# Rain in a train wagon - speed, momentum and KE

1. May 28, 2006

### Bonulo

We have a physics exam tomorrow, and have been presented with a previous exam for practice. One of the assignments have puzzled us quite a bit. Here goes:

Rain in a train wagon
Consider a train wagon whose upper part is a lidless box. The train wagon moves without friction. There is water in the train wagon. In the following the system consisting of the train wagon and the water in it is considered. In 3 different situations (i, ii and iii) decide what happens with each of the three variables speed $v$, momentum $p$ and the kinetic energi $K$.

(i) It's raining vertically, and water is accumulated in the wagon.
(ii) As in (i) with the addition that there's now a hole in the bottom where the water's running out. The amount of water running out of the wagon is equal to the amount of water raining into the wagon.
(iii) The rain stops, but there is still a hole in the bottom of the wagon, where the water's running out.

Since we've been given the results (the correct answers, no explanations), I'm not interested in that information. Only the explanation.

(i) Result: Momentum is conserved, $p$ is not changed. I guess the system is a closed system. The speed $v$ decreases - because mass $m$ increases. Thus the kinetic energy $K$ drops too, since $K$ is proportional to the square of $v$.

(ii) Result: All three variables, $p, v, K$ drops. It's clear that since the mass is the same, a speed decrease causes the momentum and energy decrease. But why does the speed $v$ drop?

(iii) Result: Here, water is running out, and thus the mass $m$ decreases. This causes a decrease in $p$ and $K$, since the speed $v$ doesn't change. But why does the speed $v$ change in (i) and not in (iii) ?

2. May 28, 2006

### Andrew Mason

i) $$m_{car}v_i = m_{car+rain water}v_f$$

Since $$m_{car} < m_{car+rain water}$$,

$$v_f < v_i$$

ii) Let $v_{i2} = v_f$ from i) above.

$$m_{car+water}v_{i2} = m_{car+water-escaped water}v_{f2} + m_{escaped water}v_{f2}$$

Since $$m_{car+water} = m_{car+water-escaped water} + m_{escaped water}$$

$$v_{f2} = v_{i2}$$

iii) Same as ii) but since the car is not slowing down due rain capture, it keeps going at the same rate.

AM

3. May 28, 2006

### Bonulo

What I wrote to ii) was the result as stated in the solutions we got. There, the speed decreases, as does the momentum and kinetic energy.

4. May 28, 2006

### Staff: Mentor

It should be no surprise that the speed (and thus momentum and kinetic energy) decreases in case ii): for every $\Delta m$ of zero speed rain that you add, you lose $\Delta m$ of $v_i$ speed rain. Thus, the momentum of car+water constantly decreases.

I'd say that after gaining/losing $\Delta m$ of rain/water:
$$m_{car+water}v_{f} = m_{car+water-escaped water}v_{i}$$

Thus:
$$v_{f} < v_{i}$$

5. May 28, 2006

### Bonulo

Yup, that's more like it! Or, it's exactly it, rather.

We did have some notion about this actually, though not a clarified one.

6. May 28, 2006

### Andrew Mason

That is what I said. In ii) the problem is the same as in i) with the added hole in the bottom letting water out. The added hole in the bottom does not change the velocity. Break it into two parts: 1. rain is captured - this is the same as i) and the speed slows ($v_f<v_i$. You then take that reduced speed ($v_{i2} = v_f$) and see if the escape of water changes it. It doesn't. So the car keeps that slower speed $v_{i2} = v_f < v_i$.

AM

7. May 29, 2006

### Staff: Mentor

Sorry if I misinterpreted what you had said. (I guess I didn't understand your notation.)