Throwing masses to gain velocity. A question of efficiency I was thinking the other day of a system in which you eject mass. And then I wonder whether it was more efficient to throw a big mass or many little. Let me put it in other terms.
Suppose you are moving at a constant speed $v_0$ in a car with a total mass $M$. Then you decide to throw a ball of mass $m$ with an exit velocity of $v_e$. And you throw the ball in opposite direction of the movement. (The problem is in one dimension) The ball will be moving at $v_\mathrm{ball}= v_0 - v_e$. By conservation of momentum, $$P_1 = Mv_0$$
But then after you throw the ball the momentum is  $$P_2= m(v_0-v_e) + (M-m)V$$ where $V$ is the new velocity. Lets write $V= v_0 + \Delta V$ because we will be interested in $\Delta V$. Then we can write 
\begin{aligned}
P_2&= mv_0 -mv_e +Mv_0 -mv_0 + (M-m)\Delta V\\ &= Mv_0 -mv_e + (M-m)\Delta V
\end{aligned}
If we assume no external forces $P_1=P_2$ we will have an equation:
$$\Delta V= \frac{m}{M-m}\ v_e$$ 
And here is were it gets weird. If we throw to masses $m$. But at the same time we will get a first $\Delta V_0$
$$\Delta V_0=\frac{2m}{M-2m}\ v_e$$
But if we throw the first mass and then the second mass we will get to $\Delta V$'s with:
\begin{aligned}
\Delta V_1&=\frac{m}{M-m}\ v_e\\
\Delta V_2&=\frac{m}{(M-m)-m}\ v_e
\end{aligned}
The total chance in velocity is the sum of 1 and 2
Since $M-2m<M-m$ then $\frac{1}{M-2m} > \frac{1}{M-m}$ then
$\frac{2}{M-2m}\ m\ v_e > (\frac{1}{M-m}+ \frac{1}{M-2m})\ m\ v_e $
This -I think- suggest that you gain more velocity if you throw a big mass in one shot that if you throw them in two pieces. My teacher told me that you will get more speed if you throw them apart. But it's seems that she is wrong. I have some doubts on this problem. What do you think? What is the more "efficient" way of gaining velocity assuming you can throw them at the same velocity $V_e$.
 A: All of your math is correct, but your result conflicts with our intuition as well as the reality of how rockets work.  Here is why:
You assume that "The ball will be moving at $v_\mathrm{ball}=v_0-v_e$". This approximation is valid only in the limit where $m$ is much smaller than $M$.  (i.e. the limit of a continuous stream of tiny balls, like a normal rocket)  In the case of larger chunks of mass, the process of accelerating the ball will accelerate the car significantly enough to alter the exhaust velocity meaningfully.
This is final velocity of the ball in reality:
$$v_\mathrm{ball}=(v_0+\Delta V)-v_e$$
Working with this assumption rather than yours should lead to the opposite result, vindicating your teacher.  It is more efficient to throw many little masses rather than one big one.
A: This is a quite subtle problem. You have to be careful about three different situations. A ball can be thrown with velocity (relative to the ground):
a) $v_0-v_e$.
b) $v(t)-v_e$, where $v(t)$ is the velocity of the car just after the ball is thrown.
c) $v(t)-v_e$, where $v(t)$ is the velocity of the car just before the ball is thrown.
You actually stated the problem satisfying a) but solved it using c). Your solution is correct as long as you rephrase how the balls are thrown. Your teacher probably was considering case b).
Case b): Let us call $v_i$ the velocity of the car after throwing the ith ball. Conservation of momentum just after throwing the first ball is
$$Mv_0=(M-m)v_1+m(v_1-v_e),$$
which gives
$$v_1=v_0+\frac{mv_e}{M}.$$
Throwing the second ball will give
$$(M-m)v_1=(M-2m)v_2+m(v_2-v_e)$$
$$v_2=v_1+\frac{mv_e}{M-m}.$$
The final velocity being
$$v_2=v_0+\frac{mv_e}{M}+\frac{mv_e}{M-m}.$$
If you throw both balls at same time you get
$$Mv_0=(M-2m)v_2'+2m(v_2'-v_e),$$
giving
$$v_2'=v_0+\frac{2mv_e}{M}.$$
Therefore $v_2>v_2'$.
