How do I calculate the velocity of a two-ended rocket?

[haruspex]

Now that you say it, it being 3 m/s starts to make sense. So $v_R=\frac{9}{4}\frac{m}{s}$?.

Yes.

 

[Eclair_de_XII]

Okay, so I tried expressing $\vec p_C$ as $\vec p_C=\vec p_R - \vec p_L$, but I'm not getting any results.

$\vec p_C=(m_C)(\frac{1}{3}v_R-v)$
$\vec p_R=(m_R)(v+v_R)$
$\vec p_L=(m_L)(u-v)$
$\vec p_C = \vec p_R - \vec p_L = m_Rv+m_Rv_R-m_Lu+m_Lv$
$\frac{1}{3}m_Cv_R-m_Cv=m_Rv+m_Rv_R-m_Lu+m_Lv$
$m_Rv+m_Lv+m_Cv=\frac{1}{3}m_Cv_R-m_Rv_R+m_Lu$
$v(m_R+m_L+m_C)=\frac{1}{3}m_Cv_R-m_Rv_R+m_Lu$
$v=(\frac{\frac{1}{3}m_Cv_R-m_Rv_R+m_Lu}{m_R+m_L+m_C})$
$v=(\frac{(\frac{9}{2}-\frac{9}{2}+6)(N⋅s)}{10kg})=\frac{3}{5}\frac{m}{s}$

 

[Eclair_de_XII]

Okay, I thought I the best approach would be to start with the equations for step (1) and (2).

$(1)$ $(m_L)(u-v)=(m_C+m_R)(v)$
$(2)$ $(m_R)(v_R+v)=(m_C)(v-\frac{m_R}{m_C}v_R)$

Rearranging (1) gives me:

$v=(\frac{m_L}{m_C+m_R})(u-v)$
$v(1+\frac{m_L}{m_C+m_R})=(\frac{m_L}{m_C+m_R})(u)$
$v=\frac{(\frac{m_L}{m_C+m_R})(u)}{(1+\frac{m_L}{m_C+m_R})}=(\frac{m_Lu}{m_C+m_L+m_R})=\frac{3}{5}\frac{m}{s}$

Rearranging (2) gives me:

$(v_R+v)=\frac{(m_C)(v-(\frac{m_R}{m_C})v_R)}{m_R}$
$v=(\frac{(m_C)(v-(\frac{m_R}{m_C})(v_R))}{m_R})-v_R$
$v(1-\frac{m_C}{m_R})=(\frac{-m_Rv_R}{m_R})-v_R$
$v=\frac{m_R}{m_R-m_C}((\frac{-m_Rv_R}{m_R})-\frac{m_Rv_R}{m_R})$
$v=((\frac{-m_Rv_R}{m_R-m_C})-\frac{m_Rv_R}{m_R-m_C})=-2(\frac{v_R}{m_R-m_C})=-2(\frac{\frac{9}{4}\frac{m}{s}}{-4kg})=\frac{9}{8}\frac{m}{s}$

So now I just have to express this in terms of the time parameter, I think?

 

[haruspex]

I'm not sure where you are going with your last two posts.
In post #30 you wrote that v_R=9/4m/s. I confirmed that, but omitted to check exactly how you are defining v_R there. From your post #28, it seems to be the final speed of the rightmost block relative to v, i.e. relative to its speed after the first explosion. Ok?
That would make your 9/4 correct.
So what is the speed of the central block relative to v after the second explosion, and in what direction?
What does that give you for the final speed of the central block in the lab frame?

 

[Eclair_de_XII]

Whoa, I just figured out the answer. The velocity of $m_C$ isn't $v$, it's actually $v-(\frac{m_R}{m_C})(v_R)$.

$v_{C}=v-(\frac{m_R}{m_C})(v_R)=\frac{3}{5}\frac{m}{s}-(\frac{2kg}{6kg})(\frac{9}{4}\frac{m}{s})=-\frac{3}{20}\frac{m}{s}$

 

[haruspex]

Whoa, I just figured out the answer. The velocity of $m_C$ isn't $v$, it's actually $v-(\frac{m_R}{m_C})(v_R)$.

$v_{C}=v-(\frac{m_R}{m_C})(v_R)=\frac{3}{5}\frac{m}{s}-(\frac{2kg}{6kg})(\frac{9}{4}\frac{m}{s})=-\frac{3}{20}\frac{m}{s}$

Yes!

 

[Eclair_de_XII]

$t=2s$
$x_0=vt_1=\frac{12}{25}m$
$x_f=x_0+v_Ct=\frac{12}{25}m+(-\frac{3}{20}\frac{m}{s})(2s)=\frac{48-30}{100}m=\frac{9}{50}m$

Well, thank you very much for helping me through this, haruspex. I literally do not believe I could have figured this out without any sort of help. Thank you for always helping me.

 

[haruspex]

$t=2s$
$x_0=vt_1=\frac{12}{25}m$
$x_f=x_0+v_Ct=\frac{12}{25}m+(-\frac{3}{20}\frac{m}{s})(2s)=\frac{48-30}{100}m=\frac{9}{50}m$

Well, thank you very much for helping me through this, haruspex. I literally do not believe I could have figured this out without any sort of help. Thank you for always helping me.

You are welcome. But you could do it without help next time, right?

 

[Eclair_de_XII]

I think so. I'll see next weekend.