Given m_1 = 4m_2. How much extra distance (in | vedclass

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(B) For the given system,the acceleration of $m_1$ is $a_1 = \frac{m_1g - 2T}{m_1}$ and $m_2$ is $a_2 = \frac{T - m_2g}{m_2}$.From constraint motion,$

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$40$

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(B) For the given system,the acceleration of $m_1$ is $a_1 = \frac{m_1g - 2T}{m_1}$ and $m_2$ is $a_2 = \frac{T - m_2g}{m_2}$.From constraint motion,$a_1 = a_2/2$. Let $a_2 = a$. Then $a_1 = a/2$.Using $m_1 = 4m_2$,the equations of motion are:$4m_2g - 2T = 4m_2(a/2) = 2m_2a \implies 2m_2g - T = m_2a$$T - m_2g = m_2a$Adding these,$m_2g = 2m_2a \implies a = g/2 = 5 \, m/s^2$.$m_1$ moves up by $20 \, cm = 0.2 \, m$. The velocity $v$ of $m_1$ when it hits the top is $v^2 = u^2 + 2as = 0 + 2(5)(0.2) = 2 \implies v = \sqrt{2} \, m/s$.Since $a_1 = a_2/2$,the velocity of $m_2$ is $v_2 = 2v = 2\sqrt{2} \, m/s$.When $m_1$ stops,$m_2$ continues to move under gravity. The extra distance $h$ covered by $m_2$ is $h = v_2^2 / (2g) = (2\sqrt{2})^2 / (2 	imes 10) = 8 / 20 = 0.4 \, m = 40 \, cm$.

Given $m_1 = 4m_2$. How much extra distance (in $cm$) does $m_2$ travel before coming to rest?

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