How Does Sand Transfer Affect the Speed of a Freight Car?

[geoffrey159]

Homework Statement

(Kleppner & Kolenkow - Introduction to Mechanics - 3.12)
A sand-spraying locomotive sprays sand horizontally into a freight car situated ahead of it. The locomotive and freight car are not attached. The engineer in the locomotive maintains his speed so that the distance to the freight car is constant. The sand is transferred at a rate dm/dt = 10 kg/s with a velocity of 5 m/s relative to the locomotive. The car starts from rest with an initial mass of 2000 kg. Find its speed after 100 s.

Homework Equations

Conservation of momentum

The Attempt at a Solution

Let $x_l$ be the horizontal position of the locomotive, $x_f$ the horizontal position of the freight, and $x_s$ the horizontal position of a piece of sand about to fall into the freight.
We have the relationships:
1 - $x_f - x_l = cst$ so that the locomotive and the freight have the same speed: $v_l = v_f$
2 - $x_{s} = x_l + x_{s/l} \Rightarrow v_{s} = v_l + v_{s/l} = v_f + 5$ meters per second.

B - Momentum change
Call $m(t)$ the changing mass of the freight, and $\triangle m$ a small amount of sand falling into the freight in $\triangle t$ seconds.
$P(t) = m(t) v_f(t) + \triangle m v_s(t)$
$P(t+\triangle t) = (m(t) + \triangle m) v_f(t+\triangle t)$

Since there are no external forces:
$0 = \frac{dP}{dt} = m(t) \frac{dv_f}{dt} + \frac{dm}{dt} (v_f - v_s) = m(t) \frac{dv_f}{dt} - 5 \frac{dm}{dt}$

Using the constraint $m(t) = 2000 + 10 t$, we get
$\frac{dv_f}{dt} = \frac{ 50 } {2000 + 10 t}$

So,
$v_f(t) = v_f(t) - v_f(0) = 5 ( \ln(2000 + 10t) - \ln(2000) ) = 5 \ln(\frac{2000+ 10t}{2000 })$

And after 100 seconds, $v_f(100) = 5\ln(1.5) \approx 2$ meters per second.

Is that correct ?

 

[TSny]

Your work looks very good to me. I think your answer is correct.

 

[geoffrey159]

:-) Thank you very much