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Two 3kg masses have velocities barv_(1)=2hat(i)+3hat(j)m//s and barv_(2)=4hat(i)-6hat(j)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 bar(F)=24hat(i)N, d) position of centre of mass after 5s if it is at the origin at t=0 |
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Answer» Solution :(a) `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 `vecV_(c )=3hati-1.5hatjms^(-1)` (b) The momentum of the system `=Mv_(c )=6kg(3hati-1.5hatj)ms^(-1)=18hati-9hatjkgms^(-1)` (c ) To FIND the velocity of centre of mass after 5s 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 ).=(23hati-1.5hatj)ms^(-1)` (d) 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 )` and `t=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 5s of application of the force `vecF` are `(65m,-7.5m)` |
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