question, yep sure is one

[Sarahisme]

hi all,

here is the question, i will post my answer in the next post:

 

Sarah

[Sarahisme]

for part (a)

use the equation for adding relativistic velocites together

[math]

u=\frac{u^{'}+v^{'}}{1+\frac{u^{'}v^{'}}{c^{2}}}

[/math]

 

So u = the speed of one ship relative to the other.

[math]

u=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}

=\frac{1.2c}{1.36}

=0.88c = 0.9c

[/math]

 

oh by the way, how do you make an equation go to the next line using latex?

eg.

 

u=42x

= ...

=....

etc.

[Sarahisme]

for part (b)

again use the equation for adding relativistic velocites together

[math]

u=\frac{u^{'}+v^{'}}{1+\frac{u^{'}v^{'}}{c^{2}}}

[/math]

 

So u = the speed of one ship relative to the other.

[math]

u=\frac{30000+30000}{1+\frac{(30000)^{2}}{c^{2}}}

=\frac{60000}{1\times10^{-8}}

=59999.9994(m/s)

=6.0000\times10^{4}

[/math]

 

and the final answer should be too 5 sig. figures, as the information given in the question (for part b) is to 5 sig. figs.

 

so what did people think of all that? i am pretty damn sure its right!

[timo]

[math] \begin{array}{rcl}

a &=& b + c \\

&=& (b + d) - (d-c)

\end{array} [/math]

 

Source: \begin{array}{rcl}

a &=& b + c \\

&=& (b + d) - (d-c)

\end{array}

 

Your answers look good. Especially part b) which gives the classical result as one would expect at such low velocities.

[Sarahisme]

[math] \begin{array}{rcl}

a &=& b + c \\

&=& (b + d) - (d-c)

\end{array} [/math]

 

Source: \begin{array}{rcl}

a &=& b + c \\

&=& (b + d) - (d-c)

\end{array}

.

 

what this about?

[DQW]

It's about this :

 

oh by the way' date=' how do you make an equation go to the next line using latex?[/quote']

[Sarahisme]

oh right, oops , thanks

[Sarahisme]

oh so you have to have all this array stuff...?