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Question: The centre of mass of a system of three particles of masses \(1g,2g\) and \(3g\) is taken as the ori...

The centre of mass of a system of three particles of masses 1g,2g1g,2g and 3g3g is taken as the origin of a coordinate system. The position vector of a Fourth particle of mass 4g4g such that the centre of mass of the four particle system lies at the point (1,2,3)(1,2,3) is α(i^+2j^+3k^)\alpha (\hat i + 2\hat j + 3\hat k) where α\alpha is a constant. The value of α\alpha is:
(A) 103\dfrac{{10}}{3}
(B) 52\dfrac{5}{2}
(C) 12\dfrac{1}{2}
(D) 25\dfrac{2}{5}

Explanation

Solution

Centre of mass is a point of a system where whole mass can be considered to be concentrated. We will use the general formula of centre of mass in order to find the value of α\alpha . Formula is given as XC.M=m1x1+m2x2m1+m2{X_{C.M}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}} where x1,x2{x_1},{x_2} are the positions of masses m1,m2{m_1},{m_2} in a coordinate system.

Complete step by step answer:
Since, it’s given that the centre of mass of 1g,2g1g,2g and 3g3g having total mass of the system is 6g6g can be considered to be concentrated at point origin (0,0,0)(0,0,0) .
Let us imagine this total mass as m1=6g{m_1} = 6g with position x1=y1=z1=0{x_1} = {y_1} = {z_1} = 0 .
Now, imagine mass m2=4g{m_2} = 4g is placed at position (x2,y2,z2)({x_2},{y_2},{z_2}) then this combine system has a Centre of mass given as (1,2,3)(1,2,3)
Now, putting values in formula XC.M=m1x1+m2x2m1+m2{X_{C.M}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}} we get,
1=6(0)+4(x2)101 = \dfrac{{6(0) + 4({x_2})}}{{10}}
x2=104{x_2} = \dfrac{{10}}{4}
Similarly, For Y-axis
2=4y2102 = \dfrac{{4{y_2}}}{{10}}
y2=204{y_2} = \dfrac{{20}}{4}
For Z-axis
3=4z2103 = \dfrac{{4{z_2}}}{{10}}
z2=304{z_2} = \dfrac{{30}}{4}
So, given position vector is α(i^+2j^+3k^)\alpha (\hat i + 2\hat j + 3\hat k)
Putting values of x2=104{x_2} = \dfrac{{10}}{4} y2=204{y_2} = \dfrac{{20}}{4} z2=304{z_2} = \dfrac{{30}}{4} as in vector form we get
Position vector is 52(i^+2j^+3k^)\dfrac{5}{2}(\hat i + 2\hat j + 3\hat k)
On comparing the both position vectors we get, α=52\alpha = \dfrac{5}{2}
Hence, the correct option is (B) α=52\alpha = \dfrac{5}{2}.

Note: Whenever we have given the position of centre of mass of a system having number of individual particles each of having own mass then, the combined total mass of the system can be considered as at a position of their centre of mass and can be taken as a single system having position of centre of mass and mass equals to total mass of the system.