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Tension in a String

Find the acce...

Question

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 $A$ .

A

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

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B

g,g,g

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C

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

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D

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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