IIT-JEE 2007 · Paper 1 · Physics Q3

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PhysicsSingle correct+3 / -1Question 3 of 66

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

Two masses m on a horizontal line, each at distance a from the midpoint P of the string joining them; a vertical force F acts upward at P.

\dfrac{F}{2m}\dfrac{a}{\sqrt{a^2-x^2}}

\dfrac{F}{2m}\dfrac{x}{\sqrt{a^2-x^2}}

\dfrac{F}{2m}\dfrac xa

\dfrac{F}{2m}\dfrac{\sqrt{a^2-x^2}}{x}

Answer and worked solution

Official answer: B

Geometry. With the masses $x$ from the centre and each half of the string of length $a$ , the string makes an angle θ with the vertical where $\sin\theta=\frac xa$ , $\cos\theta=\frac{\sqrt{a^2-x^2}}a$ .

At the midpoint (massless): $2T\cos\theta=F$ , so $T=\dfrac F{2\cos\theta}$ .

Each mass feels the horizontal component $T\sin\theta$ (the vertical part is balanced by the normal reaction for a small F):
$a_m=\frac{T\sin\theta}m=\frac F{2m}\tan\theta=\frac F{2m}\frac x{\sqrt{a^2-x^2}}.$

(B)

Topics:Newton's laws Free body diagrams

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IIT-JEE 2007 · Paper 1, Physics question 3