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# Help w/ Momentum question Watch

1. Hi can someone explain how to do part (i) please? I tried using the principle of momentum but as you can see from what I have done so far, the velocity will cancel itself out, so i've probably done something wrong.

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2. Is the initial velocty of the astronaut zero? She's moving along with the spacecraft before moving off, so both need to have the same initial velocity.
3. (Original post by Pangol)
Is the initial velocty of the astronaut zero? She's moving along with the spacecraft before moving off, so both need to have the same initial velocity.
So would you be left with 130V = (130 X 1.8) which then gives the velocity as 1.8 ms-1?
4. (Original post by jessyjellytot14)
So would you be left with 130V = (130 X 1.8) which then gives the velocity as 1.8 ms-1?
Common sense tells us that this can't be right - the spacecraft wouldn't move off with the same speed as the astronaut.

If the spacecraft is moving at a constant speed, then Newton's first law tells us that there is no resultant force acting on it. So let's regard it as stationary - that is, the spacecraft and the astronaut can both be taken to have an initial velocity of zero. (If you don't like this, then everything will work out if you assume that they both have an initial velocity of v, but the maths is a bit more awkward.)

There are a couple of problems with your first way of doing this. One is that you are treating the spacecraft as having a different initial velocity to the astronaut. But the other is that you are using the same symbol, v, for the velocity of the spacecraft before and after the astronaut pushes off.

Since momentum is conserved, it is easier to think about this as the change of momentum of the astronaut being equal to the change of momentum of the spacecraft, so that the total change in momentum is zero.
5. (Original post by Pangol)
Common sense tells us that this can't be right - the spacecraft wouldn't move off with the same speed as the astronaut.

If the spacecraft is moving at a constant speed, then Newton's first law tells us that there is no resultant force acting on it. So let's regard it as stationary - that is, the spacecraft and the astronaut can both be taken to have an initial velocity of zero. (If you don't like this, then everything will work out if you assume that they both have an initial velocity of v, but the maths is a bit more awkward.)

There are a couple of problems with your first way of doing this. One is that you are treating the spacecraft as having a different initial velocity to the astronaut. But the other is that you are using the same symbol, v, for the velocity of the spacecraft before and after the astronaut pushes off.

Since momentum is conserved, it is easier to think about this as the change of momentum of the astronaut being equal to the change of momentum of the spacecraft, so that the total change in momentum is zero.
I don't have common sense when it comes to physics lol. But what you've said does make sense.
Would doing 0 = (130 X 1.8) + (4.0 X 104)V and then solving for V give the correct answer?

Edit: That gives V= 0.00585 ms-1. That doesn't seem right.

Edit 2: Oh wait are you supposed to use the mv - mu thing?
6. (Original post by jessyjellytot14)
I don't have common sense when it comes to physics lol. But what you've said does make sense.
Would doing 0 = (130 X 1.8) + (4.0 X 104)V and then solving for V give the correct answer?

Edit: That gives V= 0.00585 ms-1. That doesn't seem right.

Edit 2: Oh wait are you supposed to use the mv - mu thing?
You would expect the velocity of the spaceship to be much smaller than the velocity of the astronaut, considering that they have the same change in momentum and the spacecraft is much more massive than the astronaut. So this is correct.
7. (Original post by Pangol)
You would expect the velocity of the spaceship to be much smaller than the velocity of the astronaut, considering that they have the same change in momentum and the spacecraft is much more massive than the astronaut. So this is correct.
Oh okay, thank you!

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