A gun is mounted on a railroad car. The combined mass of the car,...

Let the initial mass of the system be $M_0 = 50m$ and the velocity of the shell relative to the gun prior to firing be $u = 200 \text{ m/s}$.

For the first shot, the initial momentum of the system is zero. Since the gun is initially at rest, the velocity of the first shell relative to the ground is $u$. The mass of the remaining system (car, gun, operator, and remaining shells) is $49m$. Let $V_1$ be the velocity of the car after the first shot. By conservation of linear momentum:

$0 = 49m V_1 + m u$

$\Rightarrow V_1 = -\dfrac{u}{49}$

For the second shot, the initial momentum of the system is $49m V_1$. The velocity of the second shell relative to the ground is $V_1 + u$, as its velocity relative to the gun prior to firing is $u$. The mass of the remaining system is $48m$. Let $V_2$ be the velocity of the car after the second shot. By conservation of linear momentum:

$49m V_1 = 48m V_2 + m(V_1 + u)$

$\Rightarrow 48m V_1 = 48m V_2 + m u$

$\Rightarrow V_2 = V_1 - \dfrac{u}{48}$

Substituting the value of $V_1$:

V_2 = -\dfrac{u}{49} - \dfrac{u}{48} = -u \left(\dfrac{1}{49} + \dfrac{1}{48} \right)

The recoil speed is the magnitude of $V_2$. Substituting $u = 200 \text{ m/s}$:

|V_2| = 200 \left(\dfrac{1}{49} + \dfrac{1}{48} \right) \text{ m/s}