Question

Two particles of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $a$ from the center P (as shown in the figure). Now, the midpoint of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface. Neglecting the earth's gravitation, the magnitude of acceleration, when the separation between them becomes $2x$, is :

• A. $\frac{F}{2m}\frac{a}{\sqrt{a^2-x^2}}$
• B. $\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$
• C. $\frac{F}{2m}\frac{x}{a}$
• D. $\frac{F}{2m}\frac{\sqrt{a^2-x^2}}{x}$

Answer 1

The correct option is B $\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$

As shown in the FBD resolving the forces along vertical and horizontal axis in the final position we have $tan\theta =\frac{\sqrt{a^2-x^2}}{x}$

$2\times T\times sin\theta =F$

$T=\frac{F}{2}\times cosec\theta$

$T\times cos\theta =m\times a$

Putting in the values and solving we get $a=\frac{F\times cot\theta }{2\times m}$

Putting in the value of $cot\theta$ as $\frac{x}{\sqrt{a^2-x^2}}$ we have the option B as the correct one.