d
According to conservation of momentum

\({m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)\,v,\)

where \(v\) is common velocity of the two bodies.

\(\begin{array}{l}
{m_1} = 0.1\,kg\,{m_2} = 0.4\,kg\\
{v_1} = 1\,m/s,\,{v_2} =  - \,0.1\,m/s,\\
\therefore 0.1 \times 1 + 0.4 \times \left( { - 0.1} \right) = \left( {0.1 + 0.4} \right)v\\
or\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,01. - 0.04 = 0.5\,v,\\
v = \frac{{0.06}}{{0.5}} = 0.12\,m/s
\end{array}\)

Hence, distance covered \(= 0.12\) \( \times \,10 = 1.2\,m.\)