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

[Eclair_de_XII]

Homework Statement

"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?"

Homework 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##

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.

 

[Simon Bridge]

Take (a)
This is normal conservation of momentum like you realize ... 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?

 

[Eclair_de_XII]

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.

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.

 

[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.

 

[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.

 

[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##.

 

[Eclair_de_XII]

perhaps it would be better if you expleined your ideas behind how you did (1)

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##.

 

[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?

 

[Eclair_de_XII]

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?

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.

 

[haruspex]

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

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

 

[Eclair_de_XII]

So write an equation which expresses that.

Okay. I'll try:

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

 

[haruspex]

Okay. I'll try:

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

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?

 

[Eclair_de_XII]

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?

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

 

[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.

 

[haruspex]

I don't think it does.

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

 

[Eclair_de_XII]

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

 

[haruspex]

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

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?

 

[Eclair_de_XII]

2 m/s?

 

[haruspex]

2 m/s?

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

 

[Eclair_de_XII]

Is it ##u=v_L+v##?

 

[haruspex]

Is it ##u=v_L+v##?

Yes.

 

[Eclair_de_XII]

So then the equation describing momentum of the two bodies in step (1) would be: ##(m_L)(u-v)=(m_C+m_R)v##?

 

[haruspex]

So then the equation describing momentum of the two bodies in step (1) would be: ##(m_L)(u-v)=(m_C+m_R)v##?

Yes.

 

[Eclair_de_XII]

Now I'm trying to think of a way to describe the relative velocity in step (2). Since block C (the second term in the below equation) has three times more mass than block R, it should experience one-third times the velocity as block R. So:

##u_1=(v_R)+[(\frac{1}{3}v_R)-(v)]=\frac{4}{3}v_R-v##

And since ##v_R=v+3##, ##u_1=\frac{1}{3}v+4##.

Verify?

 

[haruspex]

So:

##u_1=(v_R)+[(\frac{1}{3}v_R)-(v)]##

I do not understand how you get that. Please explain your reasoning, and confirm or deny that u₁ is the relative velocity (3m/s) after the second explosion.

 

[Eclair_de_XII]

Please explain your reasoning, and confirm or deny that u₁ is the relative velocity (3m/s) after the second explosion.

Okay, so I basically have ##m_R## moving to the right at ##v_R##, and ##m_C## moving to the left at ##\frac{1}{3}v_R##; then there's the leftover velocity from step (1) that moves ##m_C## to the right, which is subtracted from ##\frac{1}{3}v_R##. So in this equation, ##u_1## is the net relative velocity, if that makes any sense. But if I were to describe ##u_1## as ##3\frac{m}{s}##, it would be ##u_1=v_R-v##, since ##v_R## and ##v## are moving in the same direction, I think?

 

[haruspex]

Okay, so I basically have ##m_R## moving to the right at ##v_R##, and ##m_C## moving to the left at ##\frac{1}{3}v_R##; then there's the leftover velocity from step (1) that moves ##m_C## to the right, which is subtracted from ##\frac{1}{3}v_R##. So in this equation, ##u_1## is the net relative velocity, if that makes any sense. But if I were to describe ##u_1## as ##3\frac{m}{s}##, it would be ##u_1=v_R-v##, since ##v_R## and ##v## are moving in the same direction, I think?

The v_R and v_R/3 relationship is right relative to their reference frame prior to the second explosion, i.e. as though they had been at rest before that. But they were both moving right at speed v. Your equation is as though only one of them had been moving at speed v.

 

[Eclair_de_XII]

Oh, so is it ##u_1=|(v+v_R)-(v-\frac{1}{3}v_R)|=\frac{4}{3}v_R=\frac{4}{3}v+4##? The first term is the velocity of ##m_R##, which is ##v+v_R##, and not just ##v_R## like you pointed out. The second term is the velocity of ##m_C##, which is moving in the positive direction at ##v##, and in the negative direction at ##\frac{1}{3}v_R##. So its velocity is ##v-\frac{1}{3}v_R##.

 

[haruspex]

Oh, so is it ##u_1=|(v+v_R)-(v-\frac{1}{3}v_R)|=\frac{4}{3}v_R##? .

Yes, where u₁ is 3m/s, right?

##=\frac{4}{3}v+4##? .

No, where did that come from?

 

[Eclair_de_XII]

Yes, where ##u_1## is 3m/s, right?

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

No, where did that come from?

I thought ##v_R=v+3=v+u##. Sorry; old habits.

 

[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}##