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

1. Nov 5, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Figure 9-55 shows a two-ended "rocket" that is initially stationary on a friction-less floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass $m_C=6kg$) and blocks L and R (each of mass $m_L=2kg$ and $m_R=2kg$) on the left and right sides. Small explosions can shoot either of the side blocks away from block C and along the x axis. Here is the sequence:

(1) At time $t_0=0$, block L is shot to the right with a speed of $3\frac{m}{s}$ relative to the velocity that the explosion gives the rest of the rocket.

(2) at time $t_1=\frac{4}{5}s$, block R is shot to the right with a speed of $3\frac{m}{s}$ relative to the velocity that block C then has.

At $t_f=\frac{14}{5}s$, what are (a) the velocity of block C and (b) the position of its center?"

2. Relevant equations
$p_i=p_f$ at two different points in time, in a closed and isolated system

$m_C=6kg$
$m_L=2kg$
$m_R=2kg$

3. The attempt at a solution
Sequence (1):

$v_L-v_C=-3\frac{m}{s}$
$v_C-v_L=3\frac{m}{s}$

Sequence (2):

$v_R-v_C=3\frac{m}{s}$

$p_C-p_L=(m_C)(v_C)-(m_L)(v_L)=p$
$p_R-p_C=(m_R)(v_R)-(m_C)(v_C)=p$
$(m_C)(v_C)-(m_L)(v_L)=(m_R)(v_R)-(m_C)(v_C)$
$2(m_C)(v_C)=(m_L)(v_L)+(m_R)(v_R)$
$v_C=\frac{1}{2m_C}[(m_L)(v_L)+(m_R)(v_R)]$

So I got stuck here, not knowing what the actual velocities of the two side blocks were. These were just outlining the relationship between the momenta in the two sequences described. I've yet to come up with an equation for what the problem actually asks: the velocity and center of mass at $t=\frac{14}{5}s$. I'm just guessing here, but I'm thinking that the momenta are dependent on time, since velocity is dependent on time. So, I'm just re-expressing the last equation in terms of t.

$v_C(t)=\frac{1}{2m_C}[m_Lv_L(t)+m_Rv_R(t-\frac{4}{5})]$

And that's as far as I can go without knowing the actual velocities of the two blocks. Can anyone help me understand this problem? I'm guessing that the absence of the actual velocities imply that they are not needed to calculate the velocity of the center block.

2. Nov 5, 2016

### Simon Bridge

Take (a)
This is normal conservation of momentum like you realise ... done in the lab frame, then,
Block L shoots off to the left at speed $v_L$ across the floor. "The rest of the rocket" goes to the right with speed $v$ wrt the floor.
$m_Lv_L=(m_C+m_R)v$ ... where v is the speed you have to find.

There are two unknowns in that equation, so you need another equation that also has those two unknowns in it. There is also a quantity given that has not been used - perhaps the other equation should use it?

3. Nov 5, 2016

### Eclair_de_XII

Oh, I see it. $v_L$ is the other unknown. And the equation basically says that the momentum block L gains from the explosion is the same as the momentum the other pieces of the rocket gain from the same explosion.

4. Nov 5, 2016

### Eclair_de_XII

Okay, I couldn't derive an equation that included $v_L$, but this is what I came up with:

$m_Rv_R=m_C(v_R-v)$
$m_R(v+3)=3m_C$

If it did affect center of mass, and by extension, $m_L$, I wouldn't know how it would affect it, if at all.

Last edited: Nov 6, 2016
5. Nov 6, 2016

### Simon Bridge

Oh OK - I think I misunderstood where you got stuck.
Recap. For (1).
Conservation of momentum: $m_Lv_L = (m_C+m_R)v_0$
ie. for $0<t<0.2$s, $v_L = -3$m/s, $v_C=v_R=v_0$
... do you understand why this is the case?

So for (2)
At t=0.2s, though, block R is shot off ...
Conservation of momentum means that $(m_C+m_R)v_0 = m_Cv_C + m_Rv_R$
... where $v_R$ is the speed of block R over the ground... which you are not told.

but you are given $u_R$ which is the speed that block R is fired from block C.
So it is already going $v_0$, and it gets an additional $u_R$ ... so $v_R=u_R+v_0$
... now you have two equations and two unknowns, so you can solve it.
This what you are thinking?

Then for $(0.2 < t < 2.8)$s, $v_L=-3$m/s, $v_R=\cdots$ and $v_C=\cdots$

You are also asked about the position of block C at t=2.8s ... probably a good strategy is to say that x=0 when t=0, so you can say how far C is right or left from the start. You can do this from x=vt ... or by velocity-time diagram for the motion. Remember that velocity is a vector. Careful to make sure that C does not collide with block L before time is up.

