Question

Two particles of mass m each are tied at the ends of a light
string of length 2a. The whole system is kept on a frictionless
horizontal surface with the string held tight so that each mass
is at a distance 'a' from the center

P (as shown in the figure). Now,
the mid-point of the string is
pulled vertically upwards with a
small but constant force F. As a
result, the particles move towards
each other on the surface. The
magnitude of acceleration, when the separation between
them becomes 2x, is
​

Answer 1

Answer:

Fx / 2m√(a2–x2)

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Explanation:

As shown in the figure,

T = tension of the string

a = Acceleration of the each particles.

Solution:

Here,

i) F = 2T sinθ

=> T = F / 2 sinθ

ii) sinθ = √(a2–x2)

iii) cosθ = x

The horizontal force on each particle is

T cosθ = Fcosθ / 2sinθ = F / 2tanθ

Therefore, the magnitude of acceleration of each particle is given by ,

Acceleration = Force / Mass

= F / ( 2m tanθ )

= F / [ 2m√(a2–x2) / x ]

= Fx / 2m√(a2–x2) [ Answer ]

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Note:

tanθ

= sinθ / cosθ

= √(a2–x2) / x

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