Confusion about the conservation of momentum of a ball and an angled wall

Question

Say we have a ball that is traveling to the right at some velocity $v = v_x$. Say there is a completely immobile or infinitely heavy wall that is angled such that the normal vector of the wall is pointed 45 degrees below horizontal (ie. the wall is angled 45 degrees downward). Assuming this is a completely elastic collision, intuition says that the ball would travel straight downward after hitting the wall.

However, a simple computation of the conservation of momentum implies that since the velocity of the wall is 0 in all directions before and after the collision (granted after the collision, there should be a minuscule wall velocity that is negligible), then the $x$-velocity $v_x$ of the ball should be the same before and after. But this can't be the case based on intuition. What am I missing here?

Answer 1

Say there is a completely immobile or infinitely heavy wall

This is where you're getting tripped up. Try assigning a finite weight to the wall and then calculate it from there. You'll find that the wall changes it's momentum to cancel out the change in momentum of the ball.

As you make the wall heavier, that same amount of momentum of the wall becomes imperceptible. E.g. if we take a 100g ball at 30 m/s, and bounce it off a wall attached to the earth, then the wall/earth system's velocity change is a staggeringly tiny $5\cdot 10^{-25}$ m/s!

Answer 2

If the ball moves to the right at a constant speed, this simply means that there is either no gravity or the ball is moving on a frictionless surface. We assume it's the former.

Based on the correct intuition you raise, the x component of the ball's momentum disappears after the collision, and, moreover, it seems that some non-zero momentum is found in the -y direction. In fact, however, this apparent lack of momentum conservation is compensated for by the forces exchanged between the wall and the ball: $$\vec{F}_{\text{wall/ball}}=-\vec{F}_{\text{ball/wall}}=\frac{\Delta \vec{P}}{\Delta t}= -\,\big(m v_x \,\hat{\imath}+ m v_y \,\hat{\jmath}\big).$$

In practice, this force causes some microscopic vibrations in the wall. From a macroscopic perspective, these oscillations are understood as heat generated at and near the collision point. The amount of heat is equal to the kinetic energy lost by the ball

Nonetheless, elastic collisions are the ideal case in which such oscillations do not occur.

Answer 3

An approximation which is often made, eg molecules in a box when deriving the gas law, is that when an object rebounds elastically off a much more massive / immovable object the speed of rebound is equal to the speed of approach.
This is an approximation as shown below.

Momentum conservation: $m_{\rm b}u+0=m_{\rm b}v_{\rm b}+m_{\rm w} v_{\rm w}$

Kinetic energy conservation: u = v_{\rm w} - v_{\rm b}$
(relative velocity of approach = relative velocity of separation)

$\Rightarrow v_{\rm b}= -\dfrac {m_{\rm w}-m_{\rm b}}{m_{\rm w}+m_{\rm b}} u$

Note that if $m_{\rm w} \gg m_{\rm b}$ then $v_{\rm b} =-\dfrac {m_{\rm w}-m_{\rm b}}{m_{\rm w}+m_{\rm b}} u \to u$

Answer 4

You have to use the properties of elastic collision in this case as momentum conservation would surely fail, the vel component parallel to the wall would be unchanged but the component perpendicular to the wall would be affected. Use $$e =1= v_{sep}/v_{approach}$$ along the direction perpendicular to the wall and you would probably get the answer.