किसी कण को x-अक्ष के अनुदिश वेग v₀ से प्रक्षेपित किया गया है। इस कण पर कोई अवमंदक बल कार्य कर रहा है जो मूलबिन्दु से दूरी के वर्ग के अनुक्रमानुपाती है, अर्थात् ma = - αx² है। वह दूरी जिस पर यह कण रुक जाएगा है :

Question

Question

\({\left( {\frac{{{\rm{2m\upsilon }}_0^2}}{{{\rm{3\alpha }}}}} \right)^{\frac{1}{2}}}\)
\({\left( {\frac{{{\rm{3m\upsilon }}_0^2}}{{{\rm{2\alpha }}}}} \right)^{\frac{1}{2}}}\)
\({\left( {\frac{{{\rm{3m\upsilon }}_0^2}}{{{\rm{2\alpha }}}}} \right)^{\frac{1}{3}}}\)
\({\left( {\frac{{{\rm{2}}{{\rm{m\upsilon }}_0}}}{{{\rm{3\alpha }}}}} \right)^{\frac{1}{3}}}\)

Answer 1

CONCEPT:

The damping force is defined as the force which is proportional to the velocity of the mass but opposite to the motion of the mass.
Newton's second law of motion, states that the force which is equal to the rate of change of momentum. For a constant mass, force is equal to mass times acceleration.

F = ma

Here, F is the force, m is the mass and a is the acceleration of the given body.

CALCULATION:

It is given that the damping force of the particle is proportional to the square of the distance and it is written as,

F =- \(\alpha\)x² ----(1)

According to Newton's law, we have;

F = ma ----(2)
Now, from equation (1) and equation (2) we have;

ma = - \(\alpha\)x²

⇒ \(a=- \frac{\alpha x^2}{m}\)

⇒ \(a=v\frac{dv}{dx}\)

⇒ \(v\frac{dv}{dx}= -\frac{\alpha x^2}{m}\) ----(3)

Now, Integrating equation (3) we have;

\(v\frac{dv}{dx}= -\frac{\alpha x^2}{m}\)

⇒ \(v{dv}= -\frac{\alpha x^2}{m}dx\)

⇒ \(\int_{v_o}^{0} v{dv}= -\int_{0}^{x} \frac{\alpha x^2}{m}dx\)

⇒ \([\frac{v^2}{2}]_{v_o}^{0}=-\frac {\alpha}{m}[\frac{x^3}{3}]_{0}^{x}\)

⇒ \(-\frac{v_{o}^{2}} {2}=-\frac{\alpha}{m}\frac{x^3}{3} \)

⇒ \(x = (\frac{3mv_{0}^{2}}{2\alpha})^\frac{1}{3}\)

Hence option 1) is the correct answer.

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