In a system of three particles, the linear momenta of the three particles are (1,2,3), (4,5,6) and (5,6,7). These components are in `kgms^(-1)`. If the velocity of centre of mass of the system is `(30,39,48)ms^(-1)`, then find the total mass of the system.

Explanation

Here `vec(p_(1))=(1,2,3)kgms^(-1)`
`vec(p_(2))=(4,5,6)kgms^(-1)`
`vec(p_(3))=(5,6,7)kgms^(-1)`
`vec(v_(cm))=(30,39,48)ms^(-1)`
`Mvecv_(cm)=vecp`
`therefore Mvecv_(cm)=vec(p_(1))+vec(p_(2))+vec(p_(3))`
`therefore M(30,39,48)=(1,2,3)+(4,5,6)+(5,6,7)`
`M(30,39,38)=(1+4+5, 2+5+6,3+6+7)`
`M(30,39,48)=(10,13,16)`
Comparing respective co-efficients on both sides.
`M(30)=10," "M(39)=13," "M(48)=16`
`therefore M=(1)/(3)kg,M=(1)/(3)kg,M=(1)/(3)kg`
`therefore M=(1)/(3)kg`
Thus the total mass of the system is `(1)/(3)kg`,