Why is momentum not conserved when one body is fixed? Why is momentum not conserved when one body is fixed? Why can't I apply C.O.M with a ball hits a wall, I mean I don't get the correct answer doing so?
 A: The momentum is always conserved, but sometimes we just ignore that. There is nothing really "fixed" in the universe.
If you throw a ball against a wall that is rigidly fixed to earth (which we just assume for the sake of that argument) there will be a momentum transfer from that ball to earth. Now since the inertia of earth is some magnitudes higher than that of a normal ball, iwe tend to ignore that.
Also, the acceleration of the ball already took place in the reference frame of earth as will the deceleration afterwards. So it all cancels out anyway as long as everything stays inside the same system (in this case "earth").
A: It is conserved. Imagine a ball bouncing back from the ground on Earth.
Then per momentum conservation law :
$$ m_\epsilon~\vec u_\epsilon + m_b~\vec u_b = m_\epsilon~\vec v_\epsilon + m_b~\vec v_b$$,
where $m_\epsilon$ - Earth mass, $m_b$ ball mass; $\vec u,\vec v$ - initial and final velocities of Earth and ball according to subscript.
Then assuming Earth initial velocity zero, we can solve equation for Earth velocity change after each collision with a ball :
$$ \vec v_\epsilon = \frac {m_b} {m_\epsilon} \left( \vec u_b - \vec v_b \right)$$
Applying vector subtraction rule to the RHS equation part, we get :
$$ \vec v_\epsilon = 2\frac {m_b} {m_\epsilon} ~\vec u_b$$
Let's assume that ball has about $m_b = 0.5~kg$,- about the basketball ball mass and on collision with Earth has velocity about $1~m/s$. Earth mass is $m_\epsilon \approx 10^{24}~kg $, substituting this data we find out that on each ball collision with Earth ground, Earth changes speed in reverse direction of ricocheting ball by,
$$ || \vec v_{\epsilon} || \approx 10^{−24}~ m/s$$
Needless to say that nobody will not ever notice such a small Earth speed change. Same phenomena applies in the case of some moving object hits a firmly fixed target,- it is trying to pass it's momentum to the Earth reference frame. However because Earth speed change is such jokingly miniature,- we usually assume, that fixed target doesn't move at all, but actually it isn't. That is why you get this paradox thinking about momentum conservation. It always is conserved, the question is, does your instruments sensitive enough to spot such difference ?
A: There is an external force applied ( the reaction force from the wall ), the conservation of momentum cannot simply hold, instead  RΔt = Δp , where p is the momentum, R is the reaction force from the wall and Δt is the impact time.
