Let the initial velocities be $v_{A,i} > 0$ and $v_{B,i} < 0$. The initial kinetic energies are $K_{A,i} = \frac{1}{2}m_A v_{A,i}^2$ and $K_{B,i} = \frac{1}{2}m_B v_{B,i}^2$. Given $K_{A,i} = K_{B,i} = K$. The initial momenta are $p_{A,i} = m_A v_{A,i} = \sqrt{2m_A K}$ and $p_{B,i} = m_B v_{B,i} = -\sqrt{2m_B K}$. Since $m_A > m_B$, we have $|p_{A,i}| > |p_{B,i}|$. The total initial momentum of the system is $P_i = p_{A,i} + p_{B,i} > 0$. By momentum conservation, the total final momentum $P_f$ must also be positive.

(4) If both velocities reverse and their final kinetic energies are equal ($K_{A,f}=K_{B,f}$), their final momenta would satisfy $|p_{A,f}| > |p_{B,f}|$. The total final momentum would be $P_f = -|p_{A,f}| + |p_{B,f}| < 0$, which contradicts momentum conservation. So, (4) is impossible.

(2) If sphere B stops ($v_{B,f}=0$), then $P_f = m_A v_{A,f}$. Since $P_f > 0$, then $v_{A,f} > 0$. For a one-dimensional collision of impenetrable spheres, their relative order cannot change, which implies $v_{A,f} \le v_{B,f}$. The result $v_{A,f} > v_{B,f}$ violates this condition. So, (2) is impossible.

Both (1) and (3) are possible. For example, a perfectly inelastic collision results in the spheres sticking together and moving with a common velocity $V_f = P_i/(m_A+m_B) > 0$. This is an instance of (3). A perfectly elastic collision can result in sphere A stopping if the mass ratio is just right, which is an instance of (1). Since (3) represents the most general outcome of a collision, it is the most robust answer.