Two unequal masses are connected on two sides | vedclass

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(D) The net pulling force on the system is $F_{net} = (m_2 - m_1)g = (2 - 1) 	imes 10 = 10\, N$.The total mass being pulled is $M = m_1 + m_2 = 1 + 2

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$\frac{1}{3}\, s$

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(D) The net pulling force on the system is $F_{net} = (m_2 - m_1)g = (2 - 1) 	imes 10 = 10\, N$.The total mass being pulled is $M = m_1 + m_2 = 1 + 2 = 3\, kg$.The acceleration of the system is $a = \frac{F_{net}}{M} = \frac{10}{3}\, m/s^2$.At $t = 1\, s$,the velocity of both blocks is $v_0 = a 	imes t = \frac{10}{3} 	imes 1 = \frac{10}{3}\, m/s$.At this moment,the $2\, kg$ block is stopped (velocity becomes $0$),while the $1\, kg$ block continues to move upwards with velocity $v_0 = \frac{10}{3}\, m/s$.After the $2\, kg$ block is released,it falls freely with acceleration $g$ downwards. The $1\, kg$ block moves upwards with initial velocity $v_0$ and acceleration $-g$ (due to gravity).The string becomes tight again when the displacement of the $1\, kg$ block (upwards) equals the displacement of the $2\, kg$ block (downwards) from the point where the larger mass was stopped.Let $t'$ be the time taken for the string to become tight again.For the $1\, kg$ block: $s_1 = v_0 t' - \frac{1}{2} g (t')^2$.For the $2\, kg$ block: $s_2 = \frac{1}{2} g (t')^2$.Equating $s_1 = s_2$,we get $v_0 t' - \frac{1}{2} g (t')^2 = \frac{1}{2} g (t')^2$,which simplifies to $v_0 t' = g (t')^2$.Thus,$t' = \frac{v_0}{g} = \frac{10/3}{10} = \frac{1}{3}\, s$.

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