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 and along the Y-axis. As a result, the masses move towards each other What is the acceleration of each mass ? Assume the instantaneous position of the masses as (- x,0) and {x, 0) respectively
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a
Fm⋅xa2−x2
b
F2m⋅xa2−x2
c
2Fm⋅xa2−x2
d
2Fm⋅a2−x2x
answer is B.
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Detailed Solution
See figure.From figure F=2Tcosθ or T=F/(2cosθ)The force responsible for motion of masses on X-axis is Tsinθ∴ ma=Tsinθ=F2cosθ×sinθ =F2tanθ=F2×OBOA=F2×x(a2−x2)So, a=F2m×x(a2−x2).