A spring is compressed between two toy carts | vedclass

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(C) The spring exerts an equal and opposite impulse on both carts,so the magnitude of the momentum $p$ imparted to each cart is the same $(p_1 = p_2 =

Vedclass · Accepted Answer

$\frac{S_1}{S_2} = -\left(\frac{m_2}{m_1}\right)^2$

Vedclass · Answer

(C) The spring exerts an equal and opposite impulse on both carts,so the magnitude of the momentum $p$ imparted to each cart is the same $(p_1 = p_2 = p)$.Using the work-energy theorem,the work done by friction equals the change in kinetic energy: $\mu mg S = \frac{p^2}{2m}$.Rearranging for displacement $S$,we get $S = \frac{p^2}{2\mu g m^2}$.Since $p$,$\mu$,and $g$ are the same for both carts,the displacement $S$ is inversely proportional to the square of the mass: $S \propto \frac{1}{m^2}$.Therefore,the ratio of displacements is $\frac{S_1}{S_2} = \frac{m_2^2}{m_1^2} = \left(\frac{m_2}{m_1}\right)^2$.Since the displacements are in opposite directions,the ratio is $\frac{S_1}{S_2} = -\left(\frac{m_2}{m_1}\right)^2$.

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