Rocket: conservation of momentum

In summary, using the conservation of momentum equation, the magnitudes of the velocities v1 and v2 of the two pieces of the fireworks rocket after it breaks into two pieces can be determined. However, the exact values of v1 and v2 depend on the cause of the rocket's breakage and may not necessarily follow the expressions given in the solution attempt. A more detailed problem statement and a diagram may be needed to accurately solve this problem.
  • #1
Np14
27
2

Homework Statement


A fireworks rocket is moving at a speed of 45.0 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 and v2. What are the magnitudes of v1 and v2?

Homework Equations


Conservation of Momentum
m1v1 + m2v2 = m1vo1 + m2vo2

The Attempt at a Solution


v1 = 45cos30° = 38.97
v2 = 45cos60° = 22.55

45 m/s ≠ 61.47 m/s

I have a feeling this problem is a lot more complicated, but I am not sure how to solve it.
 

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  • #2
Np14 said:

Homework Statement


A fireworks rocket is moving at a speed of 45.0 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 and v2. What are the magnitudes of v1 and v2?

Homework Equations


Conservation of Momentum
m1v1 + m2v2 = m1vo1 + m2vo2

The Attempt at a Solution


v1 = 45cos30° = 38.97
v2 = 45cos60° = 22.55

45 m/s ≠ 61.47 m/s

I have a feeling this problem is a lot more complicated, but I am not sure how to solve it.
Your problem statement makes no mention of 30 and 60 degrees yet you introduce them in your solution. I think you should draw a diagram and do a better job of stating the problem.
 
  • #3
Answer depends on what caused the rocket to break up. If it was an internal explosion then the answer depends on how severe the explosion was. If it just broke in two, what would v1 and v2 be?
 
  • #4
Np14 said:
have a feeling this problem is a lot more complicated
Yes, your method has no logic to it. Why should the speeds be given by those expressions?
Use the conservation equation you quoted, once in the original direction and once at right angles to that.
 
  • #5
Np14 said:

Homework Statement


A fireworks rocket is moving at a speed of 45.0 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 and v2. What are the magnitudes of v1 and v2?

Homework Equations


Conservation of Momentum
m1v1 + m2v2 = m1vo1 + m2vo2

The Attempt at a Solution


v1 = 45cos30° = 38.97
v2 = 45cos60° = 22.55

p1 = 2m x 45 cos30° = 2m x 38.97
p2 = 2m x 45 cos60° = 2m x 22.55


45 m/s ≠ 61.47 m/s

I have a feeling this problem is a lot more complicated, but I am not sure how to solve it.
 

1. What is the conservation of momentum?

The conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant over time, regardless of any internal changes or interactions within the system. In simpler terms, this means that the total amount of movement in a system cannot be created or destroyed, only transferred between objects within the system.

2. How does the conservation of momentum apply to rockets?

Rockets are a classic example of the conservation of momentum in action. As the rocket engine expels hot gases in one direction, the rocket is propelled in the opposite direction with an equal and opposite force, thanks to Newton's third law of motion. This applies to all stages of a rocket's launch, as well as its movement through space.

3. What factors affect the conservation of momentum in a rocket?

The conservation of momentum in a rocket is affected by a variety of factors, including the mass of the rocket, the velocity and direction of the ejected gases, and any external forces acting on the rocket, such as gravity or air resistance. All of these factors must be taken into account when designing and launching a rocket.

4. How is momentum conserved during a rocket launch?

During a rocket launch, the conservation of momentum is maintained through the use of the rocket's engines. As the engines generate thrust and expel hot gases, the rocket is propelled in the opposite direction, conserving the overall momentum of the system. Additionally, the design of the rocket's stages and trajectory must also take into account the conservation of momentum.

5. What are some real-world applications of the conservation of momentum?

The conservation of momentum has many practical applications in everyday life, including in transportation (such as rockets, cars, and planes), sports (such as throwing a ball or jumping), and even in the motion of celestial bodies in space. It is also used in the design and operation of machines, such as turbines and engines, and is a crucial concept in the field of physics and engineering.

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