Mass of truck, $m_1=M$
Mass of car, $m_2=\frac{M}{10}$
Initial velocity of truck $=u_1$
Initial velocity of $c a r=u_2$
Momentum of car \& truck are same thus,
$$
\begin{aligned}
& \mathrm{p}_{\mathrm{T}}=\mathrm{p}_{\mathrm{C}} \\
& \mathrm{m}_1 \mathrm{u}_1=\mathrm{m}_2 \mathrm{u}_2 \\
& \mathrm{M} \mathrm{u}_1=\left(\frac{\mathrm{M}}{10}\right) \mathrm{u}_2 \\
& \therefore \frac{\mathrm{u}_1}{\mathrm{u}_2}=\frac{1}{10}
\end{aligned}
$$
Both car \& truck will come to rest, thus their final velocity is zero.
using $\mathrm{eq}^{\mathrm{n}}$ of motion, $\mathrm{v}^2-\mathrm{u}^2=2$ as
Rearranging, put $\mathrm{v}=0 \& \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}$
$$
\mathrm{S}=-\frac{\mathrm{mu}^2}{2 \mathrm{~F}}
$$
Ratio of distance travelled by truck \& car is given as
$$
\begin{aligned}
& \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{\mathrm{m}_1 \mathrm{u}_1^2}{2 \mathrm{~F}} \times \frac{2 \mathrm{~F}}{\mathrm{~m}_2 \mathrm{u}_2^2}=\frac{\mathrm{p}_1(2 \mathrm{~F}) \mathrm{u}_1}{\left(\mathrm{p}_2\right)(2 \mathrm{~F}) \mathrm{u}_2} \\
& \text { since, } \mathrm{p}_1=\mathrm{p}_2 \\
& \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{\mathrm{u}_1}{\mathrm{u}_2}=\frac{1}{10}
\end{aligned}
$$