Two spherical bodies of mass M and 5 M and radii R and 2R repectively are released in free space with initial seperation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by smaller body just before collision is .........

Options

A. 2.5R
B. 4.5 R
C. 7.5R
D. 1.5R

Answer

C

Explanation

`implies` Acceleration of sphere of mass M
`a_1 = F/M = (GM (5M))/(M(12R)^2) = (5GM)/((12R)^2)`
Acceleration of shpere of mass 5 M
`a_2 = F/(5M ) = (GM(5M))/((12R)^2xx5M) = (GM)/((12R)^2)`
`:.a_1/a_2 = 5`
`:.a_1 = 5a_2 " "....(1)`
`implies` In equation of motion `d=v_0 t +1/2 "at"^2 , v_0 = 0 `
`:. d = 1/2 "at"^2` , since t is constant
`d prop a `
`:. d_1/d_2 =a_1/a_2 = (5a_1)/(a_2) = 5`
`:. d_1 =5d_2`
`:.` Distance covered by a sphere of radius R fore collision is 5 times to the distance covered by a sphere of radius 2R.
`implies` When both spheres collide, the distance covered
`= d + 5d`
`:. 6d = 12 R - 3R`
`:. 6d = 9R`
`d 1.5 R`
`:.` Total distacne covered by sphere of radius
`R = 1.5R xx 5`
= 7.5 R