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.

In a system of three particles, the linear momenta of the three partic

1.	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.
Answer» Solution :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`,