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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 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

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{each}\:\mathrm{are}\:\mathrm{tied} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{ends}\:\mathrm{of}\:\mathrm{a}\:\mathrm{light}\:\mathrm{string}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2}{a}. \\ $$$$\mathrm{The}\:\mathrm{whole}\:\mathrm{system}\:\mathrm{is}\:\mathrm{kept}\:\mathrm{on}\:\mathrm{a}\:\mathrm{frictionless} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{the}\:\mathrm{string}\:\mathrm{held} \\ $$$$\mathrm{tight}\:\mathrm{so}\:\mathrm{that}\:\mathrm{each}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance} \\ $$$$'{a}'\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\:{P}\:\left(\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{figure}\right).\:\mathrm{Now},\:\mathrm{the}\:\mathrm{mid}-\mathrm{point}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{string}\:\mathrm{is}\:\mathrm{pulled}\:\mathrm{vertically}\:\mathrm{upwards}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{small}\:\mathrm{but}\:\mathrm{constant}\:\mathrm{force}\:{F}.\:\mathrm{As}\:\mathrm{a}\:\mathrm{result}, \\ $$$$\mathrm{the}\:\mathrm{particles}\:\mathrm{move}\:\mathrm{towards}\:\mathrm{each}\:\mathrm{other} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{surface}.\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of} \\ $$$$\mathrm{acceleration},\:\mathrm{when}\:\mathrm{the}\:\mathrm{separation} \\ $$$$\mathrm{between}\:\mathrm{them}\:\mathrm{becomes}\:\mathrm{2}{x},\:\mathrm{is} \\ $$

Commented by Tinkutara last updated on 14/Sep/17

Commented by alex041103 last updated on 15/Sep/17

Isn′t it (F/(2m)) (x/(√(a^2 −x^2 ))) = a

$${Isn}'{t}\:{it}\: \\ $$$$\frac{{F}}{\mathrm{2}{m}}\:\frac{{x}}{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:=\:{a} \\ $$

Commented by ajfour last updated on 15/Sep/17

Commented by ajfour last updated on 15/Sep/17

T=ma 2Tcos θ=F ⇒ a=(F/(2mcos θ)) =(F/(2m)).(a/(√(a^2 −x^2 ))) let/magnitude of acceleration of each towards each other be a_x , then ma_x =Tsin θ ⇒ a_x =(F/(2mcos θ))×sin θ =(F/(2m)).(x/(√(a^2 −x^2 ))) .

$${T}={ma} \\ $$$$\mathrm{2}{T}\mathrm{cos}\:\theta={F} \\ $$$$\Rightarrow\:\:\:\:\:{a}=\frac{{F}}{\mathrm{2}{m}\mathrm{cos}\:\theta}\:=\frac{{F}}{\mathrm{2}{m}}.\frac{{a}}{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\: \\ $$$${let}/{magnitude}\:{of}\:{acceleration}\:{of} \\ $$$${each}\:{towards}\:{each}\:{other}\:{be}\:{a}_{{x}} \:, \\ $$$${then}\:\:\:{ma}_{{x}} ={T}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\:\:{a}_{{x}} =\frac{{F}}{\mathrm{2}{m}\mathrm{cos}\:\theta}×\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{F}}{\mathrm{2}{m}}.\frac{{x}}{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:. \\ $$

Commented by Tinkutara last updated on 15/Sep/17

alex′s answer is correct. Explain please.

$$\mathrm{alex}'\mathrm{s}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}.\:\mathrm{Explain}\:\mathrm{please}. \\ $$

Commented by ajfour last updated on 15/Sep/17

you please check for what exactly is asked..

$${you}\:{please}\:{check}\:{for}\:{what}\:{exactly} \\ $$$${is}\:{asked}.. \\ $$

Commented by alex041103 last updated on 15/Sep/17

My solution is analogical to mr ajfour′s

$${My}\:{solution}\:{is}\:{analogical}\:{to}\:{mr}\:{ajfour}'{s} \\ $$

Commented by Tinkutara last updated on 15/Sep/17

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

$$\mathrm{Question}\:\mathrm{is}\:\mathrm{exactly}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{from} \\ $$$$\mathrm{IIT}-\mathrm{JEE}\:\mathrm{2007}. \\ $$

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