Question

Two spheres of radii R and 2R of the same material are separated by a distance of 10R
between there centers. Under mutual force of attraction, they are moving towards each other.
Find the distance moved by smaller sphere before the two spheres meet together.
(A) 50R/9
(B) 56R/9
(C) 65R / 9
(D) 30R/9​

Answer 1

Answer: OPTION B

THEY WILL MEET AT CENTRE OF MASS

LET' d ' be the density of spheres,x1 be the distance travelled by the smaller sphere and x2 be the distance travelled by the larger sphere

and they are separated by the distance of 7R

x2= 7R- x1

m1*x1=m2*x2          

d*4/3*3.14*R^3 *x1= D*4/3*3.14*8*R^3 *(7R -x1)

BY SOLVING THIS

x1=56R/9

Explanation:

Answer 2

Answer:

Now,

Here the net force is zero and also the displacement will also be zero.

So we have

$m_{1}\delta x_{1} + m_{2}\delta x_{2} = 0$

$m_{1}\delta x_{1} - m_{1}\delta x_{2} = 0$

$\delta x_{1} - \delta x_{2} = 0$  ---------------(i)

and also,

$\delta x_{1} + \delta x_{2} = 7R$ ---------------(ii)

$\delta x_{1} - \delta x_{2} = 0$ -------------------(iii)

Adding equation (iii) and (ii) we get

=> $2\delta x_{1} = 7R$

=> $\delta x_{1} = \dfrac{7R}{2}$

=>$\delta x_{1} = \dfrac{7R}{2}$

=> $\delta x_{1} = 3.5R$

Hence the distance covered by the smaller body just before the Collison will be 3.5 R ≈ 30/9 R

So I guess it will be option (D)

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