# Homework Help: Firework problem, conservation of energy?

1. Nov 15, 2016

### Jrlinton

1. The problem statement, all variables and given/known data
A firework is shot from the ground at a speed of 50 m/s and an angle of 45 degrees. At its highest point it blows up in to two pieces due to internal forces. One piece (.25 mass) falls straight down to the ground below the explosion, where does the second piece land?

2. Relevant equations

3. The attempt at a solution
So first i used the equation of finding max height of projectile.
h=Vo^2sin^2(theta)/(2g)
h=63.7105 m
So then i would take the potential energy at the time of the explosion (mgh) and set it equal to the kinetic energies of the two pieces after the explosion? What would the velocity be of the falling mass? the x component of the velocity, zero?

2. Nov 15, 2016

### Jrlinton

Am i correct in assuming that the energy at the explosion is all potential as the velocity would be instantaneously zero?

3. Nov 15, 2016

### Jrlinton

If this is indeed the path to the solution then i found the velocity of the 3/4 mass to be 13.6 m/s for 3.6 seconds (found in freefall equation) and its displacement to be 49.04 m from the mass that fell to the ground.

4. Nov 15, 2016

### Orodruin

Staff Emeritus
Is this the problem statement exactly as posed? Is there an accompanying illustration?

You cannot use conservation of energy, it is explicitly stated that there is a release of internal energy in the explosion.

5. Nov 15, 2016

### Staff: Mentor

Which velocity are you referring to? The firework was launched at an angle of 45°....

6. Nov 15, 2016

### Jrlinton

This is the entire problem:

7. Nov 15, 2016

### Jrlinton

8. Nov 15, 2016

### Jrlinton

I was understanding that because the explosion came from forces inside the system then the energy would be conserved as no external forces had acted on it.

9. Nov 15, 2016

### Orodruin

Staff Emeritus
The problem is missing vital information necessary to solve it.

Regardless, you cannot apply conservation of energy. The potential energy is the same before and after the explosion, it is just distributed as potential energy of the individual parts.

10. Nov 15, 2016

### Staff: Mentor

I suspect that we are meant to interpret "One piece (.25 mass) falls straight down to the ground" as implying it momentarily has no velocity and thus free-falls from rest. That would constrain things sufficiently.

11. Nov 15, 2016

### Jrlinton

Yes that is my understanding. As the horizontal velocity is instantaneously zero it explodes in a way that causes the .25 mass to fall straight down in a freefall and the remaining mass to continue as a projectile.

12. Nov 15, 2016

### Orodruin

Staff Emeritus
I suspect so too, but it is far from clear from the problem statement. People need to learn to write proper problem statements.

13. Nov 15, 2016

### Orodruin

Staff Emeritus
The horizontal velocity of what? The horizontal velocity of the original rocket is constant.

Hint: This is a conservation of momentum problem.

14. Nov 15, 2016

### Jrlinton

Okay so can I use the x component of the velocity?
m*50m/s*cos45=.25m*0m/s+.75m*v
with v=47.1405m/s and time being the same time of 3.6 sI found using the freefall equation?
Making distance from the fallen mass being 47.14m/s*3.6s=169.7m

15. Nov 15, 2016

### Staff: Mentor

Yes, that looks good.

Of course the question doesn't state what reference location to use to specify the position. The initial launch point? The location of the smaller piece? I guess you should give both.

16. Nov 15, 2016

### Jrlinton

I thought of that as well. So if it was in fact in reference to launch point then i could just add half of the horizontal range of the original launch right?

17. Nov 15, 2016

### haruspex

Yes. Actually, even closer to 170.

18. Nov 15, 2016

### Staff: Mentor

Sure. You've got time and velocity information that would do it, too.

19. Nov 15, 2016

### mpresic

What can you say about the center of mass of the system of particles. Can you work out the location of the center of mass as a function of time. What does this imply about the location of the mass you seek.