1. Apply conservation of momentum

Initially, the artillery piece and shell are at rest, so total momentum is zero.

After firing, let:

• velocity of artillery $= v_1$ (backward)
• velocity of shell $= v_2$ (forward)

Taking magnitudes, conservation of momentum gives $M_1 v_1 = M_2 v_2$

So, $v_1 = \frac{M_2}{M_1}v_2$

2. Write kinetic energies

Kinetic energy of artillery: $K_1 = \frac{1}{2}M_1 v_1^2$

Kinetic energy of shell: $K_2 = \frac{1}{2}M_2 v_2^2$

We need the ratio: $\frac{K_1}{K_2} = \frac{\frac{1}{2}M_1 v_1^2}{\frac{1}{2}M_2 v_2^2} = \frac{M_1 v_1^2}{M_2 v_2^2}$

Substitute $v_1 = \dfrac{M_2}{M_1}v_2$: $\frac{K_1}{K_2} = \frac{M_1 \left(\frac{M_2}{M_1}v_2\right)^2}{M_2 v_2^2}$

$= \frac{M_1 \cdot \frac{M_2^2}{M_1^2} v_2^2}{M_2 v_2^2}$

$= \frac{M_2}{M_1}$

3. Match with options

Thus, $\boxed{\frac{\text{KE of artillery}}{\text{KE of shell}} = \frac{M_2}{M_1}}$

This corresponds to Option B.