To find the kinetic energy (KE) of the other part after the bomb explodes, we need to apply the principle of conservation of momentum and the formula for kinetic energy.
Given:
The initial momentum of the bomb is zero because it is at rest. According to the conservation of momentum:
\(M \cdot 0 = m_1 \cdot v_1 + m_2 \cdot v_2\)
Substituting the known values:
\(0 = 3 \cdot 16 + 6 \cdot v_2\)
This gives:
\(48 + 6v_2 = 0\)
Solving for \(v_2\):
\(v_2 = -\frac{48}{6} = -8 \, \text{m/s}\)
The negative sign indicates that the second part moves in the opposite direction.
Now, let's calculate the kinetic energy of the other part using the kinetic energy formula:
\(\text{KE}_2 = \frac{1}{2} m_2 v_2^2\)
Substituting the known values:
\(\text{KE}_2 = \frac{1}{2} \cdot 6 \cdot (-8)^2\)
\(\text{KE}_2 = \frac{1}{2} \cdot 6 \cdot 64\)
\(\text{KE}_2 = 3 \cdot 64 = 192 \, \text{J}\)
Therefore, the kinetic energy of the other part is 192 J, which matches the correct answer.