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?


2 Answers 2


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.

  • $\begingroup$ Thanks, could you explain me one last thing, why can't the force act in the x direction? $\endgroup$
    – Suriya
    Jan 30, 2016 at 17:44
  • $\begingroup$ There's nothing to physically push in that direction. The collision is caused by contact force and contact forces are always perpendicular to the surface at the point they apply the pressure. $\endgroup$ Jan 30, 2016 at 18:40
  • $\begingroup$ I think you meant $v_{fy} = -v_{iy}$ in your last equation (the indices for $y$, not $x$). $\endgroup$
    – StefanH
    Sep 11, 2021 at 18:29

This assumes a smooth surface collision.
The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction.
Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the wall after collision is equal in magnitude but opposite in direction to the incident velocity.


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