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DLS

  • 2 years ago

If the centre of mass of three particles of masses 10,20 and 30g is at the point(1,-2,3)then where should a fourth particle of mass 40g be placed,so that the combined centre of mass of the system is (1,1,1)

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  1. DLS
    • 2 years ago
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    @mayankdevnani

  2. DLS
    • 2 years ago
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    @Yahoo!

  3. Yahoo!
    • 2 years ago
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    Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)

  4. DLS
    • 2 years ago
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    I didn't understand the co-ordinates,is it 1,0 -2,0 & 3,0?

  5. Yahoo!
    • 2 years ago
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    There are three Co-ordinates....For the 1 St one..CM is @ (1,-2,3)

  6. DLS
    • 2 years ago
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    10-40+90/60 =1?

  7. DLS
    • 2 years ago
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    i didnt get it what to do

  8. Yahoo!
    • 2 years ago
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    First Find the co-ordinates of m1 m2 and m3

  9. Yahoo!
    • 2 years ago
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    (1,-2,3 ) x , y , z x axis of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3) y axis of Cm = m1y1 + m2y2 + m3y3 / (m1+m2+m3) etc...

  10. Yahoo!
    • 2 years ago
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    u got it..)

  11. DLS
    • 2 years ago
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    thanks :D then?

  12. DLS
    • 2 years ago
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    @Mashy

  13. Mashy
    • 2 years ago
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    wait .. i know.. on call with gf :P.. 2 mins

  14. Mashy
    • 2 years ago
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    ok.. so let xcm be the centre of mass of the three particle system then \[Mxcm = m1x1+ m2x2 + m3x3 \] now .. when the fourth particle is added.. the centre of mass shifts hence \[M'xcm' = m1x1+ m2x2 + m3x3+ m4x4\] or \[M'xcm' = Mxcm + m4x4\] the only thing you have to find now is x4.. so calculate.. do the same in other 2 axes

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