Two masses m_1 and m_2 are connected by a str | vedclass

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(A) Let the masses be released from height $H$. The velocity of each mass just before the collision is $v = \sqrt{2gH}$.After the collision,the veloci

Vedclass · Accepted Answer

$\frac{5}{2}\sqrt{\frac{l^2 - a^2}{2gH}}$

Vedclass · Answer

(A) Let the masses be released from height $H$. The velocity of each mass just before the collision is $v = \sqrt{2gH}$.After the collision,the velocities of the masses $m_1$ and $m_2$ become $v_1 = e_1 v = 0.8 \sqrt{2gH}$ and $v_2 = e_2 v = 0.4 \sqrt{2gH}$ respectively,directed upwards.The relative velocity of separation is $v_{rel} = v_1 - v_2 = (0.8 - 0.4) \sqrt{2gH} = 0.4 \sqrt{2gH} = \frac{2}{5} \sqrt{2gH}$.The string becomes tight when the relative vertical displacement between the two masses equals the vertical distance required to stretch the string of length $l$ to its full extent,given the horizontal separation $a$. From the geometry,the required relative vertical displacement is $\Delta h = \sqrt{l^2 - a^2}$.Since the relative acceleration is zero (both are under gravity $g$ upwards),the time $t$ taken is given by $t = \frac{\Delta h}{v_{rel}}$.Substituting the values: $t = \frac{\sqrt{l^2 - a^2}}{\frac{2}{5} \sqrt{2gH}} = \frac{5}{2} \sqrt{\frac{l^2 - a^2}{2gH}}$.

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