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:

Question

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: (A) 2.0 cm right and 0.9 cm above 1 kg mass (B) 0.9 cm right and 2.0 cm above 1 kg mass (C) 0.6 cm right and 2.0 cm above 1 kg mass (D) 1.5 cm right and 1.2 cm above 1 kg mass

Tardigrade Admin · Accepted Answer

Take 1kg mass at origin Xcm=(1×0+1.5×3+2.5×0/5)=0.9cm Ycm=(1×0+1.5×0+2.5×4/5)=2cm

Q. Three point particles of masses $1.0 k g, 1.5 k g$ and $2.5 k g$ are placed at three corners of a right angle triangle of sides $4.0 c m$ , $3.0 c m$ and $5.0 c m$ as shown in the figure. The center of mass of the system is at a point:

2.0 cm right and 0.9 cm above 1 kg mass

0.9 cm right and 2.0 cm above 1 kg mass

0.6 cm right and 2.0 cm above 1 kg mass

1.5 cm right and 1.2 cm above 1 kg mass

Take 1kg mass at origin
$X_{c m} = \frac{1 \times 0 + 1.5 \times 3 + 2.5 \times 0}{5} = 0.9 c m$
$Y_{c m} = \frac{1 \times 0 + 1.5 \times 0 + 2.5 \times 4}{5} = 2 c m$

Q. 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 center of mass of the system is at a point: