Two 3 kg masses have velocities `barv_(1)=2hati+3hatj` m/s and `barv_(2)=4hati-6hatj` m/s. Find a) velocity of centre of mass, b) the total momentum of the system, c) The velocity of centre of mass 5s after application of a constant force `barF=24hatiN`,d) position of centre of mass after 5s if it is at the origin at t = 0

Explanation

`vecv_(c)=(m_(1)vecv_(1)+m_(2)vecv_(2))/(m_(1)+m_(2)),vecv_(c)=(3(2hati+3hatj)+3(4hati-6hatj))/(6)`
`:.` Velocity of centre of mass `barV_(c)=3hati-1.5hatjms^(-1)`
b) The momentum of the system
`=Mv_(c)=6kg(3hati-1.5hatj)ms^(-1)=18hati-9hatjkgms^(-1)`
To find the velocity of centre of mass after 5 s of application of the force `vecF=24hatiN` we first find the acceleration of the centre of mass. It is given by
`veca_(c)=(vecF)/(M)=(24hati)/(6)=4hatims^(-2)`
The velocity of centre of mass before the force is applied is `vecv_(c)`.
and from the equation `vecv_(c)=vecv_(c)+veca_(c).t`
`vecv_(c)=(3hati-1.5hatj)+(4hati)5=(3hati-1.5hatj+20hati)`
`vecV_(c)^(1)=(23hati-1.5hatj)ms^(-1)`
From the equation of the position vector
`vecr=vecr_(0)+vecv_(0)t+(1)/(2)vecat^(2)` where `vecr_(0)=vec0` (origin at t=0), `vecv_(0)=vecv_(c),veca=veca_(c)andt=5s`
`vecr=(3hati-1.5hatj)5+(1)/(2)(4hati)25,vecr=(15hati-7.5hatj+50hati)`
`vecr=(65hati-7.5hatj)m`
The coordinates of the centre of mass after 5 s of application of the force `vecF` are (65m, -7.5m)