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Question

In the figure shown, a body of mass 6 kg is at a distance of 16 m from the pulley at t=0 when it is released. If the string is cut after 2 s, then find the time after which the 6 kg block will hit the pulley. Consider the surface to be smooth (Take g=10 m/s2)

A
1 s
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B
2 s
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C
3 s
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D
4 s
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Solution

The correct option is C 3 s
FBD of 4 kg block can be drawn as

FBD of 6 kg block :

From the FBDs, we can say that the acceleration a of the system (from t=0 to t=2 s) is
a=Supporting forceTotal mass=4g4+6=4 m/s2
Distance covered in first 2 s can be found using the second equation of motion.
s=ut+12at2=12at2
(because the system was at rest initially)
s=12×4×22=8 m
So remaining distance to the pulley =168=8 m
After t=2 s, string is cut, so the block will move with uniform velocity, v=u+at=0+4×2=8 m/s
So, time taken to cover remaining distance of 8 m is
t=8 m8 m/s=1 s
Therefore, total time taken for the block to hit the pulley will be 2+1=3 s

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