Answer to Question #350359 in Classical Mechanics for Oshada

Question

A train with a mass of 500 metric tons travels at constant velocity on a horizontal line and unleashes its last compartment which is with a mass of 20 metric tons. If the engine is turned off after moving another 240 m, what would be the distance between the last compartment and the train when it stops? Note that the force generated by the engine remains unchanged throughout the motion and the resistance force applied to each part is proportional to the mass.

Answer 1

Given:

$M_0=500$ metric tons;

$M=20$ metric tons;

$d_0=240$ m.

Find:

$d$ - distance between last compartment and train after they stop.

Solution:

Let's denote some parameters:

$d_1$ - distance covered by last compartment after unleash until stop;

$d_2$ - distance covered by train after tuned off engine until stop. So,

$d=(d_2-d_1)+d_0;$

$d_2=vt_2-\frac{at_2^{2}}{2};$ here $v$ - speed of train just after turned off engine; $a$ - deceleration of train after turned off engine; if train stops after some time:

$t_2=\frac{v}{a};$ So,

$d_2=\frac{v^{2}}{2a};$ Now find $d_1$;

$d_1=v_0t_1-\frac{at_1^{2}}{2};$ here $v_0$ - speed of the last compartment just after unleash, $a$ - deceleration of the last compartment, which the same with deceleration of train after turned off engine, Because:

$Ma=kMg;$ $a=kg$ - for last compartment, here $k$ - friction coefficient. And for train:

$(M_0-M)a=k(M_0-M)g;$ $a=kg$ - for train after turned of engine.

So,

$t_1=\frac{v_0}{a};$ from here:

$d_1=\frac{v_0^{2}}{2a};$

$d_2-d_1=\frac{v^{2}}{2a}-\frac{v_0^{2}}{2a}=\frac{1}{2a}\times(v^{2}-v_0^{2});$ (1)

Let's denote $(v^{2}-v_0^{2})=\delta$ and find;

$d_0=v_0t+\frac{a_1t^{2}}{2};$ here $a_1$ - acceleration of train from just after unleash least compartment until turned off engine;

$t=\frac{v-v_0}{a_1};$ from here:

$d_0=\frac{v^{2}-v_0^{2}}{2a_1}=\frac{\delta}{2a_1};$ so, from this formula we will put $\delta$ above (1) formula and take:

$d_2-d_1=d_0\times\frac{a_1}{a};$ (2) Now we need to find $a_1$ and $a$ .

Before unleash least compartment the train had constant speed, it means magnitude engine force was equal to friction force action to train. So,

$F_e=kM_0g;$

For least compartment after unleash: $Ma=kMg;$ $a=kg;$

For train train from just after unleash least compartment until turned off engine:

$(M_0-M)a_1=F_e-kg(M_0-M);$

$a_1=kg\times\frac{M}{M_0-M};$

Now, we will put $a$ and $a_1$ (2) formula:

$d_2-d_1=d_0\times\frac{M}{M_0-M};$ So, finally:

$d=d_2-d_1+d_0=d_0\times\frac{M}{M_0-M}+d_0=d_0\times\frac{M_0}{M_0-M};$

$d=d_0\times\frac{M_0}{M_0-M}=240m\times\frac{500}{500-20}=250m;$

Answer:

$d=250m.$