NLM BYU 43 - Physics.lt

Solution 43 – Decoupled Train

Solution

Let the resistance offered to a part of the train of mass $\mu$ be $$R=k\mu,$$ where $k$ is a constant. Thus, every freely moving part of the train has the same retardation $k$.

1. Before the carriage gets decoupled

The complete train of mass $M$ is moving with uniform velocity $v_0$. Hence the engine force $F_0$ exactly balances the resistance:

$$F_0-kM=0$$ $$\boxed{F_0=kM}$$

2. Just after decoupling

← $km$

Detached carriage
mass $m$

← $k(M-m)$

Engine + train
mass $M-m$

$F_0$ →

The detached carriage is acted upon only by resistance, so its retardation is

$$a_{\text{rear}}=\frac{km}{m}=k.$$

The front portion still experiences the same engine force $F_0$. Therefore its acceleration is

$$a_{\text{front}} =\frac{F_0-k(M-m)}{M-m}.$$

Using $F_0=kM$,

$$a_{\text{front}} =\frac{kM-k(M-m)}{M-m} =\frac{km}{M-m}.$$

3. Motion until the driver notices

The driver notices the decoupling after the front part has travelled a distance $l$. Let its speed at this instant be $v_1$.

Using $v^2-u^2=2as$,

$$v_1^2-v_0^2 =2\left(\frac{km}{M-m}\right)l.$$

$$\boxed{\frac{v_1^2-v_0^2}{2k} =\frac{ml}{M-m}} \qquad (1)$$

4. After the engine is switched off

Once the engine is switched off, the front portion also slows down solely due to resistance. Its retardation is therefore also $k$.

Detached carriage: Starting with speed $v_0$, its total distance travelled from the point of decoupling until it stops is

$$s_1=\frac{v_0^2}{2k}.$$

Front portion: It first travels $l$ while the engine remains on, and then travels

$$s_2=\frac{v_1^2}{2k}$$

after the engine is switched off.

Hence the final separation between the two parts is

$$s=l+s_2-s_1$$ $$s=l+\frac{v_1^2-v_0^2}{2k}.$$

Using equation (1),

$$s=l+\frac{ml}{M-m}$$ $$s=l\left(1+\frac{m}{M-m}\right)$$ $$\boxed{s=\frac{Ml}{M-m}}.$$

Substituting

$$M=1000\text{ ton},\qquad m=200\text{ ton},\qquad l=100\text{ m},$$ $$s= \frac{1000\times100}{1000-200} =125\text{ m}.$$

$$\boxed{s=125\ \text{m}}$$

Notice that the initial speed $v_0$ and the resistance constant $k$ both cancel out. Hence the final separation depends only on the two masses and the distance travelled before the driver notices the decoupling.