Question

An internal explosion breaks an object, initially at rest, into two pieces, one of which has 2.1 times the mass of the other.If 8700 J were released in the explosion, how much kinetic energy did each piece acquire?

Answer 1

Let the mass of two pieces be m_s and m_l

Given $m_{l}=2.1*m_{s}$

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Initially object is at rest ,so initial momentum is zero

During explosion momentum is conserved

So initial momentum =final momentum =zero

Final momentum, $p_{final}=m_{s}v_{s}+m_{l}v_{l}=0$

$m_{s}v_{s}=-m_{l}v_{l}$

$m_{l}=2.1*m_{s}$

$m_{s}v_{s}=-2.1*m_{s}v_{l}$

${\color{Red} v_{s}=-2.1*v_{l}}$

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Now consider the kinetic energy of both pieces

$KE_{s}=\frac{1}{2}m_{s}v_{s}^{2}$

$KE_{l}=\frac{1}{2}m_{l}v_{l}^{2}$

find the relation between both kinetic energies

$KE_{s}=\frac{1}{2}m_{s}v_{s}^{2}$

$KE_{s}=\frac{1}{2}(\frac{m_{l}}{2.1})*(-2.1v_{l})^{2}$

$KE_{s}=\frac{1}{2}(\frac{m_{l}}{2.1})*2.1^{2}v_{l}^{2}$

$KE_{s}=\frac{1}{2}m_{l}*2.1v_{l}^{2}$

$KE_{s}=2.1*\frac{1}{2}m_{l}v_{l}^{2}$

${\color{Red} KE_{s}=2.1*KE_{l}}$

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Energy released in explosion =8700J

$KE_{s}+KE_{l}=8700J$

$2.1KE_{l}+KE_{l}=8700J$

$3.1KE_{l}=8700J$

$KE_{l}=8700/3.1$

ANSWER: ${\color{Red} KE_{l}=2806.45J}$

$KE_{s}=2.1*KE_{l}=2.1*\frac{8700}{3.1}$

ANSWER: ${\color{Red} KE_{s}=5893.55J}$

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