Explanation

(b) $a_{A} = \frac{F}{m_{A}} = \frac{4 \times 10}{20} = 2ms^{- 2}$
$a_{B} = \frac{F}{m_{B}} = \frac{4 \times 10}{5} = 8ms^{- 2}$
Given that, $K_{A} = K_{B}$
$i.e.,\ \ \ \ \ \ \ \ \ \ \frac{1}{2}m_{A}v_{A}^{2} = \frac{1}{2}m_{B}v_{B}^{2}$
Or $\ \ \ \ \ \ \ \ \ m_{A}\left( u + a_{A}t_{A} \right)^{2} = m_{B}\left( u + a_{B}t_{B} \right)^{2}\ \ \ \ \ \ \ \ \ \ (\because v = u + at)$
Or $\ \ \ \ \ \ \ \ \ \ m_{A}a_{A}^{2}t_{A}^{2} = m_{B}a_{B}^{2}t_{B}^{2}\ \ \ \ (\because u = 0)\ $
Or $\frac{t_{A}}{t_{B}} = \sqrt{\frac{m_{B}}{m_{A}} \times \frac{a_{B}^{2}}{a_{A}^{2}}}$
$= \sqrt{\frac{5}{20} \times \frac{(8)^{2}}{(2)^{2}}} = \sqrt{\frac{5 \times 64}{20 \times 4}} = 2$