# Confusion with "explosive" part, Conservation of Momentum

• tul725
The rocket is saved and you can rest easy.Now, take your super-power away. That's it. Now you're in the rocket and you see the same thing from the rocket's frame of reference. It's the same, just that now you're the rocket and the rocket is the Superman. You'll see the first stage flying away faster than you, while you'll be the one slowing down faster than the first stage (because you're heavier). In the end you'll both be flying away from the crash site at the same speed, but in opposite directions. This is the solution to the problem.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
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?

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.

tul725
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.

tul725
+1

I came back because I realized i hadn't explained that at all well but you beat me to it.

tul725
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.

## 1. What is the "explosive" part in the context of Conservation of Momentum?

The "explosive" part in the context of Conservation of Momentum refers to the release of energy in the form of an explosion during a collision or impact between objects.

## 2. How does the "explosive" part affect the conservation of momentum?

The "explosive" part can affect the conservation of momentum by introducing an additional force or impulse to the system. This force can alter the momentum of the objects involved in the collision, leading to a change in the overall momentum of the system.

## 3. Can the "explosive" part violate the law of conservation of momentum?

No, the law of conservation of momentum states that the total momentum of a closed system remains constant. This means that the momentum before and after an explosion, or any other event, must be equal. Therefore, the "explosive" part cannot violate this law.

## 4. How can the "explosive" part be accounted for in calculations involving conservation of momentum?

The "explosive" part can be accounted for by considering the additional forces or impulses involved in the collision. These forces must be included in the momentum equations to accurately calculate the change in momentum of the objects involved.

## 5. Are there any real-life examples of the "explosive" part in conservation of momentum?

Yes, there are many real-life examples of the "explosive" part in conservation of momentum. For instance, an explosion of a bomb or a rocket launch involves the release of energy and an impulse that affects the momentum of the objects involved. Similarly, collisions in sports like football or car crashes also involve an element of explosiveness that affects the conservation of momentum.

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