JEE Main 2003 — Gravitation Question with Solution

Let $$t$$ be the time taken for the two masses to collide and $$x_{5M,}{x}_M$$ be the distance travelled by the mass $$5M$$ and $$M$$ respectively.

The gravitational force acting between two sphere when the distance between them (12R - x) where x is a variable,

$$F=\frac{GM\times 5M}{{\left(12R-x\right)}^2}$$

Acceleration of mass M, $$a_M=\frac{G\times 5M}{{\left(12R-x\right)}^2}$$

Acceleration of mass 5M, $$a_{5M}=\frac{GM}{{\left(12R-x\right)}^2}$$

For mass $$5M$$

$$u=0,S=x_{5M},t=t,a=a_{5M}$$

$$S=ut+\frac{1}{2}a{t}^2$$

$$\therefore$$ $$x_{5M}=\frac{1}{2}{a}_{5M}{t}^2...\left(i\right)$$

For mass $$M$$

$$u=0,s=x_M,t=t,a=a_M$$

$$\therefore$$ $$s=ut+\frac{1}{2}a{t}^2\Rightarrow$$

$$x_M=\frac{1}{2}{a}_M{t}^2...\left(ii\right)$$

Dividing $$(ii)$$ by $$(iii)$$

$$\frac{x_{5M}}{x_M}=\frac{\frac{1}{2}{a_5}_M{t}^2}{\frac{1}{2}{a}_M{t}^2}$$

$$=\frac{a_{5M}}{a_M}=\frac{1}{5}$$

$$\therefore$$ $$5x_{5M}=x_M...(iii)$$

From the figure,

$$x_{5M}+x_M=12R-2R-R=9R...\left(iv\right)$$

From $$(iii)$$ and $$(iv)$$

$$\frac{x_M}{5}+x_M=9R$$

$$\therefore$$ $$6x_M=45R$$

$$\therefore$$ $$x_M=\frac{45}{6}R=7.5R$$

So two sphere collide when the sphere of mass M covered the distance of 7.5R.