Question

Sanity check...The last stage of a rocket is traveling at a speed of 7200m/s. The last stage has two parts, the payload (m=150kg) and the rocket case (m=290kg). A compressed spring separates them with a relative speed of 910m/s. What are the speeds of the individual pieces after they separate? Here's what I've done:

Answer 1

\[P_f = P_i = (150.0kg +290.0kg)*7200\frac{m}{s}\]
\[3,334,000\frac{kgm}{s}=[290kg(v_f-910\frac{m}{s}]+150kg*v_f\]

Answer 2

I got 8177m/s for the payload, and subtracting relative velocity from final velocity, I got 7270m/s for the rocket. The book got 7290m/s for the rocket case, and 8200m/s for the payload. Am I missing something?

Answer 3

How can both pieces speed up?

Answer 4

They're not...I'm showing the top one increasing in velocity, and the bottom one decreasing. The initial velocity of the system was 7600m/s before the spring separation.

Answer 5

Oh. It says 7200 m/s in the original problem.

Answer 6

Oops, sorry. Yeah, that should be 7600m/s. Bad typing. :P

Answer 7

I've been shopping for hard drives...I'll bet 7200(rpm) was stuck in my head. Going to run the numbers again....

Answer 8

Conserving momentum:
\[m _{c}v _{c}=m _{p}v _{p}\]
and
\[v _{c}=910-v _{p}\]

I get \[v _{c}=310 m/s \] and \[v _{p}=600 m/s\]

Those are absolute values. Add payload increase to 7600, and subtract case slowing from 7600.

Answer 9

I'm not sure how you're applying conservation to the split. Are you equating the momentum of the rocket case to the payload? Shouldn't it be: \[m_{tot}v = m_1v_1 +m_2v_2\]

Answer 10

The velocity before separation needn't be considered in the calculation. After all, it's 7600m/s relative to what?

Total momentum is unaffected by the spring's release.

Total momentum is unaffected with one change in momentum equal and opposite the other.

Answer 11

Sorry, I'm not trying to be stubborn here, I just want to be sure I understand. I *think* that \[(m_{1}+m_{2})v_i=m_{1}v_1 + m_2v_2\] *requires* that the total momentum be unaffected, because no matter what happens to the velocity of one piece, the other velocity has to compensate to balance the other side. The spring is an internal force.

It doesn't make sense to me that the momentum of the individual pieces should be equal to one another--they each have different masses and velocities. In fact, the numbers don't work out. If I were to assume your equation\[m_cv_c=m_pv_p\]I would get\[150kg*310\frac{m}{s}=290kg*600\frac{m}{s}\]Unfortunately, that leads me here:\[46,500\frac{kgm}{s} \neq 174,000\frac{kgm}{s}\] Even if I were to just go with the final speeds, it still doesn't work.\[290kg*7290\frac{m}{s} \neq 150kg*8200\frac{m}{s}\]

Answer 12

Stubborn? A physics student? Try convincing a class of 30 that there's no such thing as cetrifugal force!

It's the *changes* in momenta that are equal..

After the words "I would get..." you have the masses and speeds reversed.

Also, 440(7600)=150(8200) + 290(7290) with a rounding error.

Answer 13

Ha! Funny, @NoelGreco. As a 42-year-old autodidact, stubborn might not be the best description...but certainly equally challenging at times. :)

Answer 14

Ok, the numbers are working for me now, but I'm still trying to grasp the intuition.
It would appear that my approach, that is, using \[(m_1+m_2)v = m_1v_1+m_2v_2\]wasn't wrong *per se* (the answers were pretty close) Is this the same as saying \[p_{system} = p_1+p_2\] But the velocities are in opposing directions, so it's ok to say? \[m_1v_1=m_2v_2\]Teaching myself, I'm trying to be especially careful to not make any unwarranted assumptions.