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

[Davidllerenav]

No, 1.0198 would not be abbreviated to 1.012.

Then what would it be?

and even if it were, subtracting 1 does not yield 0.12.

Yes sorry, I meant 0.012.

 

[haruspex]

Then what would it be?

You will kick yourself when you realize the error you are making.
Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.

 

[Davidllerenav]

Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.

Oh, I think I see it know, woudl it be 1.02?

 

[haruspex]

Oh, I think I see it know, woudl it be 1.02?

Yes.

 

[Davidllerenav]

Yes.

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?

 

[haruspex]

E_k=mΔv²

Not quite... you forgot a factor.

 

[Davidllerenav]

Not quite... you forgot a factor.

Oh, yes the $\frac{1}{2}$, so it is $E_k=\frac{1}{2}m\Delta v$.

 

[haruspex]

Oh, yes the $\frac{1}{2}$, so it is $E_k=\frac{1}{2}m\Delta v$.

Now you forgot the square.

 

[Davidllerenav]

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.

 

[haruspex]

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.

 

[Davidllerenav]

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.

 

[haruspex]

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.

 

[Davidllerenav]

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?

 

[haruspex]

Yes, so simplifying I end up with $E_k=m(0.0002)v_0^2$. And that's all, right?

Perfect.

 

[Davidllerenav]

Perfect.

Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I?

 

[haruspex]

Ok, thanks. I don't need to write the equations of the momentum of both rocket, do I?

No, you already used that.

 

[Davidllerenav]

No, you already used that.

Ok, thank you so much for your help.

 

[jbriggs444]

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 $v_0$?

 

[Davidllerenav]

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?

 

[jbriggs444]

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.