A spring is compressed between two toy carts of masses $m_1$ and $m_2$. When the toy carts are released the spring exerts on each toy cart equal and opposite forces for the same time t. If the coefficients of friction $\mathrm{\mu }$ between the ground and the toy carts are equal, then the displacements of the toy carts are in the ratio :

 Let the minimum stopping distance be s. The force of friction would be p m g. Hence work done against friction

$$\begin{gathered} \mathrm{W}=\mathrm{\mu mgs} \\ \text{ Initial K.E. of toy cart }=\left({\mathrm{p}}^2/2\mathrm{m}\right) \\ \therefore \mathrm{\mu mgs}=\left({\mathrm{p}}^2/2\mathrm{m}\right) \\ \mathrm{or} \mathrm{s}=\left[{\mathrm{p}}^2/\left(2{\mathrm{\mu gm}}^2\right)\right] \\ \mathrm{so} \mathrm{s}\propto 1/{\mathrm{m}}^2 \\ [.\text{'} \mathrm{p} \mathrm{and} \mu \mathrm{ar}e \mathrm{same} \mathrm{for} \mathrm{two} \mathrm{toy} \mathrm{carts}. \mathrm{Further}, \mathrm{the} \mathrm{momentum} \mathrm{is} \mathrm{numerically} \mathrm{same}) \\ \frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}={\left(\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\right)}^2 \\ \text{ As }{\mathrm{s}}_1\text{ and }{\mathrm{s}}_2\text{ are in opposite directions, so } \\ \left(\frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}\right)=-{\left(\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\right)}^2 \end{gathered}$$