If a particle of mass $M$ spontaneously explodes breaking up into two pieces we know by energy conservation $m_1 v_1 = -m_2v_2$, so if i measure the position of $m_1$ accurately and measure the velocity $v_2$ then by this equation i could also accurately get the velocity of $m_1$ along with its position which violates the uncertainty principle. What is going on here? Is momentum conservation not applicable in this case?
Conservation means that $m_1v_1$ is exactly equal and opposite to $m_2v_2$. But how can that be true if $m_1v_1$ is uncertain?
Here’s where the idea of correlated states enters in. The final state has both equal & opposite momenta and uncertain moments because the two sides are correlated. If, due to uncertainty, $m_1v_1$ is a bit small, so will $m_2v_2$ be. Ditto is they’re a bit larger, they’re both larger.