question_answer

Two identical small masses each of mass m are connected by a light inextensible string on a smooth m horizontal floor. A constant force F is applied t at the mid point of the string as shown in the figure. The acceleration of each mass 1 towards each other is,

A) \[\frac{F}{2\sqrt{3m}}\]

B) \[\frac{\sqrt{3}F}{2m}\]

C) \[\frac{2}{\sqrt{3}}\,\frac{F}{m}\]

D) None of these            

Correct Answer: A

Solution :

[a] Let the tension in the string be T at nay angular position\[\theta \], the acceleration of each ball along x and y axes be a and \[{{a}_{1}}\]respectively. Writing the equation of motion of m, we obtain  \[\Sigma {{F}_{x}}=ma\] \[\Rightarrow T\cos \theta =ma...(i)\] \[\Sigma {{F}_{y}}=m{{a}_{1}}\] \[\Rightarrow T\sin \theta =m{{a}_{1}}...(ii)\] At point P, as it is accelerating with an acceleration a, therefore \[F-2Tcos\theta ={{m}_{p}}\] a wgere \[{{m}_{p}}=\] mass of the string at the point \[P\cong 0\] \[\Rightarrow F=2T\cos \theta ...(iii)\] (ii)\[\div \](i) \[\Rightarrow \tan \theta =\frac{{{a}_{1}}}{a}\] \[\Rightarrow {{a}_{1}}=a\tan \theta \] Where \[a=\frac{T\cos \theta }{m}\]from (iii), we obtain \[{{a}_{1}}=\frac{F}{2m}\tan \theta \] Putting\[\theta \]=\[30{}^\circ \]\[\Rightarrow {{a}_{1}}=\frac{F}{2\sqrt{3m}}\]