An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 5500 J is released in the explosion, how much kinetic energy does each piece acquire?

The linear momentum of a body is defined as the product of its mass and velocity. It is given by \(p = mv\).

As the conservation of linear momentum,

\(\begin{array}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{array}\)

Let the object be broken into two fragments: A (heavier piece) and B (lighter piece).

The mass of A is\({m_{\rm{A}}} = 1.5{m_{\rm{B}}}\).

The initial velocities of A and B are\({v_{\rm{A}}} = {v_{\rm{B}}} = 0\).

The amount of energy released in the explosion is\({E_{{\rm{released}}}} = 5500\;{\rm{J}}\).

The fragments A and B will separate and travel in opposite directions. Suppose the direction of motion of A is along the positive direction. Let the final velocities of A and B after the explosion be \({v_{\rm{A}}}^\prime \) and \({v_{\rm{B}}}^\prime \), respectively.

Applying the conservation of momentum in one dimension,

\(\begin{array}{c}{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = {m_{\rm{A}}}{v_{\rm{A}}}^\prime + {m_{\rm{B}}}\left( { - {v_{\rm{B}}}^\prime } \right)\\0 = \left( {1.5{m_{\rm{B}}}} \right){v_{\rm{A}}}^\prime + {m_{\rm{B}}}\left( { - {v_{\rm{B}}}^\prime } \right)\\{v_{\rm{A}}}^\prime = \frac{2}{3}{v_{\rm{B}}}^\prime \end{array}\) … (i)

The negative sign with the final velocity of B indicates its direction.

The mechanical energy before the collision is zero as the original object is at rest. The total kinetic energy of the fragments after the explosion must be equal to the energy released during the explosion. Therefore,

\(\begin{array}{c}{E_{{\rm{released}}}}=\frac{1}{2}{m_{\rm{A}}}{v_{\rm{A}}}{\prime2}+\frac{1}{2}{m_{\rm{B}}}{v_{\rm{B}}}{\prime2}\\=\frac{1}{2}\left( {1.5{m_{\rm{B}}}} \right){\left( {\frac{2}{3}{v_{\rm{B}}}^\prime } \right)2} + \frac{1}{2}{m_{\rm{B}}}{v_{\rm{B}}}{\prime2}\\=\frac{5}{3}\left({\frac{1}{2}{m_{\rm{B}}}{v_{\rm{B}}}{{\prime}2}}\right)\\=\frac{5}{3}K{E_{\rm{B}}}\prime\end{array}\) … (ii)

From equation (ii), the kinetic energy acquired by fragment B is as follows:

\(\begin{array}{c}K{E_{\rm{B}}}^\prime = \frac{3}{5}{E_{{\rm{released}}}}\\ = \frac{3}{5}\left( {5500\;{\rm{J}}} \right)\\ = 3300\;{\rm{J}}\end{array}\)

The kinetic energy acquired by fragment A is as follows:

\(\begin{array}{c}{E_{{\rm{released}}}} = K{E_{\rm{A}}}^\prime + K{E_{\rm{B}}}^\prime \\K{E_{\rm{A}}}^\prime = {E_{{\rm{released}}}} - K{E_{\rm{B}}}^\prime \\ = 5500\;{\rm{J}} - 3300\;{\rm{J}}\\ = 2200\;{\rm{J}}\end{array}\)