- 31,671
- 4,667
That is not the change in velocity.the change in the velocity of the second rocket, which was ##1.01v_i##?
What is √1.04?why did you said that it is rather inaccurate?
That is not the change in velocity.the change in the velocity of the second rocket, which was ##1.01v_i##?
What is √1.04?why did you said that it is rather inaccurate?
It is the new velocity. So what velocity should I use?That is not the change in velocity.
It is ##1.0198...##.What is √1.04?
Right, so what is the change in velocity?It is the new velocity.
So what would be the right way to express that to two decimal places?It is 1.0198...
it would be ##\Delta v= v_f-v_0## so the change would be ##\Delta v= 1.012v_0-v_0=0.012v_0##, right?Right, so what is the change in velocity?
1.012?Right, so what is the change in velocity?
So what would be the right way to express that to two decimal places?
No, 1.0198 would not be abbreviated to 1.012; and even if it were, subtracting 1 does not yield 0.12.it would be ##\Delta v= v_f-v_0## so the change would be ##\Delta v= 1.012v_0-v_0=0.12v_0##, right?
1.012?
Then what would it be?No, 1.0198 would not be abbreviated to 1.012.
Yes sorry, I meant 0.012.and even if it were, subtracting 1 does not yield 0.12.
You will kick yourself when you realise the error you are making.Then what would it be?
Oh, I think I see it know, woudl it be 1.02?Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.
Yes.Oh, I think I see it know, woudl it be 1.02?
So ##\Delta v=0.02## and the kinetic energy of the second rocket, the one that was at rest is ##E_k=m\Delta v^2= m(0.02)^2v_0^2##? Is t right?Yes.
Not quite... you forgot a factor.E_{k}=mΔv^{2}
Oh, yes the ##\frac{1}{2}##, so it is ##E_k=\frac{1}{2}m\Delta v##.Not quite... you forgot a factor.
Now you forgot the square.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.Now you forgot the square.
I'm sorry to say that's another arithmetic error.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.I'm sorry to say that's another arithmetic error.
Right. And you can simplify slightly combining the 1/2 with the 4.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?Right. And you can simplify slightly combining the 1/2 with the 4.
Perfect.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?Perfect.
No, you already used that.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.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.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?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.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?