An explosion blows a rock into three parts. Two parts go off at right angles to each other: a $1\text{kg}$ part moving at $12{\text{ms}}^{-1}$ and a $2\text{kg}$ part moving at $8{\text{ms}}^{-1}$. If the third part flies off with a velocity of $4{\text{ms}}^{-1}$, its mass would be:

By conservation of linear momentum, the total momentum after the explosion must be zero (${\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3=0$).Momentum of first part: $p_1=m_1{v}_1=1\times 12=12{\text{kg ms}}^{-1}$ (let's say along x-axis).Momentum of second part: $p_2=m_2{v}_2=2\times 8=16{\text{kg ms}}^{-1}$ (at $90^{\circ }$, so along y-axis).Magnitude of resultant of $p_1$ and $p_2$:
$$p_{res}=\sqrt{p_1^2+p_2^2}=\sqrt{12^2+16^2}=\sqrt{144+256}=\sqrt{400}=20{\text{kg ms}}^{-1}$$
The third part must have momentum $p_3=20{\text{kg ms}}^{-1}$ in the opposite direction.
$$p_3=m_3{v}_3⟹20=m_3\times 4$$
$$m_3=5\text{kg}$$