Conservation of linear momentum of rocket

In summary, the equation that describes the conservation of linear momentum states that the total momentum of a system is conserved. Before the launch of block L, the system has a total momentum of M. After the launch, the system has a total momentum of M + (M - v)
  • #1
mbisCool
136
0

Homework Statement



The figure below shows a two-ended "rocket" that is initially stationary on a frictionless floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6.00 kg) and blocks L and R (each of mass m = 2.50 kg) 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, block L is shot to the left with a speed of 2.60 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.80 s, block R is shot to the right with a speed of 2.60 m/s relative to the velocity that the explosion gives to block C?

(a) What is the velocity of block C at t = 2.80 s?
m/s

(b) What is the position of block C's center at that time?
m


Homework Equations





The Attempt at a Solution



I know how to conserve momentum for the different pieces of the problem; however, if someone could help me approach the problem it would be greatly appreciated. I believe I need to find the velocity of each piece after the explosion but am unsure of how to go about finding it. Are the individual velocities required, or do simply use the relative velocity between the two? Any hints on how to approach would be greatly appreciated!
 
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  • #2
Dear mbisCool,

I take it you do know the equation that describes the general case of the conservation of linear momentum. Perhaps this question may help you start: what is the total momentum of the system before L is shot? Remember, this is an explosion.

Regards,
Horatio
 
  • #3
Hmmmm...

Might it help to consider the TOTAL MOMENTUM before the explosions and the TOTAL at the end bearing in mind the 'explosions' are internal to the system.
 
  • #4
I know the total momentum for the system in conserved but i am not sure where to go from there as the individual pieces of the system do not conserve momentum. Sorry if its painfully obvious too :)
 
  • #5
mbisCool said:
I know the total momentum for the system in conserved but i am not sure where to go from there as the individual pieces of the system do not conserve momentum. Sorry if its painfully obvious too :)

It's OK... Let's approach it step-by-step. Since no other forces act on the system, overall momentum is conserved. Hence, it would right to say that the total momentum of C and the launched object will be equal to the momentum before the launch.

That brings us to the question, what is the total momentum of the system? We know that the system of C,R and L were at rest at the beginning (ie before the launches), what does that tell you about the total momentum of the system?

In each launch, we know the mass of the launcher and the object, right? We know the total momentum, and we also know the speed of the object. So, you could find the speed of the launcher after each launch. Is it clearer now?

Regards,
Horatio
 
  • #6
Well then my question is regarding the velocity. I know the relative velocity between the two is 2.6 however this is ambiguous. My question was can you simply use 2.6 m/s? this could be anywhere a range of different velocites with a difference of 2.6. I would need to solve for 8.5kg * V1 = 2.5kg * V2 and the relative velocity would be 2.6
 
  • #7
Ok, it may seem a little ambiguous that the relative velocity is given. But don't panic. Let the velocity of the launcher be v (relative to the ground), and we know for certain that the 2 bodies are moving in opposite directions. If that is so, it is apparent that the velocity of the launched object is -(2.6 - v), since the magnitude of the relative velocity must be higher than the "absolute" velocity. Substituting these 2 values into the equation will give you an equation in 1 unknown, so it's solvable.

Regards
Horatio
 
Last edited:
  • #8
Much appreciated ^^
 

FAQ: Conservation of linear momentum of rocket

What is the conservation of linear momentum of a rocket?

The conservation of linear momentum of a rocket is a fundamental principle in physics that states that the total momentum of a system remains constant as long as there are no external forces acting on it. In the context of a rocket, this means that the total momentum of the rocket and its expelled exhaust gases must remain constant.

Why is the conservation of linear momentum important for rockets?

The conservation of linear momentum is crucial for rockets as it allows them to move through space by expelling gases at high speeds in one direction, causing the rocket to move in the opposite direction with an equal and opposite momentum. Without this principle, rockets would not be able to generate the necessary thrust for space travel.

How does the conservation of linear momentum apply to rocket propulsion?

The conservation of linear momentum applies to rocket propulsion through Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In the case of a rocket, the action is the expulsion of gases at high speeds, and the reaction is the rocket's forward movement in the opposite direction.

Can the conservation of linear momentum be violated in rocket propulsion?

No, the conservation of linear momentum is a fundamental law of physics that cannot be violated. Any violation of this principle would require an external force to act on the system, which is not possible in the vacuum of space where rockets operate.

How does the mass of the rocket and its exhaust gases affect the conservation of linear momentum?

The conservation of linear momentum states that the total momentum of a system remains constant, so any change in the mass of the system will result in a change in its velocity. In the case of a rocket, as the exhaust gases are expelled and the rocket's mass decreases, its velocity increases to maintain the total momentum of the system.

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