Centre of mass of three particles of masses I 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 $$2$$ 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?

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Infinity Learn · Accepted Answer

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 at (x3$$, $$y3$$, $$z3) i.e.$$m1$$ $$=$$ $$6$$ kg, (x1$$, $$y1$$,$$z1) $$=$$ (1$$,$$2$$,$$3)m2$$ $$=$$ $$5$$ kg, $$(x2$$, $$y2$$,$$z2) $$=$$ $$(-1$$, $$3$$,$$-2)m1$$ $$=$$ $$5$$ kg, (x3$$, $$y3$$,$$z3) Given, (XCM$$, YCM, $$ZCM) $$=$$ $$(1$$, $$2$$, $$3)Using$$ XCM $$=$$ $$m1x1+m2x2+m3x3m1+m2+m31$$ $$=$$ $$6$$×$$1+5$$×$$(-1)+5x36+5+55x3$$ $$=$$ $$16-1$$ $$=$$ $$15$$ or $$x3$$ $$=$$ $$3Similarly$$, $$y3$$ $$=$$ $$1$$ and $$z3$$ $$=$$ $$8$$

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