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  • Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown in the figure. The centre of mass of the system is at a point: <img alt="" src="https://cdn.entrance3

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

Option: 1 2.0\; cm right and 0.9\; cm above 1\: kg mass
   
Option: 2 0.9\; cm right and 2.0\; cm above 1\: kg mass

Option: 3 0.6\; cm right and 2.0\; cm above 1\: kg mass  

Option: 4 1.5\; cm right and 1.2\; cm above 1\: kg mass
 

Answers (1)

best_answer

 

 

let m1=1 kg, m2=1.5 kg and m3=2.5 kg

x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4

x_{com}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} and y_{com}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}

Let point A be origin and mass m1=1.0 kg be at origin.

So, x_{com}=\frac{1\times0+1.5\times3+2.5\times0}{1+1.5+2.5}=\frac{4.5}{5}=0.9

and y_{com}=\frac{1\times0+1.5\times0+2.5\times4}{1+1.5+2.5}=\frac{10}{5}=2

so centre of mass of the system is at (0.9,2).

So from the figure we can say that the 0.9 cm right and 2 cm above the 1 kg mass. 

So option (2) is correct.

Posted by

Ritika Jonwal

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