# Homework Help: Relativistic Mass

1. Jan 20, 2005

### 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.

2. Jan 20, 2005

### dextercioby

Daniel.

3. Jan 20, 2005

### dextercioby

$$v=\frac{2\sqrt{2}}{3}c$$

Daniel.

4. Jan 20, 2005

### 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?

5. Jan 20, 2005

### vincentchan

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

6. Jan 20, 2005

### 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.

7. Jan 20, 2005

### DB

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

8. Jan 20, 2005

### 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.

9. Jan 20, 2005

### DB

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

10. Jan 20, 2005

### dextercioby

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

Daniel.

11. Jan 20, 2005

### 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 devide 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.

12. Jan 20, 2005

### Staff: Mentor

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

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

13. Jan 20, 2005

### 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.

14. Jan 20, 2005

### DaVinci

What do you get after you do the steps I had listed in my previous post? Lets 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.

15. Jan 20, 2005

### dextercioby

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

Daniel.

16. Jan 20, 2005

### DB

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

17. Jan 20, 2005

### 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.

18. Jan 20, 2005

### 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

19. Jan 20, 2005

### DaVinci

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

20. Jan 20, 2005

### 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?