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An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.3 times the mass of the other. If 8800 J is released in the explosion, all of which goes into the kinetic energy of the two pieces, how much kinetic energy does each piece acquire?

Respuesta :

Answer:

k_2 =3826.08 J

K_1 = 4973.91 J

Explanation:

from conservation of momentum principle

[tex]0 = m_1v_1+m_2v_2[/tex]

we know that kinetic energy is given as

[tex]k_1 = \frac{1}{2} m_1v_1^2[/tex]

[tex]k_2 = \frac{1}{2} m_2v_2^2[/tex]

we have [tex]v_1 =-1.3v_2[/tex], [tex]m_2 = 1.3m_1[/tex]

therefore [tex]k_1 =\frac{1}{2} (\frac{m_2}{1.3})(-1.3v_2)^2 = 1.3k_2[/tex]

now

[tex]k_1+k_2 = 8800[/tex]

[tex]1.3k_2 +k_2 =8800[/tex]

k_2 =3826.08 J

K_1 = 4973.91 J

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