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Question: The coordinates of the centre of mass of particles of mass \(10\), \(20\) and \(30\) gm are \((1, 1,...

The coordinates of the centre of mass of particles of mass 1010, 2020 and 3030 gm are (1,1,1)(1, 1, 1) cm. The position coordinates of mass 4040 gm which when added to the system, the position of combined centre of mass be at (0,0,0)(0, 0, 0) are:
(A) 32,32,32\dfrac{3}{2},\dfrac{3}{2},\dfrac{3}{2}
(B) −32,−32,−32 - \dfrac{3}{2}, - \dfrac{3}{2}, - \dfrac{3}{2}
(C) 34,34,34\dfrac{3}{4},\dfrac{3}{4},\dfrac{3}{4}
(D) −34,−34,−34 - \dfrac{3}{4}, - \dfrac{3}{4}, - \dfrac{3}{4}

Explanation

Solution

Hint
The position of centre of mass (xm)(x_m) of an object in 3-dimensions can be calculated using the formula-
xcm=m1x1+m2x2+...+mnxnm1+m2+...+mn{x_{cm}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + ... + {m_n}{x_n}}}{{{m_1} + {m_2} + ... + {m_n}}} , where m1,m2,…,mnm_1, m_2, …, m_n are the individual masses, and x1,x2,…,xnx_1, x_2, …, x_n are their x-coordinates of their positions respectively.

Complete Step By Step Solution
Given that the centre of mass of particles of mass 1010, 2020 and 3030 gm are (1,1,1)(1, 1, 1) cm. This implies that, (xcm,ycm,zcm)=(1,1,1)(x_{cm}, y_{cm}, z_{cm}) = (1, 1, 1), (So, xcm=1x_{cm} = 1), m1=10m_1 = 10 gm, m2=20m_2 = 20 gm and m3=30m_3 = 30 gm. Let, x1,x2andx3x_1, x_2 and x_3 be the coordinates of the three masses respectively.
Substitute these values in the xcmx_{cm} formula.
1=(10g)x1+(20g)x2+(30g)x3(10+20+30)g1 = \dfrac{{(10g){x_1} + (20g){x_2} + (30g){x_3}}}{{(10 + 20 + 30)g}}
Rearrange the equation.
(10x1+20x2+30x3)g=60g(10{x_1} + 20{x_2} + 30{x_3})g = 60g
Cancel ‘g’ and divide by 10 on both sides. We get,
x1+2x2+3x3=6{x_1} + 2{x_2} + 3{x_3} = 6 … eq.1
Add the fourth mass, m4 = 40 gm, with the position coordinate x4.
Given, that the position of the combined centre of mass is at (0, 0, 0). So, the x-coordinate of the centre of mass of the new system, xcm′=0{x_{cm}}' = 0.
Substitute the values in the centre of mass formula.
0=(10g)x1+(20g)x2+(30g)x3+(40g)x4(10+20+30+40)g0 = \dfrac{{(10g){x_1} + (20g){x_2} + (30g){x_3} + (40g){x_4}}}{{(10 + 20 + 30 + 40)g}} \
Rearrange the equation.
10x1+20x2+30x3+40x4=010{x_1} + 20{x_2} + 30{x_3} + 40{x_4} = 0
Divide the equation by 10.
x1+2x2+3x3+4x4=0{x_1} + 2{x_2} + 3{x_3} + 4{x_4} = 0… eq.2
Solve eq.1 and eq.2.
Subtract (1) from (2), We get-
(x1+2x2+3x3+4x4)−(x1+2x2+3x3)=0−6 ⇒(x1−x1)+(2x2−2x2)+(3x3−3x3)+4x4=(−6) ⇒4x4=−6({x_1} + 2{x_2} + 3{x_3} + 4{x_4}) - ({x_1} + 2{x_2} + 3{x_3}) = 0 - 6\\\ \Rightarrow ({x_1} - {x_1}) + (2{x_2} - 2{x_2}) + (3{x_3} - 3{x_3}) + 4{x_4} = ( - 6)\\\ \Rightarrow 4{x_4} = - 6
x4=−64∴x4=−32{x_4} = - \dfrac{6}{4}\\\\\therefore {x_4} = - \dfrac{3}{2}
Similarly, for y-coordinates of the system, the y-coordinate of the centre of mass of the system is ycm=1y_{cm} = 1. By taking the formula of the y-coordinate of centre of mass of the system as, ycm=m1y1+m2y2+m3y3m1+m2+m3{y_{cm}} = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}} and repeating the calculations performed above, we get y4=−32{y_4} = - \dfrac{3}{2}.
And for the z-coordinate system, by considering zcm=m1z1+m2z2+m3z3m1+m2+m3{z_{cm}} = \dfrac{{{m_1}{z_1} + {m_2}{z_2} + {m_3}{z_3}}}{{{m_1} + {m_2} + {m_3}}}, we get z4=−32{z_4} = - \dfrac{3}{2}.
Therefore, the position coordinates of mass 40 gm which when added to the system are (−32,−32,−32)( - \dfrac{3}{2}, - \dfrac{3}{2}, - \dfrac{3}{2}).
Correct option is (B).

Note
The centre of mass (cm) is the point relative to the system of particles in an object. This is the unique point which represents the mean position of the particles in the system.