Question

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?

Answer 1

Answer:

k_2 =3826.08 J

K_1 = 4973.91 J

Explanation:

from conservation of momentum principle

$0 = m_1v_1+m_2v_2$

we know that kinetic energy is given as

$k_1 = \frac{1}{2} m_1v_1^2$

$k_2 = \frac{1}{2} m_2v_2^2$

we have $v_1 =-1.3v_2$, $m_2 = 1.3m_1$

therefore $k_1 =\frac{1}{2} (\frac{m_2}{1.3})(-1.3v_2)^2 = 1.3k_2$

now

$k_1+k_2 = 8800$

$1.3k_2 +k_2 =8800$

k_2 =3826.08 J

K_1 = 4973.91 J