Understanding Momentum Conservation in a Shell Explosion: A Quick Question

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In a shell explosion, momentum is conserved, meaning the total momentum before the explosion equals the total momentum after. When the shell explodes into two equal mass fragments, the momentum of the system must remain constant. If one fragment has a speed of zero after the explosion, the other fragment must have a velocity that compensates for the loss of momentum, ensuring the total remains equal to the shell's momentum before the explosion. Therefore, each fragment does not have momentum P; rather, the momentum of the stationary fragment is zero, and the moving fragment must have momentum equal to the shell's initial momentum. This illustrates the principle that as mass decreases, velocity must increase to maintain momentum conservation.
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Hey folks, I have this problem as part of my weekly homework:

A shell is shot with an initial velocity of 20 m/s, at an angle of 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically...

The rest of the problem isn't important to my question, which is this: I know that momentum is conserved. If the shell has momentum P right before it splits in half, does each half now have momentum P or does each shell have momentum 1/2 P? If each shell has momentum P, does that mean that since the mass goes down in the shell that is not dropping, that the velocity must increase to make up for the difference?

Thanks!
 
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Write the equations for conservation of momentum. If the final velocity of say fragment one is 0, what can you say about fragment two's final velocity?
 
Hmm... so the total momentum of the two pieces must equal the momentum of the first piece before it exploded?
 
Yes. There are no external forces acting, only internal.
 
Thank you!
 
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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