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Step-by-Step Solution
Step 1: Apply Conservation of Momentum
The bomb is initially at rest, so its initial momentum is zero. After the explosion, the momenta of the two pieces must sum to zero. Let the masses of the two pieces be $m_1 = 18\,\text{kg}$, $m_2 = 12\,\text{kg}$ and their velocities be $v_1$ and $v_2$ respectively. Given $v_1 = 6\,\text{m/s}$:
$ m_1 \, v_1 + m_2 \, v_2 = 0
\quad \Rightarrow \quad
18 \times 6 + 12 \times v_2 = 0
\quad \Rightarrow \quad
v_2 = -\frac{18 \times 6}{12}. $
Step 2: Calculate the Velocity of the Second Piece
Simplify the expression for $v_2$:
$ v_2 = -\frac{18 \times 6}{12} = -\frac{108}{12} = -9\,\text{m/s}. $
The negative sign indicates that the second piece moves in the opposite direction to the first piece.
Step 3: Compute the Kinetic Energy of the Second Piece
The kinetic energy ($K.E.$) of the second piece of mass $m_2$ moving with velocity $v_2$ is given by:
$ K.E. = \frac{1}{2} \, m_2 \, v_2^2 = \frac{1}{2} \times 12 \times (-9)^2.
$
$ K.E. = 6 \times 81 = 486\,\text{J}.
$
Thus, the kinetic energy of the second piece is $486\,\text{J}$.
