Take a system of two particles in the xy plane: m1= 2.20 kg is at r1=(1.00i + 2.00j)m and has a velocity (3.00i + 0.500j)m/s; m2 = 3.80 kg is at r2=(-4.00i - 3.00j)m and has a velocity (3.00i -...

Take a system of two particles in the xy plane: m1= 2.20 kg is at r1=(1.00i + 2.00j)m and has a velocity (3.00i + 0.500j)m/s; m2 = 3.80 kg is at r2=(-4.00i - 3.00j)m and has a velocity (3.00i - 2.00j)m/s.

Find the position of the center of mass of the system. 

answer should be formatted as follows.

rCM = (____i + ______j

Please note that a bold i means  and a bold j means `hatj`

` `

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine the x-axis component of vector of center of mass' position, such that:

`R_x = (m_1*x_1 + m_2*x_2)/(m_1 + m_2)`

`x_1, x_2` are the coefficients of bar i vector

`x_1 = 1, x_2 = -4`

`R_x = (2.2*1 + 3.8*(-4))/(2.2 + 3.8) => R_x = -13/6 => R_x = -2.16`

You need to determine the y-axis component of vector of center of mass' position, such that:

`R_y = (m_1*y_1 + m_2*y_2)/(m_1 + m_2)`

`y_1, y_2` are the coefficients of bar i vector

`y_1 = 2, y_2 = -3`

`R_y = (2.2*2 + 3.8*(-3))/(2.2 + 3.8) => R_y = -7/6 => R_y = -1.16`

The vector of center of mass' position is the following, such that:

`bar (r_CM) = R_x*bar i + R_y*bar j`

`bar (r_CM) = -2.16 bar i - 1.16 bar j`

You need to evaluate the x-axis component of vector of center of mass' velocity, such that:

`v_x = (m_1*v_(x_1) + m_2*v_(x_2))/(m_1 + m_2)`

`v_x = (2.2*3 + 3.8*3)/(2.2 + 3.8) => v_x =18/6 => v_x = 3`

You need to evaluate the y-axis component of vector of center of mass' velocity, such that:

`v_y = (m_1*v_(y_1) + m_2*v_(y_2))/(m_1 + m_2)`

`v_y = (2.2*0.5 + 3.8*(-2))/(2.2 + 3.8) => v_y =-1.08`

The vector of center of mass' velocity is the following, such that:

`bar (v_CM) = v_x*bar i + v_y*bar j`

`bar (v_CM) = 3 bar i - 1.08 bar j`

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