A 725-kg two-stage rocket is traveling at a speed of \({\bf{6}}{\bf{.60 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{m/s}}\) away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of \({\bf{2}}{\bf{.80 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{m/s}}\)relative to each other along the original line of motion.

(a) What is the speed and direction of each section (relative to Earth) after the explosion?

(b) How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

According to the law of conservation of linear momentum, when no external force acts on the system, the momentum of a closed, isolated system remains conserved.

In this problem, a rocket moving with a certain velocity explodes into two sections. Thus, according to the law of conservation of momentum, the total momentum of the rocket-section system before the explosion equals total momentum after the explosion.

Mass of rocket, m = 725 kg.

Speed of the rocket,\(v = 6.60 \times {10^3}\;{\rm{m/s}}\).

If a rocket explodes into two sections, A and B, with velocities\({v_{\rm{A}}}\)and\({v_{\rm{B}}}\), respectively, relative to the Earth, their speed relative to each other is\({v_{{\rm{AB}}}} = 2.80 \times {10^3}\;{\rm{m/s}}\).

Let masses of sections A and B of the rocket are\({m_{\rm{A}}}\)and\({m_{\rm{B}}}\)such that\({m_{\rm{A}}} = {m_{\rm{B}}}\).

Thus, the mass of each section will be\({m_{\rm{A}}} = {m_{\rm{B}}} = \frac{m}{2}\)

Consider the direction of motion of the rocket before the explosion to be the positive direction, i.e., direction away from the Earth to be positive.

In terms of relative velocity, the velocity of section B can be written as follows:

\(\begin{array}{c}{v_{{\rm{AB}}}} = {v_{\rm{A}}} - {v_{\rm{B}}}\\{v_B} = {v_{\rm{A}}} - {v_{{\rm{AB}}}}\\\;\end{array}\) ... (i)

The momentum of the rocket before the explosion is \({p_{\rm{i}}} = mv\).

The momentum of the rocket after the explosion is calculated as follows:

\(\begin{array}{c}{p_{\rm{f}}} = {m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}}\\ = {m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}\left( {{v_{\rm{A}}} - {v_{{\rm{AB}}}}} \right)\\ = {m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{A}}} - {m_{\rm{B}}}{v_{{\rm{AB}}}}\\ = \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right){v_{\rm{A}}} - {m_{\rm{B}}}{v_{{\rm{AB}}}}\\ = m{v_{\rm{A}}} - \frac{{m{v_{{\rm{AB}}}}}}{2}\end{array}\)

On applying the law of conservation of linear momentum on the rocket-section system,

\(\begin{array}{c}{p_{\rm{i}}} = {p_{\rm{f}}}\\mv = m{v_{\rm{A}}} - \frac{{m{v_{{\rm{AB}}}}}}{2}\\{v_{\rm{A}}} = v + \frac{{{v_{{\rm{AB}}}}}}{2}\\{v_{\rm{A}}} = \left( {6.60 \times {{10}^3}\;{\rm{m/s}}} \right) + \frac{{\left( {2.80 \times {{10}^3}\;{\rm{m/s}}} \right)}}{2}\\ = 8.00 \times {10^3}\;{\rm{m/s}}\end{array}\)

Now, substitute the value of\({v_{\rm{A}}}\)in equation (i) to get the value of\({v_{\rm{B}}}\):

\(\begin{array}{c}{v_B} = {v_{\rm{A}}} - {v_{{\rm{AB}}}}\\ = \left( {8.00 \times {{10}^3}\;{\rm{m/s}}} \right) - \left( {2.80 \times {{10}^3}\;{\rm{m/s}}} \right)\\ = 5.20 \times {10^3}\;{\rm{m/s}}\end{array}\)

Thus, the speed of fragment A is \(8.00 \times {10^{ - 3}}\;{\rm{m/s}}\), and the speed of fragment B is \(5.20 \times {10^3}\;{\rm{m/s}}\). Since the magnitude of both \({v_{\rm{A}}}\) and \({v_{\rm{B}}}\) is positive, the direction of both the fragments is away from the Earth.

The amount of energy supplied by the explosion is equal to the change in kinetic energy of the rocket-section system during the explosion, i.e.,

\(\begin{array}{c}\Delta KE = K{E_{\rm{f}}} - K{E_{\rm{i}}}\\ = \left[ {\frac{1}{2}{m_{\rm{A}}}v_{\rm{A}}^{\rm{2}} + \frac{1}{2}{m_{\rm{B}}}v_{\rm{B}}^{\rm{2}}} \right] - \frac{1}{2}m{v^2}\\ = \left[ {\frac{1}{4}mv_{\rm{A}}^{\rm{2}} + \frac{1}{4}mv_{\rm{B}}^{\rm{2}}} \right] - \frac{1}{2}m{v^2}\\ = \frac{1}{4}m\left( {v_{\rm{A}}^{\rm{2}} + v_{\rm{B}}^{\rm{2}}} \right) - \frac{1}{2}m{v^2}\end{array}\)

On substituting the given values in the above expression,

\(\begin{array}{c}\Delta KE = \frac{1}{4}m\left( {v_{\rm{A}}^{\rm{2}} + v_{\rm{B}}^{\rm{2}}} \right) - \frac{1}{2}m{v^2}\\ = \frac{1}{2}\left( {725\;{\rm{kg}}} \right)\left[ {\left( {8.00 \times {{10}^3}\;{\rm{m/s}}} \right) + \left( {5.20 \times {{10}^3}\;{\rm{m/s}}} \right)} \right] - \frac{1}{2}\left( {725\;{\rm{kg}}} \right){\left( {6.60 \times {{10}^3}\;{\rm{m/s}}} \right)^2}\\ = 7.11 \times {10^8}\;{\rm{J}}\end{array}\)

Thus, \(7.11 \times {10^8}\;{\rm{J}}\) of energy was supplied by the explosion.