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Two 3 kg masses have velocities bar(v_(1))=2hat(i)+3hat(j)m//s and bar(v_(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)=24 hat(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) `vec(v)_(c )=(m_(1)vec(v)_(1)+m_(2)vec(v)_(2))/(m_(1)+m_(2)), vec(v)_(c )=(3(2hat(i)+3hat(j))+(4hat(i)-6hat(j)))/(6)` `therefore` Velocity of centre of mass `bar(V)_(c )=3hat(i)-1.5 hat(j) ms^(-1)` (b) The momentum of the system `= Mv_(c )=6kg(3hat(i)-1.5 hat(j))ms^(-1)=18 hat(i)-9 hat(j)kg ms^(-1)` (c ) To find the velocity of centre of mass after 5 s of application of the force `vec(F)=24hat(i)N` we first find the ACCELERATION of the centre of mass. It is given by `vec(a)_(c )=(vec(F))/(M)=(24hat(i))/(6)=4hat(i)ms^(-2)` The velocity of centre of mass before the force is APPLIED is `vec(v)_(c )` and from the equaion `vec(v)_(c ).=vec(v)_(c )+vec(a)_(c )`.t `vec(v)_(c ). = (3hat(i)-1.5 hat(j))+(4hat(i))5=(3hat(i)-1.5hat(j)+20hat(i))` `vec(V)_(c )^(1)=(23 hat(i)-1.5 hat(j))ms^(-1)` (d) From the equation of the position vector `vec(r )= vec(r )_(0)+vec(v)_(0)t+(1)/(2)vec(a)t^(2)`, where `vec(r )_(0)=vec(0)` (origin at t = 0), `vec(v)_(0)=vec(v)_(c ), vec(a)=vec(a)_(c )` and t = 5s `vec(r )=(3hat(i)-1.5hat(j))5+(1)/(2)(4hat(i))25 = (15 hat(i)-7.5hat(j)+50 hat(i))` `therefore vec(r )=(65 hat(i)-7.5 hat(j))m` The coordinates of the centre of mass after 5 s of application of the force `vec(F)` are (65 m, 7.5 m) |
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