Is centre of mass important for JEE? 

The x and y coordinates of the center of mass are given by:
x? = y? = 4a / 3π

$\left(\mathrm{0,3}\right)$

$\begin{array}{rr}& {\stackrel{?}{r}}_{\mathrm{c}\mathrm{m}}=\frac{1\times \left(\frac{\hat{i}}{2}+\hat{j}\right)+1\times \left(\hat{i}+\frac{5\hat{j}}{2}\right)}{2}\\ & {\stackrel{?}{r}}_{\mathrm{c}\mathrm{m}}=\frac{3}{4}\hat{i}+\frac{7}{4}\hat{j}\end{array}$

$\left(\mathrm{0,0}\right)$

From conservation of momentum:
2×4 = 2v? + mv?
Given v? = 1 m/s (interpreted from intermediate steps)
8 = 2 (1) + mv?
mv? = 6 . (i)
From coefficient of restitution (e=1 for elastic collision):
e = (v? - v? )/ (u? - u? )
1 = (v? - v? )/ (4 - 0)
-1 = (v? - v? )/ (0 - 4)  (as written in the image)
⇒ 4 = v? - 1
⇒ v? = 5 . (ii)
Put (2) in (1), m (5) = 6
m = 1.2kg

$X_c=Y_c=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=\frac{m\times 0+m\times 0+m\times 3}{3m}=1m$

$OC=X\sqrt[]{x_0^2+Y_c^2}=\sqrt[]{{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt[]{2}m$

You should know that when the shift in the centre of mass occurs, you can use the principle of moments. With this logic, we can see the shift from the original body as the combination of two parts. The one that remains part and the other that is removed. The formula for the shift is the product of the mass removed and its distance of its centre of mass from the original centre of mass, which is divided by the mass of the remaining part.