1+\Sigma \int Fdt=m_B(v_B)2$
We plug in the known values to obtain:
$(\frac{10}{32.2})(2.909)-\mu_kN\Delta t=(\frac{10}{32.2})(0)$
This simplifies to:
$t=0.226s$