Laws Of Motion Ques 55
- 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, 3M)
(a) $\frac{F}{2 m} \frac{a}{\sqrt{a^{2}-x^{2}}}$
(b) $\frac{F}{2 m} \frac{x}{\sqrt{a^{2}-x^{2}}}$
(c) $\frac{F}{2 m} \frac{x}{a}$
(d) $\frac{F}{2 m} \frac{\sqrt{a^{2}-x^{2}}}{x}$
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Answer:
Correct Answer: 55.(b)
Solution:
Formula:
- As, from $\mathrm{FBD}$ (Free Body Diagram), $2 T \cos \theta=F$
So,
$$ T=\frac{F}{2} \sec \theta $$
Acceleration of particle
$$ \begin{aligned} & =\frac{T \sin \theta}{m}=\frac{F \tan \theta}{2 m} \\ & =\frac{F}{2 m} \frac{x}{\sqrt{a^{2}-x^{2}}} \end{aligned} $$
BETA
