Spaceship constant speed problem

[student07]

Homework Statement

The proper length of a spaceship A is 60 m and the proper length of spaceship B is 120 m. The proper mass of spaceship A is 15000 kg. An observer on Earth watches the two spaceships fly past at a constant speed and determines that they have the same length. If the speed on the slower spaceship is 0.70 c, find:
a) the length of spaceship A, relative to an observer on earth.
b) the speed of spaceship B, relative to an observer on earth.
c) the mass of spaceship A, relative to an observer on earth.

2. The attempt at a solution
a) Lm = Ls√(1- V^2 / C^2)
= 60√(1- (0.70 c) ^2 / C^2
= 60√(1-0.49) = 42.85 = 43 m
b) I'm not sure how to solve this one, I have seeing some suggestions like:
120 / 43 = 2.8 but I don't see the connection
also √{1 - (v/c)²}x120, giving v=0.9341c. but the result makes no sense
any help or equation that I can use would be appreciated

c) Mm = Ms / √(1- V^2 / C^2)
= 15000 kg / √(1-0.70 c)^2 / C^2 = 21004 = 2.1 x 10^4 kg

 

[ecastro]

For b.), you just need to solve for $v$ by replacing the $L_{\text{moving}}$ by your calculated length of Spaceship A. I think it should be $\frac{43 \text{ m}}{120 \text{ m}}$.

 

[student07]

So 43 / 120 = 0.36
0.36 = 120√(1- V^2 / C^2) = 0.99 c

I'm still thinking it should be, 120 / 43 = 2.8
2.8 = 120√(1- V^2 / C^2) = 0.98 c

 

[gneill]

It would be better to work with symbols first rather than plugging in numbers. That way the relationships of the values involved will be more obvious.

Suppose the observed length of the two ships is $L$ (and is 43 m as you calculated). The proper length of the second ship (120m) is $L_B$. Write the expression relating $L$ to $L_B$.

 

[student07]

60√(1-0.49) = 42.85 = 43 m = 120√(1- V^2 / C^2) = 0.64 c is what I get now but shouldn't ship b be faster than ship A?

 

[gneill]

Can you show some details of your work? How did you rearrange the equation to find the velocity? I ask because you've found a resulting speed for the second ship that is smaller than that of the first ship (0.70 c), yet the second ship's length has contracted by more, right?

By the way, you'll want to avoid using the equal sign between items that clearly aren't equal! You've effectively equated a velocity to a length, and while I get that you mean 0.64 c results from the expression, a marker will be entirely unforgiving!

 

[student07]

well 43 m = 120√(1- V^2 / C^2)
then, 43 = 120√(1- V^2 ) - c^2 should be canceled like in part a)
43 = 120(1- V) - square root and ^2 should cancel each other
43 / 120 = (1-V) and finally v = 1 - (43/120) = 0.64 c
since 0.64 is smaller than 0.70 c that's why I said before that it should be 120 / 43 = 2.8
then 2.8 = 120√(1- V^2 / C^2) = 0.98 c

 

[gneill]

well 43 m = 120√(1- V^2 / C^2)
then, 43 = 120√(1- V^2 ) - c^2 should be canceled like in part a)

What you mean is, you've decided to use the fractional value of c for the velocity. Fine.

43 = 120(1- V) - square root and ^2 should cancel each other

Noooo! You can't cancel the square on a term inside the square root unless it's the only term inside. $\sqrt{1 - V^2} \ne (1 - V)$. Test it with some arbitrary value for V.

Try again!

43 / 120 = (1-V) and finally v = 1 - (43/120) = 0.64 c
since 0.64 is smaller than 0.70 c that's why I said before that it should be 120 / 43 = 2.8
then 2.8 = 120√(1- V^2 / C^2) = 0.98 c

No, you can't justify arbitrarily invert a term. It makes the equation meaningless. You just need to fix your algebra above.

 

[student07]

ok then 2.8 = 120√(1- V^2 )
2.8 / 120 = √(1- V^2 )
0.023333333 = √(1- V^2 )
√0.023333333 = (1-V^2)
0.152752523 = (1-V^2)
v^2 = (1 - 0.152752523)
v^2 = 0.847247476
v = 0.92 c
is this correct now?

 

[gneill]

ok then 2.8 = 120√(1- V^2 )
2.8 / 120 = √(1- V^2 )

Where does the 2.8 come from? I thought the observed length was 43 m.

0.023333333 = √(1- V^2 )
√0.023333333 = (1-V^2)

Nope. To get rid of the square root you need to square both sides, not square one side and take the square root on the other.

0.152752523 = (1-V^2)
v^2 = (1 - 0.152752523)
v^2 = 0.847247476
v = 0.92 c
is this correct now?

Interestingly, it comes close to the correct value despite the errors above! Still, you need to fix the algebra.

 

[student07]

I don't know any other way then this people got it this way https://answers.yahoo.com/question/index?qid=20120418090716AApo5pz
but I don't get the gamma part...is ok is only a few marks thanks for your help

 

[gneill]

You have an expression of the form: $a = b \sqrt{c - d^2}$ and you want to isolate d. So just rearrange to get the square root alone on one side:

$\frac{a}{b} = \sqrt{c - d^2}$

Then square both sides to remove the square root. Note that the term $d^2$ remains! Continue from there.

 

[student07]

So then, √0.023333333 = √(1-V^2) - square root of 1 is 1 and V^2 remains right ?
=
0.152752523 = (1-V^2)
v^2 = (1 - 0.152752523)
v^2 = 0.847247476
v = 0.92 c

 

[gneill]

I don't know where your getting your values from. Where did 0.02333... come from, and why are you taking its square root? This is why I suggest doing all initial work with symbols only. That way there can be no mistake about values!

The starting equation is: $L = L_o \sqrt{1 - v^2}$, where $L$ is the observed length (43 m), and $L_o$ is the proper length of the second ship (120 m). v is the fractional c velocity of the ship (i.e., it's velocity is V = v c}. So:

$L = L_o \sqrt{1 - v^2}$

$\frac{L}{L_o} = \sqrt{1 - v^2}$

$\left( \frac{L}{L_o} \right)^2 = 1 - v^2$ $~~~~~~~~~$ clearing the square root

Right? Then rearrange to isolate $v^2$ and finally find v.

 

[student07]

(43 / 120)^2 = 1-v^2
0.128402777 = 1 - v^2
v^2 = 1 - 0.128402777
v = 0.93 c

 

[gneill]

This time I can say that you have performed a correct calculation.

 

[student07]

All thanks to you, thanks for not giving up on me lol you were very supportive :)