The system is the people and the car. Total momentum is conserved. Let the direction of the car's motion be negative. The velocity of a person relative to the ground $v_p$ is related to their velocity relative to the car $v$ and the car's velocity $V$ by $v_p = v+V$.

Case 1: Simultaneous jump. The initial momentum is 0. The final momentum is for the two people (mass $2m$) and the car (mass $M$). Let the car's final velocity be $V_1$.

$$0 = (2m)(v+V_1) + MV_1$$ $$V_1 = - \frac{2m}{M+2m}v$$

The final speed is $|V_1| = \frac{2m}{M+2m}v$.

Case 2: Sequential jump. Step A (1st person jumps): The system is (car + 2 people). Let the velocity of (car + 1 person) be $V_A$.

$$0 = m(v+V_A) + (M+m)V_A \implies V_A = -\frac{m}{M+2m}v$$

Step B (2nd person jumps): The initial system is (car + 1 person) with velocity $V_A$. Let the car's final velocity be $V_2$.

$$(M+m)V_A = m(v+V_2) + MV_2$$ $$(M+m)V_A = mv + (M+m)V_2 \implies V_2 = V_A - \frac{m}{M+m}v$$

Substituting $V_A$:

$$V_2 = -\frac{m}{M+2m}v - \frac{m}{M+m}v = -mv \left(\frac{1}{M+2m} + \frac{1}{M+m}\right)$$

The final speed is $|V_2| = mv \left(\frac{1}{M+2m} + \frac{1}{M+m}\right) = mv \frac{2M+3m}{(M+2m)(M+m)}$.

Comparison: We compare $|V_1| = \frac{2m(M+m)}{(M+2m)(M+m)}v$ with $|V_2| = \frac{m(2M+3m)}{(M+2m)(M+m)}v$. Comparing the numerators $2m(M+m) = 2mM+2m^2$ and $m(2M+3m)=2mM+3m^2$. Since $2mM+3m^2 > 2mM+2m^2$, we have $|V_2| > |V_1|$.