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01rocket_eqn_momentum07

01rocket_eqn_momentum07 - since this happened in a time t...

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Thrust Equation - Momentum Transfer Consider a vehicle made up of mass m Δ m + moving at velocity v at time t. At time t Δ t + mass m Δ has been ejected from the vehicle with a relative velocity 2 v so that m Δ has a velocity 2 v v + . The velocity of mass m is v Δ v + The momentum of the system at time t is ( 29 v m m + The momentum of the system at time t Δ t + ( 29 ( 29 2 v v m v v m + + + Since momentum is conserved for this isolated system we have ( 29 ( 29 ( 29 v m Δ m v v m Δ v Δ v m 2 + = + + + We can rewrite this as ( 29 ( 29 ( 29 0 v m m v v m v v m 2 = + - + + + Expanding 0 v m v m v m v m v m v m 2 = - - + + + Simplifying 0 v m v m 2 = + This can be written 0 v t m t v m
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Unformatted text preview: + since this happened in a time t . Taking the limit as t becomes infinitesimally small v dt dm dt v d m 2 = + Rewriting 2 v dt dm dt v d m -= Denoting dt dm m-= as the mass rate of flow leaving the vehicle of mass m we can write 2 v m dt v d m = This ejection of mass looks like a force acting on the vehicle of mass m. We call this the thrust: 2 T v m F = t t + 2 v v + m m v v + m M t m m v t...
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