Calculate the final velocity of the cart and students

[TheShapeOfTime]

Three Physics 111 AP students, each having a mass of 60kg, climb onto a large flatbed cart that has a mass of 120 kg. Standing at one end and taking turns they run to the opposite end and jump off, one immediately following the other, each with a velocity of 10 m/s with respect to the cart. Calculate the final velocity of the cart and students with respect to the earth.

I know I need to use $m_1v_1 + m_2v_2 = m_1v_1 + m_2v_2$, and I have done a bunch of simpler problems with no trouble. I'm pretty sure this has to be done in steps but I'm not sure how to incorporate the answer of each step into the next. I have something like this for the first step:

$$60 \cdot 0 + 120 \cdot 0 = 60 \cdot 10 + 120(v + 10)$$

 

[Skomatth]

Read integral's post.

 

[Integral]

I believe that this problem is a bit more subtle, it is essentially a rocket problem. The mass of the "cart" does not remain constant. As each student jumps it becomes less massive, thus each successive student will cause a larger change in velocity.
after the first student jumps the carts velocity will change by:
$$v_c = \frac {m_1 V_1} {m_2 + m_3 + m_c}$$

Repeat for each student and sum the changes to get the total change.