Did you believe it's from NTSE #2

Classical Mechanics Level 3

Two point masses A and B having masses in the ratio 4:3 are separated by a distance of 1 m . when another point mass C of mass M is placed in between A and B,the force between A and C IS one-third of the force between B and C.then the distance of C from A (in meters) is of the form P Q \frac{P}{Q} Find P+Q

h i n t : a p p l y g r a v i t a t i o n a l l a w hint: apply \ gravitational \ law


The answer is 5.

Harshi Singh by Harshi Singh , 20, India

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1 solution

Given,

Mass of A= M A M_A

Mass of B= M B M_B

And, M A M B = 4 3 \frac{M_A}{M_B}=\frac{4}{3} o r , 3 M A = 4 M B or, 3M_A=4M_B

Mass of C= M M

Distance, d = 1 m d=1m

Assume,

Distance between A and C= x x

Then, Distance between B and C= d − x d-x

By Newton's awesome Law of Gravitation, G M A M x 2 = 1 3 G M B M ( d − x ) 2 \frac{GM_A M}{x^2}=\frac{1}{3}\frac{GM_B M}{(d-x)^2} o r , 3 M A x 2 = M B ( d − x ) 2 or, \frac{3M_A}{x^2}=\frac{M_B}{(d-x)^2} o r , 3 M A x = M B d − x or, \frac{\sqrt{3M_A}}{x}=\frac{M_B}{d-x} o r , x = 3 M A 3 M A + M B . d or, x=\frac{\sqrt{3M_A}}{\sqrt{3M_A}+\sqrt{M_B}}.d = 4 M B 4 M B + M B . d =\frac{\sqrt{4M_B}}{\sqrt{4M_B}+\sqrt{M_B}}.d = 2 M B 3 M B . d =\frac{2\sqrt{M_B}}{3\sqrt{M_B}}.d = 2 3 × 1 =\frac{2}{3}\times 1 So, A B = 2 3 \frac{A}{B}=\frac{2}{3} Finally, A + B = 2 + 3 = 5 A+B=2+3=\boxed{5}

Done!

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