The correct answer is $\boxed{\text{B}}$.
The total momentum of the system is conserved, so the momentum of the first ball before the collision must equal the momentum of the second ball after the collision. The momentum of the first ball is $p_1 = mv_1 = 1 \text{ kg} \times 2 \text{ m/s} = 2 \text{ kg m/s}$. The momentum of the second ball after the collision is $p_2 = mv_2$. We know that $v_2$ is positive, so the second ball must be moving to the right after the collision.
To conserve momentum, we must have $p_1 = p_2$. Solving for $v_2$, we get $v_2 = \frac{p_1}{m_2} = \frac{2 \text{ kg m/s}}{2 \text{ kg}} = 1 \text{ m/s}$.
Option A is incorrect because the second ball cannot be moving to the left after the collision. Option C is incorrect because the second ball cannot be moving at a speed of 2 m/s after the collision. Option D is incorrect because the second ball cannot be moving at a speed of 0 m/s after the collision.