Home/ Questions/Q 475470
Next
In Process
Rehman Manoj Ravel
Asked: 2022-11-11T15:14:18+05:30 In:

Centre of mass of three particles of masses `1 kg, 2 kg and 3 kg` lies at the point `(1,2,3)` and centre of mass of another system of particles `3 kg and 3 kg` lies at the point `(-1, 3, -2)`. Where should we put a particle of mass `5 kg` so that the centre of mass of entire system lies at the centre of mass of first system ?
A. `(0, 0,0)`
B. `(1, 3,2)`
C. `(-1, 2,3)`
D. `(3, 1, 8)`

Centre of mass of three particles of masses `1 kg, 2 kg and 3 kg` lies at the point `(1,2,3)` and centre of mass of another system of particles `3 kg and 3 kg` lies at the point `(-1, 3, -2)`. Where should we put a particle of mass `5 kg` so that the centre of mass of entire system lies at the centre of mass of first system ?
A. `(0, 0,0)`
B. `(1, 3,2)`
C. `(-1, 2,3)`
D. `(3, 1, 8)`
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. Correct Answer – D
    (d) According to the definition of centre of mass, we can imagine one particle of mass `(1 + 2 + 3) kg at (1,2,3)` , another particle of mass `(2 + 3) kg at (-1, 3, -2)`
    Let the third particle of mass `5 kg` put as `(x_3,y_3,z_3)` i.e.,
    `m_1 = 6kg, (x_1,y_1,z_1) = (1,2,3)`
    `m_2 = 5 kg(x_2,y_2,z_2) = (-1, 3, – 2)`
    `m_1 = 5 kg(x_3,y_3, z_3) = (1,2,3)`
    Given, `(X_(CM), Y_(CM), Z_(CM) = (1,2,3)`
    Using `X_(CM) = (m_1 x_1 + m_2 x_2 + m_3 x_3)/(m_1 + m_2 + m_3)`
    `1 = (6 xx 1 + 5 xx(-1)+ 5x_3)/(6 + 5 + 5)`
    `5 x_3 = 16-1 = 15` or `x_3 = 3`
    Similarly, `y_3 = 1 and z_3 = 8`.

    • 0