A moving object of mass 3 kg and velocity (4 i + 4 j ) m/s breaks...

By the principle of conservation of linear momentum:
$\vec{P}_{initial} = \vec{P}_{final}$
$M\vec{v} = m_1\vec{v}_1 + m_2\vec{v}_2$

Substituting the given values:
$3(4\hat{i} + 4\hat{j}) = 1(4\hat{i} - 2\hat{j}) + 2\vec{v}_2$
$12\hat{i} + 12\hat{j} = 4\hat{i} - 2\hat{j} + 2\vec{v}_2$
$2\vec{v}_2 = (12 - 4)\hat{i} + (12 + 2)\hat{j}$
$2\vec{v}_2 = 8\hat{i} + 14\hat{j}$
$\vec{v}_2 = 4\hat{i} + 7\hat{j}\text{ m/s}$

Now, calculate the initial kinetic energy ($K_i$):
$K_i = \dfrac{1}{2} M |\vec{v}|^2 = \dfrac{1}{2} \times 3 \times (4^2 + 4^2) = \dfrac{1}{2} \times 3 \times 32 = 48\text{ J}$

Calculate the final kinetic energy ($K_f$):
$K_f = \dfrac{1}{2} m_1 |\vec{v}_1|^2 + \dfrac{1}{2} m_2 |\vec{v}_2|^2$
$K_f = \dfrac{1}{2} \times 1 \times (4^2 + (-2)^2) + \dfrac{1}{2} \times 2 \times (4^2 + 7^2)$
$K_f = \dfrac{1}{2}(16 + 4) + 1(16 + 49)$
$K_f = 10 + 65 = 75\text{ J}$

The kinetic energy released in the explosion is the increase in kinetic energy:
$\Delta K = K_f - K_i = 75\text{ J} - 48\text{ J} = 27\text{ J}$.

Answer: $27\text{ J}$