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 $s=\left(\frac{p^2}{2\mu g{m}^2}\right)$
For the two toy carts, momentum is numerically the same. Further $\mu$. and $g$are the same for the toy carts.
So, $\frac{S_1}{S_2}={\left(\frac{m_2}{m_1}\right)}^2$