Moving Target

# Moving Target - M Δ M u Δ u Multiplying out and canceling...

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The moving target as described in Class 10 At time t , the target has mass M and is moving with velocity u . As time goes on, its mass increases, and its velocity increases also. t ´¿ "! - - - v u u + Δ u Δ M M M + Δ M = Time t Time t + Δ t A stream of tiny particles, each with velocity v , is hitting the target and sticking to it. In time Δ t the mass of these particles that arrives is Δ M . In terms of the rate of arrival, R particles per second, and the mass of each particle, m , we have Δ M = Rm Δ t Thus, in time Δ t , the mass of M changes to M + Δ M , and we would expect that its velocity changes also. Let the change – which will be small – be Δ v . Since momentum is conserved each time that a particle arrives, the momentum at time t and at time t + Δ t is the same: Δ Mv + Mu = (

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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.

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Moving Target - M Δ M u Δ u Multiplying out and canceling...

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