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

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 kinetic energy of the toy cart = $\left(p^2/2m\right)$

$\therefore \mu mgs=\left(p^2/2m\right)$
$s\propto \frac{1}{m^2}$
Hence,
$\frac{s_1}{s_2}={\left(\frac{m_2}{m_1}\right)}^2$

$\therefore \left(c\right)$