Saturday, November 7, 2015

newtonian mechanics - Does the law of reflection hold for balls bouncing inelastically on inclined planes?


There is a 2-part problem in my physics book in which part I gives some information about a ball being dropped vertically onto a flat surface, and we are asked to deduce the coefficient of restution $e$. 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 which 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), lets take a simple example.


Ball moving in 2 dimmensions, no gravity


Consider a ball moving frictionlessly in the horizontal plane which impacts a wall at an angle of incidence $\alpha$ and speed $v_0$. Lets 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.


Lets 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 $e=1$ iff 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 incline plane living in the $xz$ plane with 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 off an inclined plane?



P.S sorry for the verbose question




No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...