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Unformatted text preview: M + Δ M )( u + Δ u ) Multiplying out, and canceling the terms Mu which occur on both sides: Δ Mv = Δ Mu + M Δ u + Δ M Δ u The last term is“second order”, and so can be dropped. Thus, Δ M/M = Δ v/ ( v-u ) or, taking the limit as Δ t → 0, Z dM/M = Z du/ ( v-u ) Integrating, and choosing the constant of integration so that M = M when u = u = 0: M = M • v v-u ‚ 1 or ln M/M =-ln [( v-u ) /v ] or u = v-v ( M /M ) From this equation we see that the maximum value of u is indeed v , obtained when M ± M . What is the velocity as a function of time? Since Δ M = Rm Δ t , we have that M = Rmt + M , and u = v-v [ M / ( Rmt + M )] 2...
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This note was uploaded on 04/29/2008 for the course PHYS 230 taught by Professor Harris during the Fall '07 term at McGill.
- Fall '07