Speed of Object with 3x Rest Mass

[DaVinci]

Could someone check my solution and ensure I did this right? I am pretty sure it is... but this whole relativity thing is new to me and I want to make sure I didnt screw something up somwhere.

The question is: What is the speed of an objects whose mass is three times its rest mass.

$m = 3m_0$

I used the equation
$m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$

Using $c=3x10^8$

I got $v=1.633x10^4$ m/s.

 

[dextercioby]

The answer is wrong,very wrong...Check your numbers,once again...Post them here.

Daniel.

 

[dextercioby]

The exact answer is
$$v=\frac{2\sqrt{2}}{3}c$$

Daniel.

 

[DaVinci]

Nah, wasnt very wrong... just slightly wrong.

Even though I had the c^2 in there, I forgot to square it.

Now that I got it right, I end up with $2.828x10^8$ which equals the value of what you posted as well.

Thanks for the equation. Is that a 'standard' problem?

 

[vincentchan]

if you are in 6th grade, yes... if you are in high school, NO...
(this is the situation in hongkong, I am not sure about US... I didn't receive any high school education in US)

 

[DaVinci]

Never took physics in high school. First time I have seen anything dealing with relativity is in this class; Physics 3.

And it goes to show my Physics 1 & 2 teacher was right. He had his Undergrad and Masters in Physics in China and came to the US to get his PhD. He said his Masters in China was twice as hard as his PhD in the US.

 

[DB]

I feel stupid asking this, but how did you work out the problem?

 

[DaVinci]

Plugged $m = 3m_0$ into the equation of $m_0$ and solved for v.

If you were referring to my question in whether that was a standard equation he left, what I meant by that is if it was a simple equation where a number, say the denominator (3 in this case), could be changed to represent the relationship between mass and rest mass. Since the question was $m=3m_0$ it just looked too easy.

After deriving a few others for personal gratification, I found that it was not the case.

 

[DB]

ok...
$$m=\frac{3m_0}{\sqrt{1-v^2/9*10^{16}}}$$
then what?

 

[dextercioby]

U got them mixed up.Reread the initial problem...It will come to you...

Daniel.

 

[DaVinci]

m becomes $3m_0$. The $m_0$ on the right hand side stays the same.

Then multiply both sides by the denominator on the left hand side and them divide both sides by $3m_0$. Here the two rest masses cancel leaving you with 1/3 on the right hand side.

Then square both sides and continue to use algebra until you have solved for v.

 

[jtbell]

what I meant by that is if it was a simple equation where a number, say the denominator (3 in this case), could be changed to represent the relationship between mass and rest mass. Since the question was $m=3m_0$ it just looked too easy.

Actually the 3 in the denominator does come directly from the $$m = 3m_0$$, but the numerator is a bit messier. The general result, starting from $$m = k m_0$$, is

$$\frac {v}{c} = \sqrt {1 - \frac {1} {k^2}} = \frac {\sqrt {k^2 - 1}} {k}$$

 

[DB]

For some reason I'm really having trouble...
$$3m_0=\frac{m_0}{\sqrt{1-v^2/9*10^{16}}}$$
?
I know algebra, I just can't seem to see how to figure this problem out.

 

[DaVinci]

What do you get after you do the steps I had listed in my previous post? Let's make sure you got that far and then we will look at it from there. Post the answer you get after you square both sides.

 

[dextercioby]

Simplify through $m_{0}\neq 0$ first...

Daniel.

 

[DB]

m becomes $3m_0$. The $m_0$ on the right hand side stays the same.

Then multiply both sides by the denominator on the left hand side and them divide both sides by $3m_0$. Here the two rest masses cancel leaving you with 1/3 on the right hand side.

Then square both sides and continue to use algebra until you have solved for v.

$$3m_0=\frac{m_0}{\sqrt{1-v^2/9*10^{16}}}$$
There is no denominator or the left side?
Am I suppose to divide:
$$\frac{\frac{m_0}{\sqrt{1-v^2/9*10^{16}}}}{3m_0}$$
How do i do that?

 

[DaVinci]

Thats the hard way.

Start with the original equation. Multiply both sides of the equation by the denominator on the left hand side. This will move the entire square root to the left hand side with the 3Mo. Then, divide both sides of the equation by 3Mo. So on the right hand side you have Mo / 3Mo. The Mo cancel and leaves you with 1/3 on the RHS. Now square both sides...

Keep going.

 

[DB]

Im getting there.

$$m=\frac{m_0}{\sqrt{1-v^2/c^2}}$$

$$3m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}$$

$$3m_0*(\frac{m_0}{\sqrt{1-v^2/c^2}})=\frac{m_0}{3m_0}$$

$$\sqrt{1-v^2/c^2}=1/3$$

Im lost now

 

[DaVinci]

Simple algebra... what do you do to remove a square root sign?

 

[DB]

$$\sqrt{(1-v^2/c^2)}^2=(1/3)^2$$

$$1-v^4/c^4=1/9$$

$$-v^4/8.1*10^{33}=-8/9$$

$$-v^4=(-8/9*8.1*10^{33})=-7.2*10^{33}$$

$$v=\sqrt[4]{7.2*10^{33}$$

$$v=291295063$$

Did I do it right?

 

[DB]

5th line should say the fourth root of 7.2*10^33

 

[DaVinci]

$$1-v^4/c^4=1/9$$

That is incorrect.

$$\sqrt{(1-v^2/c^2)}^2$$

Take a look at that step...

 

[DB]

$$\sqrt{(1-v^2/c^2)}^2=(1/3)^2$$

$$1-v^2/c^2=1/9-1$$

$$-v^2/9*10^{16}=-8/9$$

$$-v^2=-8*10^{16}$$

$$v=\sqrt{8*10^{16}}$$

$$v=282842712.5$$

I assume I did it right??

 

[DaVinci]

$$1-v^2/c^2=1/9-1$$

Is an incorrect statement. You need to have this:

$$1-v^2/c^2=1/9$$

Then subtract 1 from both sides simultaniously.

You got the right answer though ... good job.

 

[DB]

O that was just a typo i meant to have $$-v^2/c^2=1/9-1$$, they both become negative and can cancel out the negatives. Thank you for your help!

 

[DB]

Is it safe to say...

After all this,
Is it safe to say that the velocity of a particle/object is equal to:

$$v=\sqrt{\mid\frac{{m_0}^2}{{X_{m_0}}^2}-1\mid c^2$$

Where $$X$$ is equal to how many times larger it's relativistic mass is compared to it's rest mass.

? just curious...