Question

Two particles of mass m each are at the ends of a light 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 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)

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

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

From balancing the forces we get:-

$F=2Tsin\theta$

Let $a_t$ be the acceleration of the particle

$Tcos\theta =m{a}_t$

On solving we get:-

$⟹a_t=\frac{Fcot\theta }{2m}$

$⟹a_t=\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$

Hence, option $\left(B\right)$ is correct option.