JEE Main 2022 — Center of Mass Momentum and Collision Question with Solution

Let the masses of the three parts be $m\text{'}, m\text{'}$ and $2m\text{'}$.

Momentum of first part $p_1=m\text{'}\times 30=30m\text{'}$ and momentum of second part $p_2=m\text{'}\times 40=40m\text{'}$.

Third part will fly in a direction opposite to the direction of resultant of $p_1$ and $p_2$, so that total momentum of the system after explosion remains zero.

Momentum of the third part is $p_3=\sqrt{p_1^2+p_2^2}=\sqrt{{\left(30m\text{'}\right)}^2+{\left(40m\text{'}\right)}^2}=50m\text{'}$

Velocity of third part, $v=\frac{50m\text{'}}{2m\text{'}}=$$25 \mathrm{m} {\mathrm{s}}^{-1}$