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Show that the total momentum of system of particles is equal to the product of total mass of system and velocity of centre of mass. |
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Answer» Solution :According to equation (2) as in QUESTION 19. `MvecV=m_(1)vec(v_(1))+m_(2)vec(v_(2))+...m_(n)vec(v_(n))""...(1)` According to chapter 4, `vecp=mvecvand(dvecp)/(dt)=VECF` Now, the linear momentum of system of n particle is equal to the vector sum of individual linear momentum of all particles of the system. `THEREFORE vecP=vec(p_(1))+vec(p_(2))+...vec(p_(n))` `vecP=m_(1)vec(v_(1))+m_(2)vec(v_(2))+...m_(n)vec(v_(n))""...(2) [because vecp=mvecv]` COMPARING equation (1) and (2), `vecP=MvecV""...(3)` Hence, the linear momentum of system of particle is equal to the product of total mass of system and velocity of centre of mass. |
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