When a ball is thrown to a wall without spin and bounces, supposing it's an elastic collision, why does the angle of incidence equal the angle "reflection" (relatively to the normal), why isn't it any other angle? Does it have to do with the conservation of linear momentum?
[Physics] When a ball bounces on a wall at an angle, why does the angle of incidence equal angle of reflection
collisionconservation-lawsmomentumnewtonian-mechanics
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Best Answer
Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular.
In this simple situation, the force on the ball can only act in the Y direction. Which has the following implications
$$ F_{x} = 0 \rightarrow a_{x} = 0 \rightarrow v_{ix} = v_{fx} $$
The second assumption is that the collision is elastic, which means that energy is conserved. In this case we only need to worry about kinetic energy:
$$ T_{i} = T_{f} \rightarrow \frac{1}{2}m(v_{ix}^2 + v_{iy}^2) = \frac{1}{2}m(v_{fx}^2 + v_{fy}^2) $$
From the first condition, we know the x velocity is unchanged, so consequently the squares of the y velocity also must stay consistent. We can see from the bounce that the sign reverses.
$$ v_{fx} = - v_{ix} $$
Since the magnitudes of the velocity are unchanged then, we know the angle of reflection has to match the angle of incidence.