Quick physics question about angle of reflection

[ronburgundy]

If you hit a pool ball against the side, it will ricochet at the same angle it hit at, assuming a perfectly hard surface. But what if the side is "soft" and has some give? Does that increase or decrease the angle of reflection.

My intuition is it decreases the angle, but its surprisingly hard to google a clear answer.

 

[Malintent]

I thought this was going to be an optics question...

a pool table DOES have a soft edge that the hard ball bounces off of. The dynamics of the deformation of the soft surface would dictate the change in trajectory from the theoretical "hard body collision" laws.

My intuition says that the angle of deflection DECREASES, not increases. Assuming the deformation is consistent across the entire impacted area, the rebound would be constrained by the concave shape of the deformed side... on the other hand, the speed at wich the deflection occurs versus the speed at which the material that was deformed returns to its 'flat' state would make a difference... like in the case of the rubber rails on a pool table, the rubber rebounds faster than the speed of the ball, so the concave sides of the impact area "gets out of the way" of the ball faster than the ball can change direction...

I guess that's why they call it "hard body collisions"... soft body ones have too many variables to predict the error in deflection angle...

<shrug>

 

[Jimmy Higgins]

What does "decrease the angle mean"? Decrease from what origin?

You take a cue ball and roll it to the edge, if the edge isn't there, it continues along the same trajectory. It wants to go that way. But you fucked up its day, and put that edge there. So what is going on? Poor cue ball.

In the end, it is all vectors. The ball is moving in the x-axis so much and y-axis so much. A perfect deflection just reverses the direction of the y-axis movement, hence, the angle appears to be about the same because the absolute value of the ratio x to y remains the same.

If the edge is "soft", the question is what is the impact on the y-axis and x-axis motion. And then we can ask, "what the fuck does 'soft' mean"?! Soft as in fluffy, bouncy, collapsing?

My presumption is that it depends on the angle of the ball. Assuming the angle of impact from straight on is 90 degrees, an impact from 45 to 89 degrees would result in deflected angle that is less than initial impact. An impact from 1 to 45 would lead to a deflected angle that is greater.

The reasoning being that the greater axis energy will be absorbed more in the impact. So closer to 90 would be y, closer to 0 would be x.

 

[Speakpigeon]

If you hit a pool ball against the side, it will ricochet at the same angle it hit at, assuming a perfectly hard surface. But what if the side is "soft" and has some give? Does that increase or decrease the angle of reflection.

My intuition is it decreases the angle, but its surprisingly hard to google a clear answer.

Break down the problem in two. A ball coming in against the side can be seen as a mix between a virtual one coming at a right angle and another virtual one coming parallel to, or sliding against, the side, the two with different speeds to reflect the angle of incidence of the actual ball.

Then it's easier to think about what happens.

In case of a hard side, further assuming no friction, the first virtual ball will just bounce back at the same speed. The second virtual ball will keep going along the side without loosing speed (because zero friction), so the angle of reflection of the actual ball will be identical to the angle of incidence.

Now in the case of a soft side, the first virtual ball coming at a right angle will loose some or all its speed while the one sliding against the side will keep its original speed so the angle of reflection of the actual ball will indeed be less than the angle of incidence.

Except in case there is some friction. Depending on the relative values of softness and friction, the angle of reflection can get to be greater than the angle of incidence.
EB

 

[Kharakov]

Take into account the spin of the cue ball, coefficient of friction between the felt and the cue ball, and the linear and angular momentum of the felt. No. I mean the linear and angular momentum of the cue ball. Really.

 

[Bronzeage]

After many years of empirical observations of this phenomena, on billiard tables of varying states of condition, I can give an answer to this question.

The rails of a billiard table are made a hard rubber, which is intended to deform as it absorbs the kinetic energy of ball, and return the ball at about the same velocity as it struck the rail. There is a lost of kinetic energy while the ball is in contact with the rail, which some players can use to their advantage, by forcing the cue ball to spin in one direction or another. The ball can be forced to spin along any axis. A horizontal forward spin will lose less kinetic energy than a backward spin. At the same time, the cueball maybe spinning backwards as it slides across the felt, although this is very difficult to observe. All of these factors figure in how much energy is lost at contact.

The angle that the cueball or object ball leaves depends upon the angle at which it struck, with the adjustments allowed for the spinning of the ball. The velocity of the ball does not seem to matter in actual practice. It's well observed that on a table with "dead" rails, the exit velocity is noticeably decreased, but not the exit angle.

 

[barbos]

If you hit a pool ball against the side, it will ricochet at the same angle it hit at, assuming a perfectly hard surface. But what if the side is "soft" and has some give? Does that increase or decrease the angle of reflection.

My intuition is it decreases the angle, but its surprisingly hard to google a clear answer.

It's recoil angle. And depending on many factors, friction, moment of inertia, initial angular momentum. elasticity. So no single answer.

 

[Tigers!]

If you hit a pool ball against the side, it will ricochet at the same angle it hit at, assuming a perfectly hard surface. But what if the side is "soft" and has some give? Does that increase or decrease the angle of reflection.

My intuition is it decreases the angle, but its surprisingly hard to google a clear answer.

How much money did you lose on that game?

 

[Speakpigeon]

I think the money would decrease in line with the angle.

But, hey, I'm no specialist here. Just a hunch.
EB

 

[ryan]

Assume a frictionless path that the ball travels on, but not a frictionless rail.

Let's first think about what happens on the x axis, perpendicular to the rail. We see that the ball comes back with almost the same momentum, obviously, so let's assume an ideal spring in the x direction as well as a conservative force pushing it in the positive direction (towards the cue). This is ultimately an inelastic collision giving up some momentum to the Earth, but very little kinetic energy in terms of linear velocity. The main laws we can use here are the conservation of momentum m1vi = m1v1f +m2v2f and conservation of energy 1/2mv^2 = 1/2mv^2 + 1/2mv^2. So we can see that the magnitude of the velocity of the ball coming back will be very close to what it was in the negative direction.

Now let's see if the change in magnitude of velocity in the y direction is as negligible. We don't have to worry about an impulse with the rail (assuming negligible impulse from the rubber to the negative y direction). But we do have an obvious non-conservative force from friction in the negative y direction. We know this because of the spin that the ball gains after contacting the rail. The frictional force must have a negative y component to send the cue ball into a spin.

We can look at this frictional force in terms of applied angular acceleration. This translates to a considerable loss of linear kinetic energy from the ball's new rotational kinetic energy. And ultimately amounts to a loss of velocity in the y direction.

So |Vx| (instantaneous velocity) going toward the rail roughly equals |Vx| coming back. But the Vy loses some velocity due to the temporary angular acceleration in the negative y direction.

A slower Vyf velocity than Vyi with |Vx| staying constant means that the x final distance to y final distance ratio will be larger than the initial |-x/y| distance ratio. So the angle to the rail should be larger after impact.