Quick Q (energy) (1 Viewer)

[bunsniper]

http://puu.sh/9NpAQ/83c4304345.jpg

Thanks!

 

[Demento1]

Is the answer B? I am not sure.

 

[bunsniper]

My study group has been talking about it and we are not sure to take Energy as a vector or not..

 

[Squar3root]

Its B. Both energies tend towards zero as r gets big

 

[Squar3root]

My study group has been talking about it and we are not sure to take Energy as a vector or not..

Not sure how this is relevant

 

[Demento1]

My study group has been talking about it and we are not sure to take Energy as a vector or not..

I'm treating it as the rocket being launched infinitely far away (as r increases) and so, kinetic energy (ie it is decreasing) should be inputted to give the rocket its gravitational potential energy (ie which is increasing) by the conservation of energy.

 

[Squar3root]

I'm treating it as the rocket being launched infinitely far away (as r increases) and so, kinetic energy (ie it is decreasing) should be inputted to give the rocket its gravitational potential energy (ie which is increasing) by the conservation of energy.

But the question says it is being launched from the surface of the earth

 

[Demento1]

But the question says it is being launched from the surface of the earth

You're right. My mistake!

 

[someth1ng]

The answer is A.

 

[Squar3root]

The answer is A.

Why does kinetic energy increase in an unbounded manner?

 

[Demento1]

The answer is A.

Could you explain why it is A please? It would be greatly appreciated!

 

[someth1ng]

Why does kinetic energy increase in an unbounded manner?

It says that "the rocket's engines are fired continuously" as you go further away from the Earth.

Consider: E=1/2 mv² and F=ma

As the rocket is flying:
m decreases and hence, a increases as F should be relatively constant (based on F=ma).

As a is increasing, and already positive:
v also increases at an increasing rate - and it's squared! So you end up with the curve shown in A.

As distance increases, E_p increases which doesn't really involve any more maths apart from the given formulae in the formula sheet.

 

[Squar3root]

It says that "the rocket's engines are fired continuously" as you go further away from the Earth.

Consider: E=1/2 mv² and F=ma

As the rocket is flying:
m decreases and hence, a increases as F should be relatively constant (based on F=ma).

As a is increasing, and already positive:
v also increases at an increasing rate - and it's squared! So you end up with the curve shown in A.

As distance increases, E_p increases which doesn't really involve any more maths apart from the given formulae in the formula sheet.

Oops; didn't read the "continuously" part. Makes sense now, thanks!

 

[anomalousdecay]

The answer is A.

That's exactly what I thought. Kinetic energy increases as the velocity increases (Ek = (mv^2)/2 and Ep = -GmM/r so if you get further away then it tends to zero from the negative.

Also, if you add up the two energies, you will find a steady increase in the total energy of the rocket, and this is due to the chemical energy of the fuel.

Why does kinetic energy increase in an unbounded manner?

BELOW IS NOT HSC LEVEL OR TERMINOLOGY:

As I stated above, mechanical energy is not conserved because of the chemical energy of the rocket (which isn't non-conservative).

So it doesn't make sense for the total change in energy (derivatives) to add up to zero on the graphs either.

 

[Squar3root]

BELOW IS NOT HSC LEVEL OR TERMINOLOGY:

As I stated above, mechanical energy is not conserved because of the chemical energy of the rocket (which isn't non-conservative).

So it doesn't make sense for the total change in energy (derivatives) to add up to zero on the graphs either.

Ohh yeah: deltaE = deltaM + deltaP because it is a nonconserative force. Soz 1121 slipped straight our my head momentarily after the exam

 

[bunsniper]

Thanks guys!

 

[anomalousdecay]

Ohh yeah: deltaE = deltaM + deltaP because it is a nonconserative force. Soz 1121 slipped straight our my head momentarily after the exam

I've never used p before and initially I was like "w0t do you mean by momentum it should be U".

Then again we use so many damn notations in phys its all relative haha.

Well the fuel is adding a non-conservative force to the system, so M is not conserved.

Inb4 I fail phys and you pass.

(NOTE NONE OF THE ABOVE IS HSC IN THIS POST lol).

 

[Squar3root]

I've never used p before and initially I was like "w0t do you mean by momentum it should be U".

Then again we use so many damn notations in phys its all relative haha.

Well the fuel is adding a non-conservative force to the system, so M is not conserved.

Inb4 I fail phys and you pass.

(NOTE NONE OF THE ABOVE IS HSC IN THIS POST lol).

i just got stuck the way wolfe started writing it. I used to use U because it looked more nicer but the tutor was also like "nahh don't, caz it don't look nice". i just want to get good marks lol. inb4failphysandrepeatagainnextsem

also keep in mind that d/dv(0.5*mv^2) = mv which is momentum

 

[anomalousdecay]

i just got stuck the way wolfe started writing it. I used to use U because it looked more nicer but the tutor was also like "nahh don't, caz it don't look nice". i just want to get good marks lol. inb4failphysandrepeatagainnextsem

also keep in mind that d/dv(0.5*mv^2) = mv which is momentum

I don't need to keep that in mind for anything lol I use whatever notation I want.

If the markers don't like my notation then they should do a course on understanding other people's notations because there are many notations out there for everything.

 

[Squar3root]

I don't need to keep that in mind for anything lol I use whatever notation I want.

If the markers don't like my notation then they should do a course on understanding other people's notations because there are many notations out there for everything.

good point. i can just imagine the shear outrage when people are marked down for using a different letter to the one the professor likes