A railroad car of length L and mass m 0 when emptyis moving freely on a horizont

Question

A railroad car of length L and mass m0 when emptyis moving freely on a horizontal track while being loaded with sandfrom a stationary chute at a rate dm/dt = q. Knowing that the carwas approaching the chute at a speed v0, determine (a)the mass of the car and its load after the car has cleared thechute, (b) the speed of the car at that time.

Explanation / Answer

Let at any instant while clearing the chute , the speed & massof the railroad car be v & m respectively. So in a infinitesimally small time dt , amount of extra mass added(mass of falling sand) = q dt. change in speed be dv. (dv is negative as velocity is decreasingwith time) By conservation of linear momentum , we get , mv = (m + qdt)(v+ dv) expanding, mv = mv +mdv +qvdt +qdvdt neglecting the last term( as dv , dt being very small , dvdt willbe 0) mdv= -qvdt now m0 + qt = m (t being time elapseduptil now) so( m0 + qt) dv = -qvdt so , dv/v = -qdt/ (m0 + qt) integrating both sides ln (vfinal / vinitial) = -ln((m0 + qt)/m0 ) (given v =v0 at t =0) => vfinal =v0m0/(m0 +qt)--------------- equation(1) Or more generally v= v0 m0/(m0 +qt) (atany time t during which railroad car is under hopper) so dx/dt = v0 m0/(m0 +qt) integrating both sides after taking dt to right hand side of theequation , or, x = (v0 m0 ln((m0 +qt)/ m0 ) ) / q (as x=0 at t=0 , hence thelimits) L = (v0 m0 ln((m0 +qt)/ m0 ) ) / q or,m0 e^(qL/(v0 m0 )) =m0 +qT calculating T i.e the time taken to traverse the hopper T=(m0 e^(qL/(v0 m0 ))-m0 ) / q putting T in equation (1) vfinal = v0 e^-(qL/(v0m0 )) mfinal = m0 e^(qL/(v0m0 ))

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