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Question

A block of mass m kg is suspended by a string attached with a 2m kg mass block. The 2m mass block is suspended by a spring as shown in the figure. The system is initially at equilibrium and rest. The spring and string are massless. Now the string is cut. The accelerations of mass 2m and m just after the string is cut will be


A
g2 upwards, g downwards
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B
g downwards, g2 upwards
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C
g downwards, 2g downwards
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D
2g upwards, g downwards
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Solution

The correct option is A g2 upwards, g downwards
Initially, when system is at equilibrium and rest,
F.B.D of m kg and 2m kg block:


From the FBDs,
T=mg ... (1)
and 2mg+T=kx ... (2)

Putting (1) in (2)
kx=3mg ... (3)

As we know, force due to spring does not change instantanously. (After cutting the string, initially it will exert the same force)
But after cutting the string, T=0

F.B.D for m kg and 2m kg block after cutting the string:

For m kg block, only gravitational force is acting. Hence, acceleration of block m is g downward after cutting the string.

Assume acceleration of 2m block, a upwards
Now, kx2mg=2ma
(kx=3mg) (from (3))
3mg2mg=2ma
a=g2
It is positive means assumed direction of acceleration is correct.
Acceleration of 2m block is g2 upwards.

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