Question

Three point particles of masses $1.0kg$, $1.5kg$ and $2.5kg$ are placed at three corners of a right angle triangle of sides $4.0cm,3.0cm$ and $5.0cm$ as shown in the figure. The centre of mass of the system is at a point :

• A. $1.5cm$ right and $1.2cm$ above $1kg$ mass
• B. $2.0cm$ right and $0.9cm$ above $1kg$ mass
• C. $0.6cm$ right and $2.0cm$ above $1kg$ mass
• D. $0.9cm$ right and $2.0cm$ above $1kg$ mass

Answer 1

The correct option is D $0.9cm$ right and $2.0cm$ above $1kg$ mass

Refer image.

From the question,

we have $(0,0),(3,0)$ and $(0,4)$ to be the co-ordinates of mass $1kg,1.5kg$ and $2.5kg$ (assuming $1kg$ at origin)

$X_{com}$ is given by $\frac{\sum mixi}{\sum mi}=\frac{1\times 0+1.5\times 3+2.5\times 0}{1+1.5+2.5}$

$=\frac{4.5}{5}=0.9cm$

$Y_{com}$ is given by $\frac{\sum miyi}{\sum mi}=\frac{1\times 0+1.5\times 0+2.5\times 4}{1+1.5+2.5}$

$=\frac{2.5\times 4}{5}=\frac{4}{2}=2cm$

$\therefore$ The centre of mass is $0.9$ cm to the right and $2$ cm above the mass of $1kg$

option (D) is correct