Distance between train and its carriage

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Let's say that the last carriage gets uncoupled at time t = 0s and at this time the train has moved a distance ##l_0##.The resistive force could be given as ## F_r = AM##, where A is the appropriate constant and M is the mass of the system.

The pulling force is constant and since the train before getting uncoupled was moving with constant velocity, ## F_p = MA##.

After uncoupling,

The resistive force on the train gets reduced by mA, but the pulling force remains same. So, it has an acceleration ## a_t = \frac { mA } { M-m } ##.While the carriage has an acceleration ## a_c = -A## .

Let's say that the train stops at time t. According to the question, by this time, the train has moved a distance l.

So,

## l = l_0 +vt+ \frac { mA } {2( M-m) } t^2## ……….(1)

Let's say that the carriage stops at time ##t_1## after traveling a distance ##l_c ## after t=0s.

## l_c = l_0 + vt_1 - \frac { A{t_1}^2 } { 2 } ##...(2)

## 0 = v– At_1##...(3)

## t_1 = \frac {v } {A } ## ……….(4)

Putting (4) into (2), I will get ##l_c## in terms of A.

And I have l in terms of A and t.

So, I need one more equation to calculate d in terms of known quantities.

What to do now?
I have to calculate ## d = l-l_c##.

 

[haruspex]

at this time the train has moved a distance ##l_0##.

Might as well make that 0.

Let's say that the train stops at time t. According to the question, by this time, the train has moved a distance l.

No, the engine is shut off after distance l. It decelerates from there at the same rate as the carriage (unless that has already stopped).

 

[Pushoam]

Might as well make that 0.

No, the engine is shut off after distance l. It decelerates from there at the same rate as the carriage (unless that has already stopped).

So, the stopping of engine does not mean the stopping of train.
Using ## s = s_0 + vt + \frac { at^2 } { 2 } ## made the calculation too long.

So, I thought of using ## v^2 = u^2 + 2as##. Is there anyway to know it beforehand that which eqn I should use for a given problem.

The distance traveled by the carriage after uncoupling, ## l_c = \frac { v^2 } {2A } ##.

The speed of the train at the time of the stopping of engine ## v_t = \sqrt{ \left( v^2 + \frac {2ml } { M-m} \right)}##.

The distance traveled by the train after the engine stops, ## l_e = \frac { \left( v^2 + \frac {2ml } { M-m} \right) } { 2A } ##.

So, ## d = l + l_e – l_c = l\{ 1+\frac {2m } { A(M-m)}\}##Still, the value of A is unknown.

Taking A = 2, ## d= \frac {Ml } { M-m}## , which is the option(b).

Using, common sense, the distance has to be greater than and only the option (b) shows this.

Is this correct?

 

[haruspex]

which is the option(b).

That is what I got.
The result is so much simpler than the equations along the way I feel there must be an easier approach.

 

[Pushoam]

That is what I got.

Did you solve using some other approach?
Or you also took A = 2 or simply eliminated the unreasonable options?

 

[haruspex]

Did you solve using some other approach?

No, it was essentially the same as yours.

 

[Pushoam]

Thanks haruspex.
It feels so good to study with your help.
Thanks a lot.