Two cars of masses m_1 and m_2,joined back to | vedclass

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(A) Let the initial velocities of the cars after the spring is released be $v_1$ and $v_2$. By the law of conservation of momentum,$m_1 v_1 = m_2 v_2$

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(A) Let the initial velocities of the cars after the spring is released be $v_1$ and $v_2$. By the law of conservation of momentum,$m_1 v_1 = m_2 v_2$,which implies $v_1 / v_2 = m_2 / m_1$.Since the frictional force $f$ is the same for both cars,the deceleration $a$ for each car is $a_1 = f / m_1$ and $a_2 = f / m_2$.The time $t$ taken to come to rest is given by $v = u + at$,where final velocity is $0$. Thus,$t = v / a$.For the first car,$t_1 = v_1 / a_1 = v_1 / (f / m_1) = (m_1 v_1) / f$.For the second car,$t_2 = v_2 / a_2 = v_2 / (f / m_2) = (m_2 v_2) / f$.Since $m_1 v_1 = m_2 v_2$ (from conservation of momentum),we have $t_1 = t_2$.Therefore,the ratio of their stopping times is $t_1 / t_2 = 1$.

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