Question

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 small time $t$. If the coefficients of friction $\mu$ between the ground and the toy carts are equal, then the magnitude of displacements of the toy carts are in the ratio

• A. $\frac{S_{1}}{S_{2}} = \frac{m_{2}}{m_{1}}$
• B. $\frac{S_{1}}{S_{2}} = \frac{m_{1}}{m_{2}}$
• C. $\frac{S_{1}}{S_{2}} = \left(\frac{m_{2}}{m_{1}}\right)^{2}$
• D. $\frac{S_{1}}{S_{2}} = \left(\frac{m_{1}}{m_{2}}\right)^{2}$

Answer 1

Answer: (C)

Minimum stopping distance $= s$
Work done against the friction $= W= \mu mgs$
Initial momentum gained by both toy carts will be same because same force acts for same time nitial kinetic energy of the toy cart $= \left(\frac{P^{2}}{2m}\right)$
Therefore, $\mu$ mgs$=\frac{P^{2}}{2m}$ or $s = \left(\frac{p^{2}}{2 \mu gm^{2}}\right)$
For the two toy carts, momentum is numerically the same. Further $\mu$. and $g$are the same for the toy carts.
So, $\frac{S_{1}}{S_{2}} = \left(\frac{m_{2}}{m_{1}}\right)^{2}$