Consider a large block placed on a smooth horizontal surface, with a man standing at one end of the block. The man walks to the other end, relative to the block. The distances (absolute) moved by the man and the block are:

Question

Infinity Learn · Accepted Answer

Let m and M be the respective masses of the person and the block, and let l be its length. Now, if $$x1$$ and $$x2$$ be the absolute displacement of the person and the block and x that of the centre of mass, then x $$=$$ $$mx1+mx2m+M$$ ⇒∆x $$=$$ m∆$$x1+m$$∆$$x2m+M$$ Since horizontal forces are absent, so  ∆x $$=$$ $$0m$$∆$$x1+M$$∆$$x2$$ $$=$$ $$0$$    ⇒ $$m(-l)+(m+M)$$∆$$x2$$ $$=$$ $$0$$∆$$x2$$ $$=$$ $$mlm+MSimilarly$$, ∆$$x1$$ $$=$$ $$Mlm+M$$

Consider a large block placed on a smooth horizontal surface, with a man standing at one end of the block. The man walks to the other end, relative to the block. The distances (absolute) moved by the man and the block are:

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