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JerJason

  • 2 years ago

An internal explosion breaks an object initially at rest into two pieces, one of which has 1.6 times the mass of the other. If 5900J is released in the explosion, how much kinetic energy does each piece acquire?

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  1. JerJason
    • 2 years ago
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    I know that one of the pieces has to exert kinetic energy that is 1.6 greater than the other, and that this amount plus that amount should equal 5900J. I'm unsure on how to obtain this answer though, I could just plug in random values until I come up with the answer, but that wouldn't be the best approach...

  2. shubham
    • 2 years ago
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    Explosion is an internal phenomenon, forces involved are internal and hence, form action-reaction force pair. So, Forces on both the pieces ( A and B) are equal in magnitude, opposite in direction i.e. Fa = Fb [ eq.1] Here, you know the amount of energy released = 5900 J So, 1/2* Ma* Va^2 + 1/2*Mb*Vb^2 = 5900 [ 4 variables ] and you know relation b/w Ma and Mb [ Ma = 1.6*Mb] You can also find a relation b/w Va and Vb using eq.1 Hope you know how to proceed further !!

  3. salini
    • 2 years ago
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    first what energy does the bomb(with all parts intact )have?

  4. JerJason
    • 2 years ago
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    potetial energy, there is also no momentum on the bomb yet.

  5. salini
    • 2 years ago
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    yeah no momentum beautiful now as theree was only an INTERNAL explosion(force) momentum will be conserved? so initial momentum of system=final momentum of system?

  6. JerJason
    • 2 years ago
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    Yes the total system of the bomb would be conserved since adding each individual piece after collision would equal the same momentum.

  7. experimentX
    • 2 years ago
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    apply conservation of momentum the total momentum is zero. m1 v1 = m2 v2 and 1/2 m1 v1^2 + 1/2 m2 v2^2 = 5900 also m1 = 6 m2

  8. salini
    • 2 years ago
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    yeah can u do that? system is always BOMB as awhole

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