Physics Question

aleatorio

Im having a lot of trouble working this conservation of momentum question out, I seem to be overlooking something simple Id say Can anyone give me a hand?

A space shuttle of mass 30000kg travelling at 10ms-1 approaching a docking bay needs to readjust its direction to land in the docking bay as shown in the diagram. The space shuttle can expel gas either out the front, back or perpendicularly at either side.
The space shuttle expels 50kg of gas perpendicularly to one side. With what speed and in what direction should the gas be expelled in order for the space shuttle to dock successfully with the docking station? (The loss in mass of the space shuttle due to the expulsion of gas may be ignored)

The diagram shows the space shuttle positioned 100m away from the docking station (to the left) and 8m above it. Kind of like a right angled triangle.

(SS)

---

10m

---

...................................|8m
...................................|(Dock)

Peregrine

If it's from a book or an exam paper, which one?

Having some trouble picturing what they're saying :pac:

aleatorio

Nimrod 7 wrote: »

If it's from a book or an exam paper, which one?

Having some trouble picturing what they're saying :pac:

A mock paper, but I don't have the actual diagram here with me, I just took it all down off the board - the text is word for word from the question

LoveLamps

I'm afraid I can't be of help, I think that was on our mock - not a single person in the class attempted it!

Mr Pseudonym

The question is incredibly unclear! We know that, as it has to "readjust" its course, the velocity is not in the direction of the bay. It doesn't give the conditions necessary for docking. Must it be stationary at the point of docking? Also, in space there are no reference points, so the shuttle can't be "above" the bay.

The fact that it says it can expel gas either front, back or sideways (grammatically incorrect, because "either" should only be used when there are two choices) implies that only one expulsion of gas occurs. As this happened sideways, it does not seem necessary to cancel the forward momentum - merely change shuttle's direction. Which gives rise to another problem: in which direction is the velocity? Using the diagram in the OP, along the ten-metre side of the triangle, or along the eight-metre one?

Maybe I've missed something. I'll have a look again in a while.

Edit: it asks in what direction the gas was perpendicularly expelled. Do they mean which side?

aleatorio

Mr Pseudonym wrote: »

The question is incredibly unclear! We know that, as it has to "readjust" its course, the velocity is not in the direction of the bay. It doesn't give the conditions necessary for docking. Must it be stationary at the point of docking? Also, in space there are no reference points, so the shuttle can't be "above" the bay.

The fact that it says it can expel gas either front, back or sideways (grammatically incorrect, because "either" should only be used when there are two choices) implies that only one expulsion of gas occurs. As this happened sideways, it does not seem necessary to cancel the forward momentum - merely change shuttle's direction. Which gives rise to another problem: in which direction is the velocity? Using the diagram in the OP, along the ten-metre side of the triangle, or along the eight-metre one?

Maybe I've missed something. I'll have a look again in a while.

Edit: it asks in what direction the gas was perpendicularly expelled. Do they mean which side?

Yes, the initial velocity is along the 10m side :P

I agree that it's a horribly badly phrased question

I think they mean which side but to be honest I haven't a clue

Mr Pseudonym

Then I think it's quite an easy question (though very unrealistic).

These questions always require that you examine momentum. Based on your diagram:

Momentum in x-direction = (30 000)(10)
Momentum in y-direction = |(50)(V)|

Taking the shuttle as a single point, it is required that it travel in a direction -arctan(8/10) (to the x-axis).

So... [Latex] \displaystyle \frac{(50)(V)}{(30 000)(10)} = \frac{8}{10}[/Latex]

V should be easy to get from there.

I'm not particularly comfortable that that is entirely right! But, it's what I would submit.

Edit: obviously the gas is expelled on the left-side of the shuttle.

krekshi

Is this question from the DEB physics higher exam?

aleatorio

krekshi wrote: »

Is this question from the DEB physics higher exam?

Yeah, I think so