A spring is compressed between two toy carts | vedclass

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(C) When the spring is released,it exerts an impulse $J = F \cdot t$ on both carts. Since the forces are equal and act for the same time,both carts ac

Vedclass · Accepted Answer

$\frac{s_1}{s_2} = \left( \frac{m_2}{m_1} \right)^2$

Vedclass · Answer

(C) When the spring is released,it exerts an impulse $J = F \cdot t$ on both carts. Since the forces are equal and act for the same time,both carts acquire the same magnitude of momentum $p = m_1 v_1 = m_2 v_2$.The frictional force acting on each cart is $f = \mu m g$.Using the work-energy theorem,the work done by friction equals the change in kinetic energy: $W = \Delta K$.$f \cdot s = \frac{p^2}{2m} \implies \mu m g s = \frac{p^2}{2m}$.Solving for displacement $s$: $s = \frac{p^2}{2 \mu m^2 g}$.Since $p$,$\mu$,and $g$ are the same for both carts,$s \propto \frac{1}{m^2}$.Therefore,the ratio of displacements is $\frac{s_1}{s_2} = \frac{m_2^2}{m_1^2} = \left( \frac{m_2}{m_1} \right)^2$.

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