Firework problem, conservation of energy?

[Jrlinton]

Homework Statement

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 into 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?

Homework Equations

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?[/B]

 

[Jrlinton]

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

 

[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.

 

[Orodruin]

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.

 

[gneill]

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

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

 

[Jrlinton]

This is the entire problem:

 

[Jrlinton]

 

[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.

 

[Orodruin]

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.

 

[gneill]

The problem is missing vital information necessary to solve it.

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.

 

[Jrlinton]

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.

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.

 

[Orodruin]

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.

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

 

[Orodruin]

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.

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

Hint: This is a conservation of momentum problem.

 

[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

 

[gneill]

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.

 

[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?

 

[haruspex]

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

Yes. Actually, even closer to 170.

 

[gneill]

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?

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

 

[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.