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Not quite... you forgot a factor.Davidllerenav said:Ek=mΔv2
Not quite... you forgot a factor.Davidllerenav said:Ek=mΔv2
Oh, yes the ##\frac{1}{2}##, so it is ##E_k=\frac{1}{2}m\Delta v##.haruspex said:Not quite... you forgot a factor.
Now you forgot the square.Davidllerenav said:Oh, yes the ##\frac{1}{2}##, so it is ##E_k=\frac{1}{2}m\Delta v##.
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.haruspex said:Now you forgot the square.
I'm sorry to say that's another arithmetic error.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 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.haruspex said:I'm sorry to say that's another arithmetic error.
Right. And you can simplify slightly combining the 1/2 with the 4.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##.
Yes, so simplifying I end up with ##E_k=m(0.0002)v_0^2##. And that's all, right?haruspex said:Right. And you can simplify slightly combining the 1/2 with the 4.
Perfect.Davidllerenav said:Yes, so simplifying I end up with ##E_k=m(0.0002)v_0^2##. And that's all, right?
Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I?haruspex said:Perfect.
No, you already used that.Davidllerenav said:Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I?
Ok, thank you so much for your help.haruspex said:No, you already used that.
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.Davidllerenav said:Yes, so simplifying I end up with ##E_k=m(0.0002)v_0^2##. And that's all, right?
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?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##?
Yes, that is what I get as well.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?