Consider a two particle system with particles having masses  m1 and m2 . If the first particle is pushed  towards the centre of mass through a distance  d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position ?

Option 1)

d

Option 2)

\frac{m_{2}}{m_{1}}d

Option 3)

\frac{m_{1}}{m_{1}+m_{2}}d

Option 4)

\frac{m_{1}}{m_{2}}d

Answers (1)
A Aadil

As we learnt in

Centre of Mass of a system of N discrete particles -

x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}.........}{m_{1}+m_{2}.......}

y_{cm}=\frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}.........}{m_{1}+m_{2}+m_{3}.......}

z_{cm}=\frac{m_{1}z_{1}+m_{2}z_{2}+m_{3}z_{3}.........}{m_{1}+m_{2}+m_{3}.......}

- wherein

m1, m2 ........... are mass of each  particle  x1, x2 ..........y1, y2 ............ z1, z2 are respectively x, y, & z coordinates of particles.

 

 \Delta x_{cm}=\frac{m_{1}.\Delta x_{1}+m_{2}.\Delta x_{2}}{m_{1}+m_{2}}

Let m_{2} be moved by x so as to keep the centre of mass at the same position

\Delta x_{1}=-d and \Delta x_{CM}=0

\therefore \; \; \; m_{1}(-d)+m_{2}(x_{2})=0,

or\; \; \; m_{1}d=m_{2}x_{2}; \; \; or\; \; \; x_{2}=\frac{m_{1}}{m_{2}}d


Option 1)

d

Option 2)

\frac{m_{2}}{m_{1}}d

Option 3)

\frac{m_{1}}{m_{1}+m_{2}}d

Option 4)

\frac{m_{1}}{m_{2}}d

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