(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