Question

consists of three particles each of mass M located at.( 1,1),( 2,2) and (3,3) the coordinates of the centre of mass are

Answer 1

Answer:

Centre of Mass of a system in a coordinate plane is given by:

$x_{COM} = \dfrac{ m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3 }$

$y_{COM} = \dfrac{ m_1y_1 + m_2y_2 + m_3y_3}{m_1+m_2+m_3}$

According to the Question,

⇒ x₁ = 1, x₂ = 2, x₃ = 3 ; y₁ = 1, y₂ = 2, y₃ = 3 ; m₁ = m₂ = m₃ = M

Calculating the x coordinate of Centre of Mass, we get,

$\implies x_{COM} = \dfrac{ M(1) + M(2) + M(3)}{M+M+M}\\\\\\\implies x_{COM} = \dfrac{ M + 2M + 3M }{ 3M } \implies \dfrac{6M}{3M} \implies 2$

Hence x coordinate of the centre of mass is 2.

Calculating the y coordinate of Centre of Mass we get,

$\implies y_{COM} = \dfrac{ M(1) + M(2) + M(3)}{M+M+M}\\\\\\\implies y_{COM} = \dfrac{ M + 2M + 3M }{ 3M } \implies \dfrac{6M}{3M} \implies 2$

Hence y coordinate is 2.

Hence the coordinates of Centre of Mass is ( x,y ) = ( 2,2 )

Answer 2

Answer:see the attachment

Explanation:HOPE IT HELPS YOU