Question

if the certre of mass of three particles of mass 40 g , 20g,30g is at the point (1,-2,3) then where should a fourth particle of mass 80g be placed , so the combined certre of mass of the system is at (1,1,1)?​

Answer 1

Let,

• the position of 40 g mass = $\sf{(x_1,\ y_1,\ z_1)}$

• the position of 20 g mass = $\sf{(x_2,\ y_2,\ z_2)}$

• the position of 30 g mass = $\sf{(x_3,\ y_3,\ z_3)}$

• the position of 80 g mass (to find out) = $\sf{(x_4,\ y_4,\ z_4)}$

The x coordinate of the center of mass,

(i) before adding 80 g mass,

$\longrightarrow\sf{\overline{x}=1}\\\\\\\longrightarrow\sf{\dfrac {40x_1+20x_2+30x_3}{40+20+30}=1}\\\\\\\longrightarrow\sf{40x_1+20x_2+30x_3=90}\quad\quad\dots(1)$

(ii) after adding 80 g mass,

$\longrightarrow\sf{\overline{x}'=1}\\\\\\\longrightarrow\sf{\dfrac{40x_1+20x_2+30x_3+80x_4}{40+20+30+80}=1}\\\\\\\longrightarrow \sf{40x_1+20x_2+30x_3+80x_4=170}\\\\\\\longrightarrow \sf{90+80x_4=170\quad\quad\Longleftarrow(1)}\\\\\\\longrightarrow \sf{\underline {\underline {x_4=1}}}$

The y coordinate of the center of mass,

(i) before adding 80 g mass,

$\longrightarrow\sf{\overline{y}=-2}\\\\\\\longrightarrow\sf{\dfrac {40y_1+20y_2+30y_3}{40+20+30}=-2}\\\\\\\longrightarrow\sf{40y_1+20y_2+30y_3=-180}\quad\quad\dots(2)$

(ii) after adding 80 g mass,

$\longrightarrow\sf{\overline{y}'=1}\\\\\\\longrightarrow\sf{\dfrac{40y_1+20y_2+30y_3+80y_4}{40+20+30+80}=1}\\\\\\\longrightarrow \sf{40y_1+20y_2+30y_3+80y_4=170}\\\\\\\longrightarrow \sf{-180+80y_4=170\quad\quad\Longleftarrow(2)}\\\\\\\longrightarrow \sf{\underline {\underline {y_4=4\ \dfrac{3}{8}}}}$

The z coordinate of the center of mass,

(i) before adding 80 g mass,

$\longrightarrow\sf{\overline{z}=3}\\\\\\\longrightarrow\sf{\dfrac {40z_1+20z_2+30z_3}{40+20+30}=3}\\\\\\\longrightarrow\sf{40z_1+20z_2+30z_3=270}\quad\quad\dots(3)$

(ii) after adding 80 g mass,

$\longrightarrow\sf{\overline{z}'=1}\\\\\\\longrightarrow\sf{\dfrac{40z_1+20z_2+30z_3+80z_4}{40+20+30+80}=1}\\\\\\\longrightarrow \sf{40z_1+20z_2+30z_3+80z_4=170}\\\\\\\longrightarrow \sf{270+80z_4=170\quad\quad\Longleftarrow(3)}\\\\\\\longrightarrow \sf{\underline {\underline {z_4=-1\dfrac {1}{4}}}}$

Thus, the position of 80 g mass is,

$\sf{\underline {\underline{(x_4,\ y_4,\ z_4)=\left (1,\ 4\ \dfrac{3}{8},\ -1\ \dfrac{1}{4}\right)}}}$