Confusion with "explosive" part, Conservation of Momentum

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The discussion centers on a two-stage rocket problem involving conservation of momentum during stage separation due to an explosive charge. The initial momentum of the rocket is calculated, and the velocities of the two stages after separation are determined. It is clarified that momentum is conserved because the explosion does not exert an external force; it merely separates the two stages. The confusion arises from the reference frame used to analyze the problem, where velocities must be transformed correctly to maintain consistency. Ultimately, the key takeaway is that while one stage slows down, the other accelerates, leading to both stages moving in the same direction post-separation.
tul725

Homework Statement



A two-stage rocket is traveling at 4500 m/s before the stages separate. The 3000-kg first stage is pushed away from the second stage with an explosive charge, after which the first stage continues to travel in the same direction at a speed of 3000 m/s . How fast is the 1900-kg second stage traveling after separation?[/B]

Rocket's mass = 4900 kg
Velocity of rocket = 4500 m/s

m1= 3000kg
v1 = 3000m/s

m2=1900kg
v2 = ?

Homework Equations


Conservation of Momentum

The Attempt at a Solution


The way I tried imagining it was by changing the inertial reference frame to the zero momentum frame, the frame would be going at 4500 m/s. In this situation m1v1 + m2v2 = 0. I then plugged in numbers to get v2 = -4736.84 m/s. Then to get the velocity relative to the Earth I added the velocity of the frame relative to the earth, which was 4500m/s. I then get 236.84 m/s. Which is incorrect.
After researching the answer, I came across a similar problem where the conservation of momentum equation was set up:
(MassofRocket)(VelocityofRocket) = m1v1 + m2v2
The answer I got for this, v2 = 6868 m/s was correct.

What I am struggling to understand is why is momentum conserved even if there is the explosive part of the collision. Wouldn't that cause an impulse to the situation and momentum would be changed? And why is my original method not proper?
 
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Show your working for the first method as that does give the right answer as well.

Momentum is conserved because the explosion doesn't provide an external force/impulse. In this set up the explosion doesn't "push" like a rocket would, it just pushes the two parts apart.
 
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CWatters said:
Momentum is conserved because the explosion doesn't provide an external force/impulse. In this set up the explosion doesn't "push" like a rocket would, it just pushes the two parts apart.
Just a nitpick here: The fuel and the rocket push against each other, in much the same way as the explosive charge makes the two stages push against each other in this problem. These are equivalent situations, providing we remember that the ejected fuel is part of the system.

(I started writing when I saw your initial comment about pushing against the atmosphere, which you later rightly edited out, but it's still not entirely correct as it is)

tul725 said:
And why is my original method not proper?
Looks like you said you would change the reference frame to stationary with the rocket, but you did not transform all the velocities into that new frame.
 
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+1

I came back because I realized i hadn't explained that at all well but you beat me to it.
 
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CWatters said:
Show your working for the first method as that does give the right answer as well.

Momentum is conserved because the explosion doesn't provide an external force/impulse. In this set up the explosion doesn't "push" like a rocket would, it just pushes the two parts apart.
Bandersnatch said:
Looks like you said you would change the reference frame to stationary with the rocket, but you did not transform all the velocities into that new frame.

Oh I understand my issue. I did not change the velocities in reference to the Earth frame to the velocities in reference to the zero momentum frame. The only final thing that confuses me is that when I do this, I get v1 = -1500 m/s and when I do conservation of momentum:
0 = (-1500)(3000) + v2 (1900)
I get a positive number for v2. I then add 4500 to the answer to v2 to get 6868 m/s, which is the right answer. So my understanding is that the both stages are moving in the same direction, which makes sense, but the part that separated from the back is moving faster than the front. I'm having trouble visualizing this and when I try to imagine it it doesn't make sense. Should I not even worry about visualizing it? Thanks for the help though I understand my original error. For the most part I understand that the explosion is not an external force and therefore the momentum is conserved.
 
No the bottom 1st stage slows from 4500 to 3000 and the top 2nd stage accelerates from 4500 to 6868.
 
tul725 said:
I'm having trouble visualizing this
Imagine you're the Superman. The rocket is at risk of crashing, because its explosive charges had failed and the entire thing is moving too slow.

So you catch up with the rocket and match its speed (you are now stationary w/r to the rocket).
You grab the seam between the two stages in your frankly rather huge hands, and you push them apart.
With respect to you, both stages now fly in opposite directions, the less massive one faster than the other.
But you're still flying at the 4500 km/s you needed to match the rocket. So when you finally stop and land, you see that both stages are flying in the same direction. But one has been slowed down due to your push, and will soon crash into the ocean, while to other has been accelerated and is happily continuing on its planned trajectory.
 

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