Last edited: Nov 6, 2016
6. Nov 6, 2016

### Eclair_de_XII

Oh, I was already doing second part of the problem. Well, relative to block L, block C would be moving at a speed $u+v_L$.

7. Nov 6, 2016

### Eclair_de_XII

It was basically just vector addition, though. The rest of the work I did was attempting to equate the momentum from parts (1) and (2) together. But looking back on it now, it doesn't really seem to make any sense. I'll just try to start from $m_Lv_L=(m_C+m_R)v$.

8. Nov 6, 2016

### Eclair_de_XII

$(1)$ $m_Lv+m_Lv_L=(m_C+m_R)u$ where $u=3\frac{m}{s}$
$(2)$ $m_Lv_L=(m_C+m_R)v$

Subtracting (2) from (1), I get...

$m_Lv=(m_C+m_R)(u-v)=(u)(m_C)+(u)(m_R)-(v)(m_C)-(v)(m_R)$
$(m_L)(v)+(m_C)(v)+(m_R)(v)=(u)(m_C)+(u)(m_R)$
$(v)(m_L+m_C+m_R)=(u)(m_C+m_R)$
$v=(u)(\frac{m_C+m_R}{m_L+m_C+m_R})=(3\frac{m}{s})(\frac{6 kg+2 kg}{2kg+6kg+2kg})=\frac{12}{5}\frac{m}{s}$

That's part (1), I think. Can someone check my work?

9. Nov 6, 2016

### Eclair_de_XII

I think I can solve it, but I don't exactly feel comfortable using equations that I don't understand. I mean, I get that $v_0=v_R=v_C$ because the right and center blocks are still connected and move in unison. But I am confused on why you wouldn't need a separate equation to describe the velocity of the right block with respect to the center block when it shoots off from it.

Last edited: Nov 6, 2016
10. Nov 6, 2016

### haruspex

I don't understand this equation. Please confirm:
vL is the speed to the left of mL, in the lab frame.
v is the speed to the right of mC+mR, in the lab frame.
u is their relative speed.
Right? So write an equation which expresses that.

11. Nov 6, 2016

### Eclair_de_XII

Okay. I'll try:

$m_L(v+v_L)=(m_C+m_R)u$

12. Nov 6, 2016

### haruspex

That is the same as you wrote before.
You have two speeds in the rest frame and their relative speed. if Paul runs at 13km/h and you run 1 km/h faster, what is your speed? Is it in any way affected by the masses of you and Paul?

13. Nov 6, 2016

### Eclair_de_XII

$14\frac{km}{h}$, and I don't think it does. Let me try again: $(m_L)(v+u)=(m_C+m_R)(v)$.

Last edited: Nov 6, 2016
14. Nov 6, 2016

### Eclair_de_XII

For (2), I believe it is: $(m_R)(v+u)=(m_C)(v)$. So I have here:

$(1)$ $m_Rv+m_Ru=m_Cv$
$(2)$ $m_Lv+m_Lu=m_Cv+m_Rv$

Adding (1) to (2), I get:

$(3)$ $m_Lv+m_Rv+m_Lu+m_Ru=2m_Cv+m_Rv$
$(4)$ $m_Lv+m_Lu+m_Ru=2m_Cv$
$(5)$ $2m_Cv-m_Lv=v(2m_C-m_L)=u(m_L+m_R)$
$(6)$ $v=\frac{u(m_L+m_R)}{2m_C-m_L}=\frac{6}{5}\frac{m}{s}$

But it doesn't take into account the momentum in the negative x-direction that block C gains when block R shoots off from it.

15. Nov 6, 2016

### haruspex

So why do you keep bringing masses into the relationship between vl, v and u? There is an equation which relates just those three quantities, nothing else.

16. Nov 6, 2016

### Eclair_de_XII

Let's see... $v_L=v+u$.

17. Nov 6, 2016

### haruspex

If you are defining the three speeds as I described in post #10, no.
Bob walks E at 1m/s, Alice walks W at 1m/s. What is their relative speed?

18. Nov 6, 2016

### Eclair_de_XII

2 m/s?

19. Nov 6, 2016

### haruspex

Of course. So what is the relationship between vL, the leftward speed of the left hand mass, v, the rightward speed of the right hand masses, and u, their relative speed?

20. Nov 6, 2016

### Eclair_de_XII

Is it $u=v_L+v$?