A canon, mass 500 kg, fires a shell, mass 1 kg, horizontally to the right at 500 m·s⁻¹. What is the magnitude and direction of the initial recoil velocity of the canon?

Let $m_c$ be the mass of the cannon and $m_s$ be the mass of the shell.
Let $v_{ci}$ and $v_{si}$ be their initial velocities, and $v_{cf}$ and $v_{sf}$ be their final velocities.
Given:

• Mass of cannon, $m_c = 500 \text{ kg}$
• Mass of shell, $m_s = 1 \text{ kg}$
• Initial velocity of cannon, $v_{ci} = 0 \text{ m/s}$ (stationary)
• Initial velocity of shell, $v_{si} = 0 \text{ m/s}$ (stationary)
• Final velocity of shell, $v_{sf} = 500 \text{ m/s}$ (horizontally to the right)

According to the principle of conservation of momentum, the total momentum before firing is equal to the total momentum after firing:

$$m_c v_{ci} + m_s v_{si} = m_c v_{cf} + m_s v_{sf}$$

Since both the cannon and the shell are initially stationary, their initial momentum is zero.

$$(500 \text{ kg})(0 \text{ m/s}) + (1 \text{ kg})(0 \text{ m/s}) = m_c v_{cf} + m_s v_{sf}$$ $$0 = m_c v_{cf} + m_s v_{sf}$$

Let's define the direction to the right as positive.

$$0 = (500 \text{ kg}) v_{cf} + (1 \text{ kg})(500 \text{ m/s})$$ $$0 = 500 v_{cf} + 500$$ $$500 v_{cf} = -500$$ $$v_{cf} = \frac{-500}{500}$$ $$v_{cf} = -1 \text{ m/s}$$

The magnitude of the recoil velocity is the absolute value of $v_{cf}$. The negative sign indicates that the direction is opposite to the direction of the shell's motion. Since the shell moves to the right, the cannon recoils to the left.

The magnitude of the initial recoil velocity of the cannon is $\boxed{1 \text{ m/s}}$.
The direction of the initial recoil velocity of the cannon is $\boxed{\text{to the left}}$.