Question 283: Laws of Motion

Question

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' form 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, with the separation between them becomes 2x, is

Accepted Answer

The acceleration of mass 'm' is due to the force $$T\cos 	heta $$, from 2
nd
law of motion we have,
$$T\cos 	heta  = ma$$
$$ \Rightarrow a = {{T\cos 	heta } \over m}$$ ..... (i)
and also, $$F = 2T\sin 	heta $$
$$ \Rightarrow T = {F \over {2\sin 	heta }}$$ ..... (ii)
From (i) & (ii), we have,
$$a = \left( {{F \over {2\sin 	heta }}} ight)\,.\,{{\cos 	heta } \over m}$$
$$a = {F \over {2m	an 	heta }}$$
$$	herefore$$ $$a = {F \over {2m{{\sqrt {{a^2} - {x^2}} } \over x}}}$$
$$	herefore$$ $$a = {F \over {2m}}\,.\,{x \over {\sqrt {{a^2} - {x^2}} }}$$

Answer

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

Answer

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

Answer

$${F \over {2m}}{x \over a}$$

Answer

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

Laws of Motion

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