Question

The centre of mass of a system of three particles of masses 1g, 2g and 3g is taken as the origin of a coordinate system. The position vector of a fourth particle of mass 4g such that the centre of mass of the four particle system lies at the point (1, 2, 3) is α(î + 2Ä + 3), where α is a constant. The value of α is :- (1)10/3(2)5/2(3)1/2(4)2/5

Answer 1

b)5/2
COM of y coordinate=(m1×y1 +m2×y2 +m3×y3 +m4×y4)/m1+m2+m3+m4
but the first three terms In the numerator is equal to zero,hence
COM=4×2a/10=2
therefore a=5/2

Answer 2

Answer:

The value of α is $\frac{5}{2}$

Explanation:

From the above question,

They have given :

The centre of mass of a system of three particles of masses 1g, 2g and 3g is taken as the origin of a coordinate system. The position vector of a fourth particle of mass 4g such that the centre of mass of the four particle system lies at the point (1, 2, 3) is α(i + 2j + 3k),

where α is a constant.

Here we need to find the value of α:

          R = (m1r1 + m2r2 + m3r3 + m4r4) / M

where R is the role vector of the core of mass, m1, m2, m3, and m4 are the hundreds of the particles, r1, r2, r3, and r4 are their role vectors relative to the origin, and M is the whole mass of the system.

We are given that the core of mass of the gadget of three particles is at the origin, so we have:

                m1r1 + m2r2 + m3*r3 = 0

We are additionally given that the core of mass of the four-particle gadget is at the factor (1,2,3), so we have:

              m1r1 + m2r2 + m3r3 + m4r4 = M*R = (1,2,3)M

Substituting the first equation into the 2d equation, we get:

m4*r4 = (1,2,3)M

where M = m1 + m2 + m3 + m4

               = 1 + 2 + 3 + 4r = 10 g.

We are given that this function vector is equal to α(iˆ+2jˆ+3kˆ), the place α is a constant. Equating the components, we get:

            $\frac{5}{2}$  = α

            5 = 2α

           $\frac{15}{4}$ = 3α

Solving for α, we get:

        The value of α is  $\frac{5}{2}$.

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