When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then

kinetic energy of body at any instatn after releasing is inversely proportional to their masses

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kinetic energy of body at any instant may or may not be inversely proportional to their masses

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\frac{K . E o f B}{K . E o f A} = \frac{m a s s o f B}{m a s s o f A}

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both (b) and (c) are correct

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Solution

The correct option is A kinetic energy of body at any instatn after releasing is inversely proportional to their masses
Force on each block =kx
Let acceleration at any block of mass m is ,
$m a = k x$
$a = \frac{k x}{m}$
$ω = \sqrt{\frac{k}{m}}$
velocity $v = a t = \frac{k x}{m} t$
$K E = \frac{1}{2} m v^{2}$
$= \frac{1}{2} m \frac{k^{2} x^{2} t^{2}}{m^{2}} = \frac{1}{2} \frac{k^{2} x^{2} t^{2}}{m}$
$K E α \frac{1}{m}$

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When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then

The correct option is A kinetic energy of body at any instatn after releasing is inversely proportional to their massesForce on each block =kxLet acceleration at any block of mass m is ,ma=kxa=kxmω=√kmvelocity v=at=kxmtKE=12mv2=12mk2x2t2m2=12k2x2t2mKEα1m