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Com in Perfectly Inelastic Collision

Two springs h...

Question

Two springs have their force constants $K_{1}$ and $K_{2}$ . Both are stretched till their elastic energies are equal. If the stretching forces are $F_{1}$ and $F_{2}$ then $F_{1} : F_{2}$ is equal to

K_{1} : K_{2}

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K_{2} : K_{1}

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\sqrt{K_{1}} : \sqrt{K_{2}}

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K_{1}^{2} = K_{2}^{2}

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Solution

The correct option is C $\sqrt{K_{1}} : \sqrt{K_{2}}$
Let for same potential energy, elongation of springs be $x_{1}$ and $x_{2}$ an their respective force constant be $k_{1}$ and $k_{2}$
then we have,
$\frac{1}{2} k_{1} x_{1}^{2} = \frac{1}{2} k_{2} x_{2}^{2}$
hence $\frac{x_{1}^{2}}{x_{2}^{2}} = \frac{k_{2}}{k_{1}} . . . . . . . . . . . . . . (1)$
Using equation (1) we can write
$\frac{F_{1}}{F_{2}} = \frac{k_{1} x_{1}}{k_{2} x_{2}} = \frac{k_{1}}{k_{2}} \times \sqrt{\frac{k_{2}}{k_{1}}} = \sqrt{\frac{k_{1}}{k_{2}}}$
So, $F_{1} : F_{2} = \sqrt{k_{1}} : \sqrt{k_{2}}$

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