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-dis

Question Number 21150 by Tinkutara last updated on 14/Sep/17

Two particles of mass m each are tied

at the ends of a light string of length 2 a .

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 2 x, is

Commented by Tinkutara last updated on 14/Sep/17

Commented by alex041103 last updated on 15/Sep/17

{I s n}^{'} t i t

\frac{F}{2 m} \frac{x}{\sqrt{a^{2} - x^{2}}} = a

Commented by ajfour last updated on 15/Sep/17

Commented by ajfour last updated on 15/Sep/17

T = m a

2 T \cos θ = F

\Rightarrow a = \frac{F}{2 m \cos θ} = \frac{F}{2 m} . \frac{a}{\sqrt{a^{2} - x^{2}}}

l e t / m a g n i t u d e o f a c c e l e r a t i o n o f

e a c h t o w a r d s e a c h o t h e r b e a_{x},

t h e n {m a}_{x} = T \sin θ

\Rightarrow a_{x} = \frac{F}{2 m \cos θ} \times \sin θ

= \frac{F}{2 m} . \frac{x}{\sqrt{a^{2} - x^{2}}} .

Commented by Tinkutara last updated on 15/Sep/17

{alex}^{'} s answer is correct . Explain please .

Commented by ajfour last updated on 15/Sep/17

y o u p l e a s e c h e c k f o r w h a t e x a c t l y

i s a s k e d . .

Commented by alex041103 last updated on 15/Sep/17

M y s o l u t i o n i s a n a l o g i c a l t o m r {a j f o u r}^{'} s

Commented by Tinkutara last updated on 15/Sep/17

Question is exactly the same as from

IIT - JEE 2007 .