The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position (xa/3π, xa/3π) where x is---------. (Round off to the Nearest integer) [a is an area as shown in the figure]

 

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

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

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

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.  

To study centre of mass, you need to follow these steps. 

• Learn the definition with examples. Most importantly, focus on visualising it. 
• Then approach the centre of mass formulas for discrete particles and continuous bodies.
• Practice finding the centre of mass for symmetrical and asymmetrical bodies. 
• Solve problems on the motion of centre of mass and its applications in collisions of particles.