Rocket: conservation of momentum

Click For Summary
SUMMARY

The discussion centers on the application of the conservation of momentum principle to a fireworks rocket breaking into two equal mass pieces. The initial speed of the rocket is 45.0 m/s, and the velocities of the fragments are denoted as v1 and v2. The proposed calculations using angles of 30° and 60° are incorrect as they lack context from the problem statement. The correct approach involves applying the conservation of momentum equation, m1v1 + m2v2 = m1vo1 + m2vo2, in both the original direction and perpendicular to it to accurately determine the velocities of the fragments.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Basic trigonometry for resolving velocities into components
  • Ability to analyze vector quantities in physics
  • Familiarity with the concept of mass and its relation to momentum
NEXT STEPS
  • Study the conservation of momentum in two dimensions
  • Learn how to resolve vectors into components using trigonometric functions
  • Explore examples of momentum conservation in explosive scenarios
  • Practice problems involving collisions and fragment velocities
USEFUL FOR

Students studying physics, particularly those focusing on dynamics and momentum, as well as educators seeking to clarify concepts related to momentum conservation in explosive events.

Np14
Messages
27
Reaction score
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.
 

Attachments

  • cf820974-e319-4fdc-a150-5e96133afb07.gif
    cf820974-e319-4fdc-a150-5e96133afb07.gif
    1.7 KB · Views: 516
Physics news on Phys.org
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.
 
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?
 
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.
 
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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
10
Views
6K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K