Two particles of mass each are tied at the ends of a light string of length . The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance from the centre P (as shown in the figure). [200

Question

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 centre 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 [2007]

Smart Achievers · Accepted Answer

(2)

$From figure, acceleration of mass m is due to the force T \cos θ$

$∴$ $T \cos θ = m a$

$\Rightarrow a = \frac{T \cos θ}{m} ...(i)$

$Also, F = 2 T \sin θ \Rightarrow T = \frac{F}{2 \sin θ}$

$Putting this value of T in eqn. (i)$

$a = (\frac{F}{2 \sin θ}) \frac{\cos θ}{m}$

$= \frac{F}{2 m \tan θ}$ $= \frac{F}{2 m} \cdot \frac{x}{\sqrt{a^{2} - x^{2}}}$ $[∵ \tan θ = \frac{\sqrt{a^{2} - x^{2}}}{x}]$

Solution

1 $\frac{F}{2 m} \frac{a}{\sqrt{a^{2} - x^{2}}}$

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

3 $\frac{F}{2 m} \frac{x}{a}$

4 $\frac{F}{2 m} \frac{\sqrt{a^{2} - x^{2}}}{x}$

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Solution

1 F2m aa2-x2

2 F2m xa2-x2

3 F2m xa

4 F2m a2-x2x

Ans. (2) From figure, acceleration of mass m is due to the force Tcosθ ∴ Tcosθ=ma ⇒ a=Tcosθm ...(i) Also, F=2Tsinθ ⇒ T=F2sinθ Putting this value of T in eqn. (i) a=(F2sinθ)cosθm =F2mtanθ=F2m·xa2-x2 [∵tanθ=a2-x2x]

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1 $\frac{F}{2 m} \frac{a}{\sqrt{a^{2} - x^{2}}}$

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

3 $\frac{F}{2 m} \frac{x}{a}$

4 $\frac{F}{2 m} \frac{\sqrt{a^{2} - x^{2}}}{x}$