Find the acceleration ratio of masses $m_{1}, m_{2}$ and $m_{3}$ (as shown in figure) just after the string is cut at point $A$ .

[0, (1 + \frac{m_{1}}{m_{2}}) g, g]

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g,g,g

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0, \frac{m_{1} g}{m_{2}}, 0

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0, g, \frac{m_{2} g}{m_{1}}

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Solution

The correct option is A $[0, (1 + \frac{m_{1}}{m_{2}}) g, g]$
Initial tension in spring
$F_{s} = m_{1} g$
When the string is cut, spring tension does not change instantly. So, acceleration of $m_{1}$ will be zero.

For block $m_{2}$ ,
$m_{2} g + F_{s} = m_{2} a$
$a = \frac{m_{2} g + m_{1} g}{m_{2}} = (1 + \frac{m_{1}}{m_{2}}) g$

After string is cut, $m_{3}$ is in free fall. So, its acceleration will be g.

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Similar questions

Q. Find the acceleration ratio of masses

m_{1}, m_{2}

and

m_{3}

(as shown in figure) just after the string is cut at point

A

Three blocks 1,2 and 3 are arranged with pulley and spring as shown in the figure. If string connecting blocks $m_{2}$ and $m_{3}$ is cut at point A, find the accelerations of masses $m_{1}$ , $m_{2}$ , and $m_{3}$ just after the string is cut at point A.