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(A) By the law of conservation of momentum,the initial momentum is zero,so $m_1 v_1 + m_2 v_2 = 0$,which implies $v_1 / v_2 = -m_2 / m_1$.When the car

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(A) By the law of conservation of momentum,the initial momentum is zero,so $m_1 v_1 + m_2 v_2 = 0$,which implies $v_1 / v_2 = -m_2 / m_1$.When the carts stop,their kinetic energy is dissipated by the work done against friction: $\frac{1}{2} m v^2 = \mu m g S$.This gives $S = \frac{v^2}{2 \mu g}$.Therefore,the ratio of distances traveled is $\frac{S_1}{S_2} = \frac{v_1^2}{v_2^2} = \left( \frac{m_2}{m_1} \right)^2$.Substituting the given values: $\frac{18}{S_2} = \left( \frac{300}{100} \right)^2 = 3^2 = 9$.Thus,$S_2 = \frac{18}{9} = 2 \ m$.

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