There is a 2-part problem in my physics book. In part I, we are given some information about a ball being dropped vertically onto a flat surface, and we are asked to deduce the coefficient of restitution $e$. In part II, we drop the same ball onto an inclined plane, and we are asked to find the location of the second point of contact of the ball with the plane. The inclined plane is presumably made of the same material as that of the flat surface, although this is not mentioned in the problem.

It's a very easy exercise, but there was something in the book's solution that confused me: the book claims (without justification) that the collisions of the ball with the plane obey the law of reflection.

To see why this doesn't make sense (to me), let's take a simple example.

Ball moving in 2 dimensions, no gravity

Consider a ball moving frictionlessly in the horizontal plane which impacts a wall at an angle of incidence $\alpha$ and speed $v_0$. Let's take our reference frame to have the wall along the $y$-axis and the ball approaches the wall from the positive $x$ direction.

During some small time $\Delta t$, the ball exerts an impulse on the ball and the impulse vector will be parallel to the $x$-axis, pointing in the positive $x$ direction. Hence during the time $\delta t$, there are no forces acting on the ball in the $y$-direction, and thus momentum is conserved in this direction.

Let us suppose the ball bounces off the wall with a new speed $v_1$, and let $e=v_1/v_0$ be the coefficient of restitution of the collision. Note that $e=1$ if the collision is perfectly elastic.

Now since momentum in the $y$-direction is conserved, the angle of reflection $\beta$ must satisfy $mv_0\sin(\alpha)=mv_1\sin(\beta)$, or $\sin(\beta)=\sin(\alpha)/e$. If we impose that the collision is perfectly elastic, we have $\alpha=\beta$, which is the famous law of reflection. But in general, the collision does not obey the law of reflection.

Ball dropped onto inclined plane

Now let's come back to part 2 of the problem, and drop our ball from some height onto an inclined plane living in the $xz$ plane with an angle of inclination $\alpha$ (gravity acts in the negative $z$ direction). The angle of incidence is thus also $\alpha$, and now we would like to calculate the angle of reflection $\beta$, knowing that the coefficient of restitution of the collision is $e$.

The problem now is that during the small time $\Delta t$ of the collision, gravity transmits an impulse of $mg\sin(\alpha)\Delta t$ in the direction parallel to the surface of the incline. Hence $mg\sin(\alpha)\Delta t=v_1\cos(\beta)-v_0\cos(\alpha)=v_0(e\cos(\beta)-\cos(\alpha))$. Unfortunately, we don't know $\Delta t$, so as far as I can see there isn't enough information to determine $\beta$.

So is the book wrong? And how would we calculate the angle of reflection of an inclined plane?

P.S Sorry for the verbose question

  • 1
    $\begingroup$ -1. Unclear. This is confusing and difficult to follow without a diagram. Also how does $\Delta t$ disappear in the 1st situation but not the 2nd? $\endgroup$ Commented Mar 13, 2017 at 20:52
  • $\begingroup$ I think it's very clear I explained everything in detail you just have to read what's written. Its not the case that Delta t disappears in the first situation, it's that it's irrelevant to the calculation of the angle of incidence because all we need is that momentum is conserved. $\endgroup$
    – math_lover
    Commented Mar 13, 2017 at 22:53

2 Answers 2


The law of reflection only holds if the collision is elastic and the object which the ball collides with does not move. If the collision is not elastic then the perpendicular component of velocity is smaller after than before, whereas the parallel component is the same. So when $e \lt 1$ the angle of reflection is greater than the angle of incidence.

The object being collided with must be very much more massive than the ball, so that it does not move. Momentum is conserved in the collision, provided the momentum of both objects is taken into account. If, for example, the ball is dropped onto and bounces from a wedge which is free to slide horizontally, the recoil of the wedge will increase the horizontal component of the velocity of the ball.

Friction and the spin of the ball will also affect the angle of reflection. During a finite collision time friction will reduce the parallel component of velocity. The spin of the ball affects whether there is relative motion between the ball and the plane during collision.

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If the plane is fixed then the perpendicular and parallel components after the collision are $eV_0\cos i, V_0\sin i$ where $i=\theta$, so the angle of reflection is given by
$\tan r=\frac{V_0\sin i}{eV_0\cos i}=\frac{1}{e}\tan i$.

If the plane is a wedge which can slide horizontally then we can conserve momentum horizontally : $mV_1\sin(i+r)=MU$.
The relative velocity of approach along the perpendicular is again $V_0\cos i$. The relative velocity of separation is $V_1\cos r+U\cos i$. Apply the Law of Restitution :
$V_1\cos r+U\cos i=eV_0\cos i$.
The relative velocity of approach along the parallel direction is initially $V_0\sin i$. If there is no friction between the ball and the wedge then this component is conserved in the collision, so
$V_0\sin i=V_1\sin r+U\sin i$.

The last 3 equations can be solved to find $V_1, r, U$.

  • $\begingroup$ What about gravity? Doesn't gravity affect the parallel component of the velocity as I described in the question details? $\endgroup$
    – math_lover
    Commented Mar 13, 2017 at 23:15
  • 2
    $\begingroup$ The reaction forces during the collision are usually much greater than the weight of the ball, so the latter can usually be ignored. $\endgroup$ Commented Mar 13, 2017 at 23:55
  • $\begingroup$ For inelastic collision snell's law applies. Right? $\endgroup$ Commented Oct 8, 2023 at 19:13
  • $\begingroup$ @JohnAlexiou No. Snell's Law applies to the refraction of light and other waves. This question relates to the reflection of bouncing objects. $\endgroup$ Commented Jan 14 at 16:36

If the surface is frictionless

The ball will slide down the slope during $\Delta t$.
Not only is the acceleration due to G still applied parallel to the slope, but the component of the ball's impact velocity parallel to the slope remains un-altered by the reflection.

With Friction

"If it doesn't slide it spins."
The sliding effect can only occur in practice if conditions are sufficient to override friction. This is dependent on the surface materials, the angle of incidence on the slope as well as the speed of the collision.
If friction is not overcome rotation will be induced.

The impulse from the collision actually deforms both the surface and the ball, but if these are hard/low-friction objects it can be a negligible effect and is thus (rather like air resistance) it is OK to neglect in for the purposes of many collision type questions.

In particular the effect of friction is observable with soft or very elastic balls as they maintain contact for a greater period of time. For example if a "bouncy ball" is dropped vertically onto a slope it will have some spin after the collision. during such collisions spin energy and kinetic energy can be exchanged.
Also just as kinetic energy in one direction can be reversed, spin can be reversed if conditions are right, such as friction being high enough.

  • $\begingroup$ This doesn't answer my question and is only tangentially related to the details I gave in the question body. My question is the one in the title. $\endgroup$
    – math_lover
    Commented Mar 13, 2017 at 14:10
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    $\begingroup$ Are you sure? $\Delta t$ is the contact time. No, the law of reflection doesn't hold, it's an approximation; often sufficiently accurate. $\endgroup$
    – JMLCarter
    Commented Mar 13, 2017 at 14:14
  • $\begingroup$ But you are talking about friction. Even assuming zero friction (as I said in the question details, no forces in the horizontal direction), the law of reflection doesn't necessarily hold (according to me). It holds iff the collision is perfectly elastic. $\endgroup$
    – math_lover
    Commented Mar 13, 2017 at 15:19
  • $\begingroup$ Oh that is something else, I added more to my answer (it won't fit in here) $\endgroup$
    – JMLCarter
    Commented Mar 13, 2017 at 15:50

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