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 equal and opposite average forces for the same time t. If the coefficient of friction $\mu$ between the ground and the carts are equal, then the displacements of the two 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

The correct option is C $\frac{S_1}{S_2}=-{\left(\frac{m_2}{m_1}\right)}^2$

Minimum stopping distance $=s$
force of friction $=\mu mg$
work done against the friction $=W=\mu mgs$
Initial K.E. of the toy cart $=\frac{p^2}{2m}$
$\therefore \mu mgs=\frac{p^2}{2m}\Rightarrow s=\frac{p^2}{2\mu g{m}^2}$
for the two toy carts, momentum is numerically the same and $\mu ,g$ are same for toy carts.
$\therefore \frac{s_1}{s_2}={\left(\frac{m_2}{m_1}\right)}^2$
As displacement $S_1$ and $S_2$ are in opposite direction.
$\therefore \frac{s_1}{s_2}=-{\left(\frac{m_2}{m_1}\right)}^2$