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

Two 3 kg masses have velocities barv_(1)=2hati+3hatj m/s and barv_(2)=

1.	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
Answer» Solution :`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)