Question

The coordinates of the centre of mass of particles of mass $10$, $20$ and $30$ gm are $(1, 1, 1)$ cm. The position coordinates of mass $40$ gm which when added to the system, the position of combined centre of mass be at $(0, 0, 0)$ are:
(A) $\dfrac{3}{2},\dfrac{3}{2},\dfrac{3}{2}$
(B) $ - \dfrac{3}{2}, - \dfrac{3}{2}, - \dfrac{3}{2}$
(C) $\dfrac{3}{4},\dfrac{3}{4},\dfrac{3}{4}$
(D) $ - \dfrac{3}{4}, - \dfrac{3}{4}, - \dfrac{3}{4}$

Answer 1

Hint
The position of centre of mass $(x_m)$ of an object in 3-dimensions can be calculated using the formula-
${x_{cm}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + ... + {m_n}{x_n}}}{{{m_1} + {m_2} + ... + {m_n}}}$ , where $m_1, m_2, …, m_n$ are the individual masses, and $x_1, x_2, …, x_n$ are their x-coordinates of their positions respectively.

Complete Step By Step Solution
Given that the centre of mass of particles of mass $10$, $20$ and $30$ gm are $(1, 1, 1)$ cm. This implies that, $(x_{cm}, y_{cm}, z_{cm}) = (1, 1, 1)$, (So, $x_{cm} = 1$), $m_1 = 10$ gm, $m_2 = 20$ gm and $m_3 = 30$ gm. Let, $x_1, x_2 and x_3$ be the coordinates of the three masses respectively.
Substitute these values in the $x_{cm}$ formula.
$1 = \dfrac{{(10g){x_1} + (20g){x_2} + (30g){x_3}}}{{(10 + 20 + 30)g}}$
Rearrange the equation.
$(10{x_1} + 20{x_2} + 30{x_3})g = 60g$
Cancel ‘g’ and divide by 10 on both sides. We get,
${x_1} + 2{x_2} + 3{x_3} = 6$ … 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, ${x_{cm}}' = 0$.
Substitute the values in the centre of mass formula.
$0 = \dfrac{{(10g){x_1} + (20g){x_2} + (30g){x_3} + (40g){x_4}}}{{(10 + 20 + 30 + 40)g}}$ \
Rearrange the equation.
$10{x_1} + 20{x_2} + 30{x_3} + 40{x_4} = 0$
Divide the equation by 10.
${x_1} + 2{x_2} + 3{x_3} + 4{x_4} = 0$… eq.2
Solve eq.1 and eq.2.
Subtract (1) from (2), We get-
$({x_1} + 2{x_2} + 3{x_3} + 4{x_4}) - ({x_1} + 2{x_2} + 3{x_3}) = 0 - 6\\ \Rightarrow ({x_1} - {x_1}) + (2{x_2} - 2{x_2}) + (3{x_3} - 3{x_3}) + 4{x_4} = ( - 6)\\ \Rightarrow 4{x_4} = - 6$
${x_4} = - \dfrac{6}{4}\\\therefore {x_4} = - \dfrac{3}{2}$
Similarly, for y-coordinates of the system, the y-coordinate of the centre of mass of the system is $y_{cm} = 1$. By taking the formula of the y-coordinate of centre of mass of the system as, ${y_{cm}} = \dfrac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + {m_3}}}$ and repeating the calculations performed above, we get ${y_4} = - \dfrac{3}{2}$.
And for the z-coordinate system, by considering ${z_{cm}} = \dfrac{{{m_1}{z_1} + {m_2}{z_2} + {m_3}{z_3}}}{{{m_1} + {m_2} + {m_3}}}$, we get ${z_4} = - \dfrac{3}{2}$.
Therefore, the position coordinates of mass 40 gm which when added to the system are $( - \dfrac{3}{2}, - \dfrac{3}{2}, - \dfrac{3}{2})$.
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.