Question

Help

Answer 1

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..

Answer 2

@mathmate

Answer 3

@shamim

Answer 4

@IrishBoy123

Answer 5

can hardly read it but they also switched U and V around to get relative velocity so i presume that it assists presentation

Answer 6

Yeah, it's very hard to read. Can you write it out using the drawing tool (the "draw" button below)?

Answer 7

No i am unable to use drawing tool because i am using tablet ..

Answer 8

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 \(\vec u\) ?

Answer 9

"very OM" = very OK

Answer 10

U is velocity of big mass where v is the velocity of ejected mass

Answer 11

I think u r right tht signs doesnt mean.. this iw due to relativity..

Answer 12

since the subsequent identity, has been used:
\[\Large {\mathbf{v}} - {\mathbf{u}} = - \left( {{\mathbf{u}} - {\mathbf{v}}} \right)\]

Answer 13

Okay thanks to all for helping me.. :)