Question

Two masses each equal to m are lying on x-axis at (-a, 0) and (+a, 0), respectively, as shown in Fig. They are connected by a light string. A force F is applied at the origin along vertical direction. As a result, the masses move towards each other without loosing contact with ground. What is the acceleration of each mass? Assuming the instantaneous position of the masses as (-x, 0) and (x , 0), respectively.

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

Answer 1

Answer: (C)

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

REF.Image.

$2Tsin\theta =F$

$T=\frac{F}{2\sin \theta },$

Component of tension force along x causes acceleration of the ball.

$a=\frac{Tcos\theta }{m}=\frac{Fcos\theta }{2msin\theta }$

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