anonymous
  • anonymous
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Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Its a derivation from momentum changes in a system of variable mass..my Question is that why a sign of velocity of big mass relative to small mass is changed in 2nd step..
anonymous
  • anonymous
@mathmate
anonymous
  • anonymous
@shamim

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anonymous
  • anonymous
@IrishBoy123
IrishBoy123
  • IrishBoy123
can hardly read it but they also switched U and V around to get relative velocity so i presume that it assists presentation
anonymous
  • anonymous
mathmate
  • mathmate
Yeah, it's very hard to read. Can you write it out using the drawing tool (the "draw" button below)?
anonymous
  • anonymous
No i am unable to use drawing tool because i am using tablet ..
IrishBoy123
  • IrishBoy123
clearly you are starting with \(\large \vec F = \frac{d}{dt}\vec p = \frac{d}{dt} (m \vec v)\) and running with that gives \(\large m \frac{d \vec v} {dt} +\frac{dm}{dt} (\vec v - \vec u)\) and then the switch to \(\large m \frac{d \vec v} {dt} -\frac{dm}{dt} (\vec u - \vec v)\) where \( (\vec u - \vec v)\ = \vec v_{rel} \) all looks very OM except we do not actually know what you are modelling! eg what is \(\vec u\) ?
IrishBoy123
  • IrishBoy123
"very OM" = very OK
anonymous
  • anonymous
U is velocity of big mass where v is the velocity of ejected mass
anonymous
  • anonymous
I think u r right tht signs doesnt mean.. this iw due to relativity..
Michele_Laino
  • Michele_Laino
since the subsequent identity, has been used: \[\Large {\mathbf{v}} - {\mathbf{u}} = - \left( {{\mathbf{u}} - {\mathbf{v}}} \right)\]
anonymous
  • anonymous
Okay thanks to all for helping me.. :)

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