How to find the energy of an object that was at rest?

In summary: Then the momentum of the gas should be the same as the momentum...of the rocket?No, the momentum of the gas is not the same as the momentum of the rocket.
  • #36
Davidllerenav said:
Ek=mΔv2
Not quite... you forgot a factor.
 
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  • #37
haruspex said:
Not quite... you forgot a factor.
Oh, yes the ##\frac{1}{2}##, so it is ##E_k=\frac{1}{2}m\Delta v##.
 
  • #38
Davidllerenav said:
Oh, yes the ##\frac{1}{2}##, so it is ##E_k=\frac{1}{2}m\Delta v##.
Now you forgot the square.
 
  • #39
haruspex said:
Now you forgot the square.
Sorry... ##E_k=\frac{1}{2}m(\Delta v)^2= \frac{1}{2}m(0.02)^2v_0^2=\frac{1}{2}m(0.004)v_0^2##. It think it is correct know.
 
  • #40
Davidllerenav said:
Sorry... ##E_k=\frac{1}{2}m(\Delta v)^2= \frac{1}{2}m(0.02)^2v_0^2=\frac{1}{2}m(0.004)v_0^2##. It think it is correct know.
I'm sorry to say that's another arithmetic error.
 
  • #41
haruspex said:
I'm sorry to say that's another arithmetic error.
I see, I made a mistake when I squared 0.02, it is 0.0004, not 0.004. so ##E_k=\frac{1}{2}m(0.0004)v_0^2.
 
  • #42
Davidllerenav said:
I see, I made a mistake when I squared 0.02, it is 0.0004, not 0.004. so ##E_k=\frac{1}{2}m(0.0004)v_0^2##.
Right. And you can simplify slightly combining the 1/2 with the 4.
 
  • #43
haruspex said:
Right. And you can simplify slightly combining the 1/2 with the 4.
Yes, so simplifying I end up with ##E_k=m(0.0002)v_0^2##. And that's all, right?
 
  • #44
Davidllerenav said:
Yes, so simplifying I end up with ##E_k=m(0.0002)v_0^2##. And that's all, right?
Perfect.
 
  • #45
haruspex said:
Perfect.
Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I?
 
  • #46
Davidllerenav said:
Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I?
No, you already used that.
 
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  • #47
haruspex said:
No, you already used that.
Ok, thank you so much for your help.
 
  • #48
Davidllerenav said:
Yes, so simplifying I end up with ##E_k=m(0.0002)v_0^2##. And that's all, right?
There is a problem with this result. Neither ##m## nor ##v_0## are given in the problem description. Instead, what is given is ##E_0##, the initial energy of the moving rocket.

Can you express the result you have in terms of ##E_0## rather than in terms of ##m## and ##v_0##?
 
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  • #49
jbriggs444 said:
There is a problem with this result. Neither ##m## nor ##v_0## are given in the problem description. Instead, what is given is ##E_0##, the initial energy of the moving rocket.

Can you express the result you have in terms of ##E_0## rather than in terms of ##m## and ##v_0##?
Well, since ##E_0=\frac{1}{2}m v_0^2##, I think I can express the result as ##E_k=(0.0004)E_0## right?
 
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  • #50
Davidllerenav said:
Well, since ##E_0=\frac{1}{2}m v_0^2##, I think I can express the result as ##E_k=(0.0004)E_0## right?
Yes, that is what I get as well.
 

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