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Q.

Two particles of equal mass m are lying on the x-axis at (–a, 0) and (+a,0) respectively as shown in the adjoining figure. They are connected by a light string. A force F is applied at the origin and along the y-axis. As a result the masses move towards each other. What is the acceleration of each mass ? Assume the instantaneous position of masses as (–x, 0) and (x, 0) respectively:

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a

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

b

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

c

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

d

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

answer is B.

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Detailed Solution

Question Image

2Tsinθ=F

T= $\frac{F}{2 sinθ}$ ,

Component of tension force along x causes acceleration of the ball.

a = $\frac{Tcosθ}{m} = \frac{Fcosθ}{2 msinθ}$

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

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Two particles of equal mass m are lying on the x-axis at (–a, 0) and (+a,0) respectively as shown in the adjoining figure. They are connected by a light string. A force F is applied at the origin and along the y-axis. As a result the masses move towards each other. What is the acceleration of each mass ? Assume the instantaneous position of masses as (–x, 0) and (x, 0) respectively: