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 $$2$$ 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) $$=$$ $$(-1$$, $$3$$,$$-2)m3=5$$ kg, let its position is (x3$$,$$y3$$.$$z3)Given$$, $$(XCM$$, YCM, $$ZCM) $$=$$ $$(1$$, $$2$$, $$3)1$$ $$=$$ $$6$$×$$1+5$$×$$(-1)+5x36+5+55x3$$ $$=$$ $$16-1$$ $$=$$ $$15$$ or $$x3$$ $$=$$ $$3Similarly$$, $$y3$$ $$=$$ $$1$$ and $$z3$$ $$=$$ $$8$$

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