Two blocks of masses 10kg and 30kg are placed on the same straight...

Initial position of centre of mass: ${x}_{CM} = \dfrac{10\left(0\right) + 30 \times x}{40} = \dfrac{3x}{4}$

Final position of the centre of mass: ${x}^{'}CM = \dfrac{10\left(6\right) + 30 \times \left(x - y\right)}{40} = \dfrac{6 + 3\left(x - y\right)}{4}$

As position of centre of mass remains same,

${x}_{CM} = {x}^{'}CM ⇒ \dfrac{3x}{4} = \dfrac{6 + 3\left(x - y\right)}{4} ⇒ 3x = 6 + 3x - 3y$ $⇒ y = 2cm$

Alternate method, $∆ {x}_{COM} = \dfrac{{m}_{1} ∆ {x}_{1} + {m}_{2} ∆ {x}_{2}}{\left({m}_{1} + {m}_{2}\right)} ⇒ 0 = 10 \times 6 + 30y ⇒ y = - 2cm$