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


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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

$Question is exactly the same as from$ $IIT - JEE 2007 .$
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