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Two 3 kg masses have velocities vecv_(1) = 2hati + 3hatj m//sand vecv_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 vecF = 24hati N d) position of centre of mass after S if it is at the origin at t = 0.

Answer» <html><body><p></p>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, therefore` Velocity of centre of mass `vecV_( c) = 3hati - 1.<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> <a href="https://interviewquestions.tuteehub.com/tag/hatj-2693584" style="font-weight:bold;" target="_blank" title="Click to know more about HATJ">HATJ</a> ms^(-1)` <br/> b) The momentum of the system = `Mv_( c)=6 kg(3hati - 1.5 hatj) ms^(-1) = 18hati - 9hatj <a href="https://interviewquestions.tuteehub.com/tag/kgms-2769162" style="font-weight:bold;" target="_blank" title="Click to know more about KGMS">KGMS</a>^(-1)` <br/> c) To find the velocity of centre of mass after 5 s of application of the force `vecF = 24hatj N` <br/>we first find the acceleration of the centre of mass. It is given by:<br/> `veca_( c) = vecF/M =(24hati)/6 = 4hati ms^(-2)` <br/> The velocity of centre of mass before the force is applied is `vecv_( c)`and from the <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>: `vecv_(c) = vecv_( c) + veca_(c).t` <br/> `vecv_(c ) =(3hati - 1.5 hatj) + (4hati)5 =(3hati -1.5 hatj + 20 hatj), V_(c )^(1) =(23 hati - 1.5 hatj) ms^(-1)` <br/> d) From the equation of the position vector `vecr = vecr_(0) + vecv_(0)t + 1/2 at^(2)`, where `vecr_(0)=<a href="https://interviewquestions.tuteehub.com/tag/vec0-3257206" style="font-weight:bold;" target="_blank" title="Click to know more about VEC0">VEC0</a>` (origin at t=0), <br/> `vecv_(0) = vecv_(c ), veca = veca_( c)` and t= 5s <br/> `vecr =(3hati - 1.5 hatj) + 5 + 1/2(4hati) 25 =(15hati - 7.5 hatj + 50hati), therefore vecr =(65 hati - 7.5 hatj)`m <br/> The coordinates of the centre of mass after 5 s of application of the force `vecF`are (65 m, - 7.5 m)</body></html>


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