Explosion of one mass into three

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A 4.2 kg object explodes into three equal mass pieces, with two pieces moving at 5.0 m/s at right angles. The conservation of momentum principle helps determine the velocity of the third piece, which is necessary for calculating the total kinetic energy. The kinetic energy of each piece can be calculated using their masses and velocities. The total kinetic energy released in the explosion is approximately 103.5 J. This calculation highlights the importance of understanding momentum and energy conservation in explosive events.
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Homework Statement


A 4.2 kg object, initially at rest, "explodes" into 3 objects of equal mass. Two of these are determined to have velocities of equal magnitudes (5.0 m/s) with directions that differ by 90 degrees. How much kinetic energy was released in the explosion? (that is what is the energy of the three object systems after the explosions)


Homework Equations



To be honest, my teacher did not cover this in class so I'm naked in the dark on this one.

The Attempt at a Solution



Since they are at 90 degrees to each other, one would be 5 j m/s and the other would be 5 i m/s. I am not sure where to go from there or where the third factors in.
 
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Since the center of mass remains the same from conservation of momentum, then first figure the velocity of the third object. (Happily you can because you know the mass of all 3 pieces are equal.)

Then you know how much KE each piece has since you know their masses and velocities.
 
Ok, thank you so much. I undertand.
 
Did you get 103.5 J?
 
Perhaps you should show your calculation?
 
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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