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Question

Two blocks ‘A’ and ‘B’ of same mass 'm' connected by a light spring are suspended by a string as shown in figure. Find the acceleration of block ‘A’ and ‘B’ just after the string is cut.

A
0,2g
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B
0,0
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C
g,g
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D
2g,0
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Solution

The correct option is D 2g,0
When block A and B are in equilibrium position


For block B,
kx=mg
For block A,
T=kx+mg=2mg
When string is cut, tension T becomes zero. But spring does not change its shape just after cutting. So spring force acting on mass B remains the same. Again drawing F.B.D. of block A and B as shown in figure.

Let aA and aB be the acceleration of the blocks A and B.
For block B,
kxmg=maB
Since kx=mg
aB=0
For block A,
mg+kx=maA
2mg=maA
aA=2g (Downwards)

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