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

Answer 1

Answer: (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.
$\therefore$ $2Tsin\theta =F$ $\dots \left(i\right)$
Also, for the mass $m$, $N+Tsin\theta -mg=0$ $\dots \left(ii\right)$
and $Tcos\theta =mA$. $\dots \left(iii\right)$
Equations (i) and (iii) give
$A=\frac{Fcot\theta }{2m}$ $=\frac{F}{2m}$ $\left(\frac{x}{\sqrt{a^2}-x^2}\right).$