Q.
Two identical small masses each of mass m are connected by a light inextensible string on a smooth horizontal floor. A constant force F is applied at the mid point of the string as shown in the figure. The acceleration of each mass towards each other is,
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a
F23m
b
3F2m
c
23Fm
d
None of these
answer is A.
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Detailed Solution
Let the tension in the string be T at any angular position θ, the acceleration of each ball along x and y axes be a and a1 respectively. Writing the equation of motion of m, we obtain ∑Fx=ma⇒ Tcosθ=ma ......(i)∑Fy=ma1⇒ Tsinθ=ma1 .......(ii)At point P, as it is accelerating with an acceleration a, therefore F−2Tcosθ=mpa where mp= mass of the string at the point P≅0⇒ F=2Tsosθ ........(iii)(2)÷(1)⇒ tanθ=a1a⇒ a1=atanθ where a=Tcosθm from (i). Putting T=F2cosθ, from (iii), we obtain a1=F2mtanθ Putting θ=30∘⇒a1=F23m
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