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A spring is compressed between two toy carts of masses $m_1$ and $m_2$. When the toy carts are released the spring exerts equal and opposite forces for the same time $t$ on each toy cart. If the coefficients of friction $\mu $ between the ground and the toy carts are equal, then the displacement of the toy carts are in the ratio
$\frac{{{s_1}}}{{{s_2}}} = \frac{{{m_2}}}{{{m_1}}}$
$\frac{{{s_1}}}{{{s_2}}} = \frac{{{m_1}}}{{{m_2}}}$
$\frac{{{s_1}}}{{{s_2}}} = {\left( {\frac{{{m_2}}}{{{m_1}}}} \right)^2}$
$\frac{{{s_1}}}{{{s_2}}} = {\left( {\frac{{{m_1}}}{{{m_2}}}} \right)^2}$
Solution
Minimum stopping distance $= S$
Force of friction $=\mu m g$
workdone against the friuction $=\mathrm{w}=\mu \mathrm{mgs}$
Initial kinetic energy of the toy cart $=\left(\mathrm{p}^{2} / 2 \mathrm{m}\right)$
${\mu \mathrm{mgs}=\left(\mathrm{p}^{2} / 2 \mathrm{m}\right)}$
${\mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mathrm{m}^{2} \mu \mathrm{g}}}$
$\mathbf{o r}$ For the two toy carts, momentum
is numerically the same. Futher $\mu$ and $g$ are the same for the toy carts.
$\frac{s_{1}}{s_{2}}=\left(\frac{m_{2}}{m_{1}}\right)^{2}$
As displacement $s_{1}$ and $s_{2}$ are in opposite directions, hence
$\frac{s_{1}}{s_{2}}=\left(\frac{m_{2}}{m_{1}}\right)^{2}$
