How Does the Law of Conservation of Momentum Apply in a Firework Explosion?

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The discussion centers on the application of the law of conservation of momentum in a firework explosion scenario. The initial momentum of the "egg" is zero since it is at rest before the explosion. After the explosion, the two pieces move in opposite directions, maintaining the total momentum at zero. The forces between the two pieces during the explosion are equal, in accordance with Newton's third law of motion. This reinforces the principle that momentum is conserved in the system.
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Homework Statement


A giant "egg" explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions.
1013816.jpg

Part A
What is the momentum PA,i of piece A before the explosion?(initial momentum)
Part B
During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A?

Homework Equations


law of conservation of momentum

The Attempt at a Solution


part a
i think as it says in the question that the egg is at rest so won't the initial momentum be zero but a bit hesitant ...cant be that easy...
part b
equal forces according to Newtons law ?
 
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That's right! just set it up lie that and remmber since they go in opposite directions, one is negative so

0 = m1v1-m2v2
 
THANKS aimslin 22 for givin positive resp
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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