Answer:
Let mass of bomb be ?m? units. Before exploding, its momentum is zero as it is at rest. After explosion, the mass of the two equal pieces is\[\frac{m}{2}\]each. Let\[{{v}_{1}}\]be the velocity of first piece and \[{{v}_{2}}\]be the velocity of second piece \[\therefore \]momentum of bomb after explosion = momentum of first piece + momentum of second piece \[=\frac{m}{2}{{v}_{1}}+\frac{m}{2}{{v}_{2}}=m\left( \frac{{{v}_{1}}+{{v}_{2}}}{2} \right)\] According to law of conservation momentum, Momentum of bomb before explosion = Momentum of bomb after explosion \[\Rightarrow 0=\frac{m({{v}_{1}}+{{v}_{2}})}{2}\Rightarrow \,\,0=m({{V}_{1}}+{{V}_{2}})\] \[\Rightarrow 0=m{{V}_{1}}+m{{V}_{2}}\,\Rightarrow \,{{V}_{1}}=-{{V}_{2}}\] (-ve sign shows that \[{{v}_{1}}\] and \[{{v}_{2}}\] are in opposite direction) Thus, two pieces of the bomb fly in opposite direction with same velocity.
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