physics help :) 9kg artillery shell explodes and splits into 2 pieces

[phatj]

this is for a quantum mechanics class... i just cant figure out the first part of this question (later i have to apply some lorentz transforms and crap) but i just cant remember general mechanics well enough to figure this one out:


A 9.0 kg artillery shell is moving to the right at 100 m/s when suddenly it explodes into two fragments, one twice as heavy as the other. Measurements reveal that 780 J of energy are released in the explosion and that the heavier fragment was in front of the lighter fragment. Take the positive direction to be to the right.

(a) First, analyze the explosion in the reference frame of the ground. Calculate the post-explosion velocities of each fragment.




so im guessing:

Einitial = Elost + E(kinetic large fragment) + E(kinetic small fragment)

.5(9kg)(100)^2 = 780J + .5(6)(v1)^2 + .5(3)(v2)^2

so now i just have 2 unknowns, v1 and v2. But i cant figure out how to take it one step further to determine the individual velocities.

Also, why the hell do they mention that the big fragment is further infront of the small fragment?

 

[Gibsons]

dunno if I really understand this, but....

Also, why the hell do they mention that the big fragment is further infront of the small fragment?

The big fragment gets propelled forwards by the explosion, the smaller fragment gets accelerated backwards (to the left), though it still might be moving to the right.... so the explosion doesn't add any ke or momentum to the system, and m1v1(one piece) will equal m2v2 (small piece + big piece). I dunno if you adjust the velocity of the small pieces by applying 780J to each piece or 390J to each. One negative (small pieces) and one positive (big piece), by my understanding.

but again I'm not sure at all about this.

 

[drinkmorejava]

Do you really know that the 780 J is taken from its momentum. If it's from the explosion itself, then some of that would go into deforming the shell rather than into changing the velocity. The question is to vague about where the energy comes from and what it goes into. If I was doing it, I'd say the 780 deformed the shell and the velocities stayed the same.

 

[phatj]

not sure... im assuming Einitial = 780 + KE1 + KE2 tho...

 

[ProviaFan]

The center of mass of the two fragments keeps moving as the projectile originally did, so the velocities that you calculate will be relative to the movement of the center of mass. I get lost with the energy part of this, but assuming that your first equation is right, then conservation of momentum should give you the second equation that you need to find v1 and v2.

 

[phatj]

should momentum really be conserved? cuz energy is lost...

 

[ProviaFan]

Originally posted by: phatj
should momentum really be conserved? cuz energy is lost...

Given only what you listed in your post, we have no way of knowing if and/or how much energy is lost. With no more information than that, I would have to assume that all of the energy released is in the form of kinetic energy of the fragments. If energy is lost, then you'll need to supply more info for anyone to be able to help, I'm afraid.

 

[phatj]

so 780 = .5(6)(v1)^2 + .5(3)(v2)^2 ?

 

[Gibsons]

I think you can do it with just MV, without using 0.5MV^2

Like ProviaFan says, the momentum of the center of mass of the two fragments is the same as that of the original fragment.

 

[phatj]

so where does the 780J come in?

 

[DrPizza]

Just a shot in the dark here, because I think it's a stupid question (how much of that energy goes to heat, etc.) But, because the action/reaction forces need to be equal, I'm just going to assume that each of the pieces receives the same impulse. I'm interpreting the 760J released to mean that some chemical energy was released, and the prof/teacher wants you to believe that all of it was converted to KE. So, from a frame of reference of the center of mass, 3kg*v1 = 6kg*v2 (velocity 1 and velocity 2) **note v1 is to the left and v2 is to the right; stick in a (-) sign if you so desire, or write it as the sum of the momenta = 0. (from the center of mass frame of reference)
.5(3)(v1)^2 + .5(6)(v2)^2 = 760. 2 equations, 2 unknowns. From the ground's frame of reference, simply add on the 100m/s to each.