How is kinetic energy and momentum conserved in an internal explosion?

[pb23me]

Homework Statement

an internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did each piece acquire.

Homework Equations

K₁+K₂=7500J
3/2mv_i+mv_i=3/2mv_f+mv_f
p²/2m=KE

The Attempt at a Solution

I am not sure exactly when kinetic energy or momentum begin and end in the conservation equations. Does kinetic energy begin at zero and end right after the explosion? Does momentum begin at zero as well? and where does it end? This is a big source of confusion for me.

 

[tiny-tim]

hi pb23me!

momentum is conserved in all collisions or explosions (reverse collisions!)

energy, obviously, is not conserved in an explosion, but you're told that the initial KE is 0, so the final KE must be 7500 J

 

[rock.freak667]

The Attempt at a Solution

I am not sure exactly when kinetic energy or momentum begin and end in the conservation equations. Does kinetic energy begin at zero and end right after the explosion? Does momentum begin at zero as well? and where does it end? This is a big source of confusion for me.

Well it is initially at rest, so the initial momentum is zero.

the final momentum will be mv₁+(1.5m)v₂. You can now form the equation for conservation of momentum

Now energy is is also conserved so 7500 = KE₁ + KE₂

KE₁ = (1/2)mv₁²

 

[pb23me]

Hi tiny tim, i just read something about this that's seems to be a contradiction... In an explosion the initial momentum is zero, the pieces fly in opposite directions so momentum is conserved. Well then how does that work out for the conservation of momentum equation m₁v_i+m₂v_i=m₁v_f+m₂v_f... it starts with zero and ends with some amount? How is that conservation. Also what is the final momentum is it right after the explosion? KE_f is that right after the explosion as well?

 

[pb23me]

or rockfreak thanx.

 

[tiny-tim]

hi pb23me!

m₁v_i+m₂v_i=m₁v_f+m₂v_f... it starts with zero and ends with some amount? How is that conservation.

momentum (unlike energy) is a vector, so the final momentum can (and does!) still add to zero

Also what is the final momentum is it right after the explosion? KE_f is that right after the explosion as well?

it doesn't matter …

momentum is always conserved anyway,

and energy will only fail to be conserved during the explosion itself

 

[pb23me]

Ok so 1/2mv₁²+1/2mv₁²/2.25=7500J
3.25mv₁²=33750
1/2mv₁²=5192.3J
1/2mv₂²=2307.7J

 

[tiny-tim]

Ok so 1/2mv₁²+1/2mv₁²/2.25=7500J
3.25mv₁²=33750
1/2mv₁²=5192.3J
1/2mv₂²=2307.7J

sorry, I'm following neither the logic nor the arithmetic

where is the conservation of momentum?

try using your KE = p²/2m equation ​

 

[pb23me]

oops... ok v₂=-3/2v₁ so [(3/2)mv₁]²/2m+(mv₂)²/2m=7500J so plugging in v₂=-3/2v₁ i get 15mv₁²/8=7500J
KE₁=2000J
KE₂=5500J

 

[tiny-tim]

[(3/2)mv₁]²/2m+(mv₂)²/2m=7500J

why are the masses on the bottom the same?

why are the momentums on the top different?

 

[pb23me]

the masses on the bottom are supposed to be 3m and 2m. I had the right masses in my calculation just forgot to write that on here. And for the momentums on the top it should be 9/4(m)v₁² for the first and (mv₂)² for the second

 

[tiny-tim]

the masses on the bottom are supposed to be 3m and 2m. I had the right masses in my calculation just forgot to write that on here. And for the momentums on the top it should be 9/4(m)v₁² for the first and (mv₂)² for the second

oh that's hilarious!

i'm going to bed now :zzz:

see if you can tidy it up by the morning! ​

 

[pb23me]

[(3/2)mv₁]²/(3/2)2m+(mv₁)²/2m=7500J
so 3mv₁²/4+mv₁²/2=7500J
15mv₁²/8=7500J
mv₁²=4000J
so 1/2mv₁²=2000J
1/2mv₂²=5500J
I don't understand where my error is at?

 

[tiny-tim]

so 3mv₁²/4+mv₁²/2=7500J
15mv₁²/8=7500J

nooo … 5mv₁²/4=7500J

 

[pb23me]

oh ok... That was from an earlier miss calculation that i didnt go back and check...thanx