Answer
The relative speed of the two parts after the detonation is $~~0.42~m/s$
Work Step by Step
Let the impulse on the part with a mass of $1200~kg$ be $J = 300~N~s$
The change in momentum of this part is equal to the impulse.
We can find the change in velocity of the part with a mass of $1200~kg$:
$\Delta p = J$
$m~\Delta v = J$
$\Delta v = \frac{J}{m}$
$\Delta v = \frac{300~N~s}{1200~kg}$
$\Delta v= 0.25~m/s$
The impulse on the part with a mass of $1800~kg$ is in the opposite direction.
Let the impulse on the part with a mass of $1800~kg$ be $J = -300~N~s$
The change in momentum of this part is equal to the impulse.
We can find the change in velocity of the part with a mass of $1800~kg$:
$\Delta p = J$
$m~\Delta v = J$
$\Delta v = \frac{J}{m}$
$\Delta v = \frac{-300~N~s}{1800~kg}$
$\Delta v= -0.167~m/s$
We can find the relative speed of the two parts:
$v_{rel} = 0.25~m/s-(-0.167~m/s) = 0.42~m/s$
The relative speed of the two parts after the detonation is $~~0.42~m/s$
