# Conservation of Momentum (Explosion Kinematics)

• jay2001
In summary: But how does that relate to the East and North references? Are you saying they are both......directly perpendicular to each other? ...directly perpendicular to each other?
jay2001

## Homework Statement

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A 3200 kg space vehicle (including a launchable lifeboat) is traveling with a velocity of 300 m/s in a straight trajectory [East]. The lifeboat (200 kg) is fired at a speed of 1000 m/s [N of original trajectory].
a) After firing it is found that the horizontal component of each object’s velocity is still 300 m/s [along original trajectory]. Explain why.

p = p'
p = mv

## The Attempt at a Solution

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More of a theory question than problem solving... but I'm stumped on this. I feel like they shouldn't both have horizontal velocity components of 300 m/s - I believe that the lifeboat should have a velocity of 1000 m/s [N] (so a horizontal velocity component of zero), and the space vehicle would be redirected slightly South of East, so the horizontal velocity component would change from 300 m/s. Am I wrong, or just misunderstanding?

jay2001 said:
I'm stumped on this
Think about what "north" and "east" etc. might mean in space.

Btw, you mean kinetics, not kinematics.

Make sure that you understand linear momentum conservation properly and come back to the question after thoroughly understanding the concept :)

ubergewehr273 said:
Make sure that you understand linear momentum conservation properly and come back to the question after thoroughly understanding the concept :)
I see no evidence that @jay2001 does not understand the concept, do you?

jay2001 said:
I believe that the lifeboat should have a velocity of 1000 m/s [N] (so a horizontal velocity component of zero)
Are you sure? Mind the lifeboat is moving along with the vehicle just before the launch.
jay2001 said:
and the space vehicle would be redirected slightly South of East, so the horizontal velocity component would change from 300 m/s.
The rocket can still redirect slightly South of East with the horizontal component of velocity fixed. What if the vertical component changes only? ##\ddot\smile##

jay2001 said:
A 3200 kg space vehicle (including a launchable lifeboat) is traveling with a velocity of 300 m/s in a straight trajectory [East]. The lifeboat (200 kg) is fired at a speed of 1000 m/s [N of original trajectory].
a) After firing it is found that the horizontal component of each object’s velocity is still 300 m/s [along original trajectory]. Explain why.
From momentum conservation
p=p1+p2,
or,p2=p12+p22+
2p1horizontalp2

From energy conservation
p2/2m=p12/2m1+p22/2m2
Where p is the momentum of the system before disintegration and p1 and p2 are the momentum of the lifeboat and space vehicle respectively.

Now p,p1 horizontal,
p1vertical and p2 is given in the question.

Apashanka said:
From momentum conservation
p=p1+p2,
or,p2=p12+p22+
2p1horizontalp2

From energy conservation
p2/2m=p12/2m1+p22/2m2
Where p is the momentum of the system before disintegration and p1 and p2 are the momentum of the lifeboat and space vehicle respectively.

Now p,p1 horizontal,
p1vertical and p2 is given in the question.
You do not appear to have understood the question. Nothing in your post helps.

haruspex said:
Think about what "north" and "east" etc. might mean in space
North and east only means the direction that are perpendicular to each other ,nothing to do with north and South in space.

haruspex said:
Yaa I have read it .
It only implies the mutually perpendicular direction ,nothing to confuse with the north and east of geography.

jay2001 said:
A 3200 kg space vehicle (including a launchable lifeboat) is traveling with a velocity of 300 m/s in a straight trajectory [East]. The lifeboat (200 kg) is fired at a speed of 1000 m/s [N of original trajectory].
a) After firing it is found that the horizontal component of each object’s velocity is still 300 m/s [along original trajectory]. Explain why.
To an observer who is sitting on the space vehicle he will see a horizontal velocity of lifeboat(being equal to that of space vehicle in the negative direction)as well as a vertical component of velocity of the lifeboat(given).
To an observer at absolute rest(reference frame) at the moment the life boat is launched he will see a horizontal component of velocity of lifeboat (being equal to that of space vehicle in the positive direction)and as well as vertical component of velocity of the lifeboat.(given).

Apashanka said:
North and east only means the direction that are perpendicular to each other ,nothing to do with north and South in space.
and how might that relate to the "horizontal" information?
(Do you think you have an answer to the question posed?)

haruspex said:
and how might that relate to the "horizontal" information?
Horizontal actually here may be referred to the plane in which the objects are moving or rather direction along the straight line of the motion as given in question ,and vertical refers perpendicular to this direction

Apashanka said:
Horizontal actually here may be referred to the plane in which the objects are moving or rather direction along the straight line of the motion as given in question ,and vertical refers perpendicular to this direction
But how does that relate to the East and North references? Are you saying they are both horizontal?

haruspex said:
But how does that relate to the East and North refernces? Are you saying they are both horizontal?
For an exm. you may take your right hand to be east direction and forehead to be north direction ,and space vehicle is moving along right hand and lifeboat is ejected along forehead.
Horizontal is along right hand and vertical is along forehead.

Apashanka said:
For an exm. you may take your right hand to be east direction and forehead to be north direction ,and space vehicle is moving along right hand and lifeboat is ejected along forehead.
Horizontal is along right hand and vertical is along forehead.
Or in 2-D Cartesian x is east(horizontal) and y is north (vertical).

Apashanka said:
For an exm. you may take your right hand to be east direction and forehead to be north direction ,and space vehicle is moving along right hand and lifeboat is ejected along forehead.
Horizontal is along right hand and vertical is along forehead.
Yes, that's the key, but there was no hint of that in post #6.
Also, I feel the question setter had something a bit more specific in mind which would lead to that interpretation of how N and E relate to horizontal and vertical.

## 1. What is conservation of momentum?

The law of conservation of momentum states that the total momentum of a closed system remains constant. This means that in a system where no external forces are acting, the total momentum before an event (such as an explosion) is equal to the total momentum after the event.

## 2. How does conservation of momentum apply to explosion kinematics?

In an explosion, the total momentum of the system before the explosion (when all particles are at rest) is equal to zero. After the explosion, the particles move in different directions with different velocities, but the total momentum of the system remains constant.

## 3. What factors affect the conservation of momentum in an explosion?

The mass and velocity of the particles involved in the explosion are the main factors that affect the conservation of momentum. The total momentum of the system will be greater if there are more particles with higher velocities involved in the explosion.

## 4. How is the conservation of momentum calculated in an explosion?

The conservation of momentum can be calculated by adding up the individual momenta of all the particles before and after the explosion. The total momentum before the explosion should be equal to the total momentum after the explosion, according to the law of conservation of momentum.

## 5. Why is conservation of momentum important in explosion kinematics?

Conservation of momentum is important in explosion kinematics because it helps us understand and predict the motion of particles involved in an explosion. It also allows us to analyze the forces and energy involved in an explosion, which is crucial for safety and engineering purposes.

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