Question

Two spherical bodies of mass M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is [AIPMT-2015] (1) 1.5R (2) 2.5R (3) 4.5R (4) 7.5R

Answer 1

Answer:

7.5R

Explanation:

force experienced by 2 spheres will be the same that is equal to F=GM5M÷144R×R

Answer 2

Question:-

Two spherical bodies of masses $\sf{M}$ and $\sf{5M}$ and radii $\sf{R}$ and $\sf{2R}$ are released in free space with initial separation between their centers equal to $\sf{12R.}$ If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is,

$\sf{(1)\quad\!1.5R\quad\quad(2)\quad\!2.5R\quad\quad(3)\quad\!4.5R\quad\quad(4)\quad\!7.5R}$

Answer:-

$\Large\boxed{\quad\sf{(4)\quad\!7.5R}\quad}$

Solution:-

Distance between the surfaces of the spherical bodies is,

$\longrightarrow\sf{12R-(R+2R)=12R-3R=9R}$

Let the smaller mass move through a distance $\sf{x}$ before collision so the larger mass move by a distance $\sf{9R-x.}$

Consider the smaller spherical body of mass M.

• Initial velocity, $\sf{u=0}$

• Displacement, $\sf{s=x}$

Let its acceleration be $\sf{a_M.}$ So by second equation of motion,

$\longrightarrow\sf{s=ut+\dfrac{1}{2}at^2}$

$\longrightarrow\sf{x=\dfrac{1}{2}a_Mt^2\quad\quad\dots(1)}$

Consider the larger spherical body of mass 5M.

• Initial velocity, $\sf{u=0}$

• Displacement, $\sf{s=9R-x}$

Let its acceleration be $\sf{a_{5M}.}$ So by second equation of motion,

$\longrightarrow\sf{s=ut+\dfrac{1}{2}at^2}$

$\longrightarrow\sf{9R-x=\dfrac{1}{2}a_{5M}t^2\quad\quad\dots(2)}$

On dividing (2) by (1),

$\longrightarrow\sf{\dfrac{9R-x}{x}=\dfrac{a_{5M}}{a_M}\quad\quad\dots(3)}$

But, since the two spheres are under same gravitational force of attraction F (say),

• $\sf{a_M=\dfrac{F}{M}\quad\quad\dots(4)}$

• $\sf{a_{5M}=\dfrac{F}{5M}\quad\quad\dots(5)}$

On dividing (5) by (4) we get,

$\longrightarrow\sf{\dfrac{a_{5M}}{a_M}=\dfrac{\left(\dfrac{F}{5M}\right)}{\left(\dfrac{F}{M}\right)}}$

$\longrightarrow\sf{\dfrac{a_{5M}}{a_M}=\dfrac{1}{5}}$

Then (3) becomes,

$\longrightarrow\sf{\dfrac{9R-x}{x}=\dfrac{1}{5}}$

$\longrightarrow\sf{\dfrac{9R-x}{x}=\dfrac{6-5}{5}}$

By rule of dividendo this implies,

$\longrightarrow\sf{\dfrac{x}{9R}=\dfrac{5}{6}}$

$\longrightarrow\sf{x=9R\cdot\dfrac{5}{6}}$

$\longrightarrow\sf{\underline{\underline{x=7.5R}}}$

Hence $\sf{(4)\quad\!7.5R}$ is the answer.