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CBSE Class 11-science Answered

<div><span style="color: #222222; font-family: arial, sans-serif; font-size: medium;">&nbsp;t</span><a class="spell" style="font-size: 18px; color: #660099; cursor: pointer; text-decoration: none; font-family: arial, sans-serif;" href="https://www.google.co.in/search?es_sm=122&amp;biw=931&amp;bih=585&amp;q=he+velocity+vectors+of+three+%22particles%22+of+masses+1kg,+2kg+and+3kg+are+respectively+(1,2,3),+(3,4,5)+and+(6,7,8).+the+velocity+vectors+are+in+m/s.+find+the+velocity+vector+of+center+of+mass+of+this+system+of+particles.&amp;spell=1&amp;sa=X&amp;ei=QQ9ZVP2oOJefugSH7oLgAQ&amp;ved=0CBgQBSgA">he velocity&nbsp;<strong><em>vectors</em></strong>&nbsp;of three "particles" of masses 1kg, 2kg and 3kg are respectively (1,2,3), (3,4,5) and (6,7,8). the velocity vectors are in m/s. find the velocity vector of center of mass of this system of particles.</a></div>
Asked by Anu | 04 Nov, 2014, 11:11: PM
Expert Answer
begin mathsize 14px style Velocity space vectors space of space three space particles space of space masses space straight m subscript 1 equals 1 space kg comma straight m subscript 2 equals 2 space kg comma space straight m subscript 3 equals 3 space kg space are space respectively space left parenthesis 1 comma 2 comma 3 right parenthesis comma space left parenthesis 3 comma 4 comma 5 right parenthesis space and space left parenthesis 6 comma 7 comma 8 right parenthesis By space defination comma space the space coordinates space of space the space velocity space vector space of space center space of space mass space of space system space of space particles space are space given space by comma stack straight v subscript xcm with rightwards arrow on top equals fraction numerator straight m subscript 1 straight v subscript straight x subscript 1 plus straight m subscript 2 straight v subscript straight x subscript 2 plus straight m subscript 3 straight v subscript straight x 3 end subscript over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction equals fraction numerator left parenthesis 1 cross times 1 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 6 right parenthesis over denominator 1 plus 2 plus 3 end fraction equals fraction numerator 1 plus 6 plus 18 over denominator 6 end fraction equals 25 over 6 space straight m divided by straight s stack straight v subscript ycm with rightwards arrow on top equals fraction numerator straight m subscript 1 straight v subscript straight y subscript 1 plus straight m subscript 2 straight v subscript straight y subscript 2 plus straight m subscript 3 straight v subscript straight y 3 end subscript over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction equals fraction numerator left parenthesis 1 cross times 2 right parenthesis plus left parenthesis 2 cross times 4 right parenthesis plus left parenthesis 3 cross times 7 right parenthesis over denominator 1 plus 2 plus 3 end fraction equals fraction numerator 2 plus 8 plus 21 over denominator 6 end fraction equals 31 over 6 space space straight m divided by straight s stack straight v subscript zcm with rightwards arrow on top equals fraction numerator straight m subscript 1 straight v subscript straight z subscript 1 plus straight m subscript 2 straight v subscript straight z 2 end subscript plus straight m subscript 3 straight v subscript straight z 3 end subscript over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction equals fraction numerator left parenthesis 1 cross times 3 right parenthesis plus left parenthesis 2 cross times 5 right parenthesis plus left parenthesis 3 cross times 8 right parenthesis over denominator 1 plus 2 plus 3 end fraction equals fraction numerator 3 plus 10 plus 24 over denominator 6 end fraction equals 37 over 6 space space straight m divided by straight s end style
Answered by Priyanka Kumbhar | 05 Nov, 2014, 09:53: AM
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