A bomb initially at rest explodes into three pieces. Two pieces,...

Let the mass of the third fragment be $m_3$ and its velocity be $\vec{v}_3$.

The first two fragments have mass $m$ and speed $v$, moving at right angles. Their momenta are $m v \hat{i}$ and $m v \hat{j}$.

By conservation of linear momentum:
$\vec{p}_{initial} = \vec{p}_{final}$
$0 = m v \hat{i} + m v \hat{j} + m_3 \vec{v}_3$
$\implies m_3 \vec{v}_3 = -m v \hat{i} - m v \hat{j}$

The magnitude of the momentum of the third fragment is:
$p_3 = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2}mv$

The kinetic energy of the third fragment is:
$K_3 = \dfrac{p_3^2}{2m_3} = \dfrac{(\sqrt{2}mv)^2}{2m_3} = \dfrac{m^2v^2}{m_3}$

The total kinetic energy released is the sum of the kinetic energies of all three fragments:
$K_{total} = \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 + K_3 = mv^2 + \dfrac{m^2v^2}{m_3}$

Given that $K_{total} = \dfrac{3}{2}mv^2$:
$mv^2 + \dfrac{m^2v^2}{m_3} = \dfrac{3}{2}mv^2$
$\implies \dfrac{m^2v^2}{m_3} = \dfrac{1}{2}mv^2$
$\implies m_3 = 2m$

The total original mass of the bomb is $m + m + 2m = 4m$.

Answer: $4m$