Two persons, each of mass m , are standing at the two extremes of a...

Let the right direction be taken as positive.

Initially, the system (car and two persons) is at rest, so the initial momentum is zero.

When the first person jumps to the left with speed $u$ relative to the initial state of the car (which is at rest), his velocity with respect to the ground is $-u$.

Let $V_1$ be the velocity of the car and the second person after the first jump. Applying conservation of linear momentum:

$0 = m(-u) + (M + m)V_1$

$\Rightarrow V_1 = \dfrac{mu}{M + m}$

The second person then jumps to the right with speed $u$ relative to the state of the car before his jump. Since the car was moving with velocity $V_1$, the velocity of the second person with respect to the ground is $V_1 + u$.

Let $V_2$ be the final velocity of the car. Applying conservation of linear momentum for the system of the car and the second person:

$(M + m)V_1 = MV_2 + m(V_1 + u)$

$\Rightarrow MV_1 + mV_1 = MV_2 + mV_1 + mu$

$\Rightarrow MV_1 = MV_2 + mu$

$\Rightarrow V_2 = V_1 - \dfrac{mu}{M}$

Substituting the value of $V_1$:

$V_2 = \dfrac{mu}{M + m} - \dfrac{mu}{M}$

V_2 = mu \left(\dfrac{M - (M + m)}{M(M + m)} \right)

$V_2 =$-\dfrac{m^2 u}{M(M + m)}

The negative sign indicates that the final velocity of the car is towards the left.