Q.
The 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 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) $\frac{F}{2m}\frac{\sqrt{a^{2} - x^{2}}}{x}$
Explanation
(b) $F = 2T\cos\theta$
$\Rightarrow T = \frac{F}{2\cos\theta}$
[image:5_image238.emf]
Magnitude of acceleration of the particle
$= \frac{T\sin\theta}{m}$
$= \frac{F\tan\theta}{2m} = \frac{F}{2m}\frac{x}{\sqrt{a^{2} - x^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.