I was doing a very old Applied Mathematics exam (from the 70s... yeah I lead a sad life) and came across the following problem.
A shell is fired from a gun mounted on a carriage which is moving forward at a speed of $2m/s$. The mass of the shell is $50\;kg$ and the combined mass of the gun and shell is $10 \;tonne$. The recoil brings the gun and carriage to rest. If the shell is fired horizontally, find the speed of the shell relative to the ground as it leaves the gun.
My solution uses the Principle of the Conservation of Momentum.
$$ m_1u_1 + m_2u_2 = m_1v_1 + m_2(v_2 + 2)$$
$m_1 =$ mass of gun and carriage and $m_2 =$ mass of shell (which is in the gun). Both $m_1$ and $m_2$ are traveling at $2\;m/s$.
$$ (10 \times 10^3)(2) + 50(2) = m_1v_1 + m_2(v_2 + 2)$$
Gun and carriage come to rest i.e. $v_1 = 0$, and the shell leaves the gun with a speed of $ (v_2 + 2)$. The $2$ is added to $v_2$ since it is originally moving with a speed of $2\;m/s$.
$$ (10 \times 10^3)(2) + 50(2) = (10 \times 10^3)(0) + (50) (v_2 + 2)$$ $$ 20,000 + 100 = 0 + (50) (v_2 + 2)$$ $$ 20,100 = (50) (v_2 + 2)$$ $$ 402 = v_2 + 2 $$ $$ v_2 = 400$$
The shell leaves the gun at $400 \;m/s$.
Is my logic correct? Or is there a twist in this question I cannot see. BTW the question was worth 6 marks out of 100 marks total.
Thanks