Q.
Three point particles of masses $1.0\, kg, 1.5\, kg$ and $2.5\, kg$ are placed at three corners of a right angle triangle of sides $4.0\, cm$, $3.0\, cm$ and $5.0\, cm$ as shown in the figure. The center of mass of the system is at a point:
Solution:
Take 1kg mass at origin
$X_{cm}=\frac{1\times0+1.5\times3+2.5\times0}{5}=0.9cm$
$Y_{cm}=\frac{1\times0+1.5\times0+2.5\times4}{5}=2cm$
