Throwing baseball at a car 
For some odd reason, you decide to throw baseballs at a car of mass $M$, which
  is free to move frictionlessly on the ground. You throw the balls at the back of
  the car at speed $u$, and at a mass rate of $σ$ kg/s (assume the rate is continuous,
  for simplicity). If the car starts at rest, find its speed and position as a function
  of time, assuming that the balls bounce elastically directly backwards off the
  back window.

I got the solution to this exercise. It comes from David Morin's book of mechanics, but I dn't understand. And it says so:


Relative to the car, the baseball hits the car in a speed $u-v$. The ball changes its direction so it's change in momentum is $2 (u-v)dm$. The rate of
    change in momentum of the car (that is, the force) is thus $\dfrac{dp}{dt}=2\sigma '(u-v)$ where $\sigma'=\sigma \dfrac{u-v}{u}$. Because although you throw the balls at speed $u$, the relative speed of the balls and the car is only $(u − v)$. So that we have the following equation to solve for $v$:
    $$M\frac{dv}{dt}=\frac{2(u-v)^2\sigma}{u}$$.


I don't understand why I need to asssign a new variable $\sigma '$? Why do I need to divide in $u$ in $\frac{u-v}{u}$?
 A: This puzzled me since I did the usual rocket equation calculations (the amount of mass hitting the car is $dm = \sigma dt$, etc.). However, the amount of mass hitting the car per unit time is not constant. As the car gains speed, less and less mass hits the car per unit time. Think about what happens when $v = u$: the stream of baseballs can't catch up to the car, so the car stops accelerating. This is similar to the Doppler shift and we can use similar math.
Imagine that we launch two baseballs at velocity $u$ with a delay of $\Delta t$ between them at the car moving at velocity $v$. After the impact of the first baseball, the second baseball requires extra time due to the movement of the car (call the time between impacts $\Delta t'$):
$$v\Delta t' = u\Delta t' - u\Delta t$$
Where $v\Delta t'$ is the distance the car travels between impacts, $u\Delta t'$ is the total distance the second ball travels after the first impact, and $u\Delta t$ is the initial separation between the baseballs at launch. So, the new delay between the baseballs is:
$$\Delta t' = \frac{u}{u-v}\Delta t$$
Inverting this gives a rate of impact:
$$f' = \frac{1}{\Delta t'} = \frac{u-v}{u}\frac{1}{\Delta t}$$
Multiplying by $dm$ and replacing $\Delta t$ with $dt$ gives the effective mass rate.
$$\sigma'=\frac{dm}{dt'} = \frac{u-v}{u}\frac{dm}{dt} = \frac{u-v}{u}\sigma$$
A: The car is essentially at rest when we subtract its velocity from that of the balls. Now the balls have to travel same distance(because car is at rest) but with differrnt speed therefore with different time. This will give you the relation of new time in terms of old one and then you can find dm/dt in this frame
