Question

In figure, two identical particles of mass $m$ are tied together with an inextensible light string. This is pulled at its centre with a constant force $F$. If the whole system lies on a smooth horizontal plane, then the acceleration of approach of particles at the instant shown will be:

• A. $\frac{\sqrt{3}F}{m}$
• B. $\frac{F}{2\sqrt{3}m}$
• C. $\frac{2F}{\sqrt{3}m}$
• D. $\frac{F}{\sqrt{3}m}$

Answer 1

The correct option is D $\frac{F}{\sqrt{3}m}$

Make free body diagram of given figure.

$2T\cos 30°=F$

$T=\frac{F}{\sqrt{3}}$

acceleration of approach of $A$

$a_A=\frac{T\sin \theta }{m}$

$a_A=\frac{F}{2m\times \sqrt{3}}$

Similarly,

$a_B=\frac{F}{2\sqrt{3}m}$

Acceleration of approach $=\frac{F}{\sqrt{3}m}$