TS EAMCET 9 May 2024 Shift 1 Solved Paper

Original system (with 4 particles)
Let coordinates of the particle as
‌A=(0,a)
‌B=(a,a)
‌C=(a,0)
‌D=(0,0)
The centre of mass of system
X‌com ‌‌=‌

mxA+mxB+mxC+mxD

4m

Y‌com ‌‌=‌

myA+myB+myC+myD

4m

Solving,
‌Xcom=‌

0+a+a+0

4

=‌

a

2

‌Ycom=‌

a+a+0+0

4

=‌

a

2

The centre of mass of original system at (‌

a

2

,‌

a

2

)
New centre of mass, when one particle (B) is removed,Xcom′=‌

m⋅xA+m⋅xC+m⋅xD

3m

Ycom′=‌

m⋅yA+m⋅yC+m⋅yD

3m

Xcon′=‌

0+a+a

3

=‌

2a

3

Ycom′=‌

a+0+0

3

=‌

a

3

Now, shift in position
√(‌

2a

3

−‌

a

2

)2+(‌

a

3

−‌

a

2

)2=‌

a

3√2

Thus, centre of mass of 3 system shift to ‌

a

3√2