Question

A small sphere of radius R is held against the inner surface of a larger sphere of radius 6 R. The masses of large and small spheres are 4 M and M respectively. This arrangement is placed on a horizontal table as shown. There is no friction between any surfaces of contact. The small sphere is now released. The coordinates of the centre of the large sphere when the smaller sphere reaches the other extreme position is

• A. (L - 2R, 0)
• B. (L + 2R, 0)
• C. (2R, 0)
• D. (2R - L, 0)

Answer 1

The correct option is B (L + 2R, 0)

In the initial position,the x-coordinate of CM is $x_i=\frac{m_1{x}_2+m_2{x}_2}{m_1+m_2}$$=\frac{4M\times L+M(L+5R)}{4M+M}=L+R$
In the final position,the x-coordinate of CM is $x_f=\frac{4M\times x+M(x-5R)}{4M+M}=x-R$
since $x_i=x_f$
$L+R=x-R$
$\therefore$ $x=L+2R$
so,coordinates of CM is ($L+2R,0$).