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Question

Two blocks A and B of same mass m attached with a light spring are suspended by a string as shown in the figure. Find the acceleration of block A and B respectively, just after the string is cut.


A
g upwards, g downwards
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B
g downwards, zero
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C
2g upwards, g upwards
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D
2g downwards, zero
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Solution

The correct option is D 2g downwards, zero
When block A and B are in equilibrium position


From FBD of B, T0=mg(i)

From FBD of A,
T=mg+T0
T=2mg

When string is cut, tension T becomes zero. But spring does not change its shape after cutting. So same spring force acts on mass B. Again drawing FBD of blocks A and B as shown.


From FBD of B,
T0mg=maB
maB=0
aB=0 (from (i))


From FBD of A,
mg+T0=maA
2mg=maA (from (i))
aA=2g (downwards)

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