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The distance between two particles of masses m_(1)andm_(2) is r. If the distance of these particles from the centre of mass of the system are r_(1)andr_(2) respectively, then show that r_(1)=r((m_(2))/(m_(1)+m_(2)))andr_(2)=r((m_(1))/(m_(1)+m_(2))) |
Answer» Solution : Position vector of c.m. `VEC(r)_(cm)=(m_(1)vec(r_(1))+m_(2)vec(r_(2)))/(m_(1)+m_(2))""....(1)` Taking c.m. at the origin `vec(r_(cm))=vec0` `therefore 0=m_(1)vec(r_(1))+m_(2)vec(r_(2))[because" From eq. " (1)]` Now c.m. is on the LINE joining `m_(1)andm_(2)` and the directions of `vec(r_(1))andvec(r_(2))` are mutually opposite. `therefore 0=-m_(1)r_(1)+m_(2)r_(2)` `therefore m_(1)r_(1)=m_(2)r_(2)""......(2)` `therefore m_(1)(r_(1))=m_(2)(r-r_(1)) [because r=r_(1)+r_(2)]` `therefore m_(1)r_(1)=m_(2)r-m_(2)r_(1)` `therefore (m_(1)+m_(2))r_(1)=m_(2)r` `therefore r_(1)=r[(m_(2))/(m_(1)+m_(2))]` Again taking `r_(1)=r-r_(2)" in "m_(1)r_(1)=m_(2)r_(2)` `therefore m_(1)(r-r_(2))=m_(2)r_(2A)` `therefore m_(1)r-m_(1)r_(2)=m_(2)r_(2)` `therefore m_(1)r=(m_(1)+m_(2))r_(2)` `therefore r_(2)=r[(m_(1))/(m_(1)+m_(2))]` |
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