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Two identical particles, each of mass, m initially located at the points (–a, 0) and (a, 0), are joined to each other by a massless string of length 2a. If a force F (= -Fj), acts continuously on the midpoint of this string, the relative (horizontal) acceleration of the two masses, when the (horizontal) distance between then becomes 4a/3 , is

(1) \(\frac{0.4\sqrt{5}F}{m}\)

(2) \(\frac{0.2\sqrt{5}F}{m}\)

(3) \(\frac{0.25\sqrt{5}F}{m}\)

(4) \(\frac{0.5\sqrt{8}F}{m}\)

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 (1) \(\frac{0.4\sqrt{5}F}{m}\)

The ‘set-up’ would be of the form shown here when the horizontal distance between the particles becomes (4a/3).

If T is the tension in each string at this instant, we would have

T cosθ + T cosθ = F

or T = F/2cosθ

The horizontal force on each particle in T sin θ . Hence the horizontal acceleration, of each particle, is \(\frac{Tsin\theta}{m}.\)

The particles are approaching each other (The particle, on the left of the origin, experiences a horizontal force, directed towards right due to the tensino in the ‘right side’ part of the string and vice-versa). 

Hence,

their relative (horizontal) acceleration is \(\frac{2Tsin\theta}{m}\)

∴ Relative (horizontal) acceleration of the two masses when the distance between them becomes 2a/3 , is

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