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 centre $P$ (as shown in the figure). 

Now, the mid-point 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. 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. $\frac{F}{2 m}$ $\frac{x}{a}$
• D. $\frac{F}{2 m}$ $\frac{\sqrt{a^{2}-x^{2}}}{x}$

Answer 1

The Correct Option is B

The arrangement is shown in the figure. The separation between the two masses is $2x$. Each mass will move in the horizontal direction as shown in the figure. Let the tension in the string be $T$. The forces acting at point $P$ and on one of the masses are shown in the figure.

Net force at point $P$ must equal zero. $\therefore\quad$ $2 T \,sin \,\theta=F$ $ \quad\ldots\left(i\right)$ Also, for the mass $m$, $N+T\, sin\, \theta-mg=0$ $ \quad\ldots\left(ii\right)$

and $T \,cos\, \theta=mA$. $\quad\ldots\left(iii\right)$ Equations (i) and (iii) give $A=\frac{F cot \theta}{2m}$ $=\frac{F}{2m}$ $\left(\frac{x}{\sqrt{a^{2}}-x^{2}}\right).$