Pool and Physics

F

Fish

AzB Silver Member
Silver Member
When I was in school, on our physics class we were discussing collision and impact, the teacher ask if the white ball was travelling at 5 mph and and hits the 1 ball at 45 degress to the left, what would be the resulting speed and angle of trajectory of the white ball after impact.

My answer " it depends on where the tip hits the white ball"

I got an F :eek:
 
lol if he didnt quanitify it with "on a frictionless table with no spin" then you definatly have the right answer :)

I'd ask him to play pool with you and show him whats what.
 
I thought it came off at like 89 degree's due to the collision not being perfect and "dragging" the object ball a bit as well as throwing it?
 
The answer I think he was looking for was a speed of Sin 45 degrees x 5mph = sqrt2/2 x 5 = approx 3.5mph

The cue ball angle would be 45 degrees from the original path.

Assuming frictionless surface on tables and balls.
 
Colin Colenso said:
Assuming frictionless surface on tables and balls.
Click to expand...

This is a standard assumption for impact-momentum studies in physics as taught in high school / secondary school, as you well know. Adding in the real-life effects of friction, air resistance, coefficient-of-restitution, a third dimemsion etc., is for much later ... like post-graduate studies later.

I'm sure that teachers always dread having someone in their class with knowledge of a topic used as an example for some leaning endeavor, although it gives them an opportunity to explain about tangents ;)

Dave
 
Colin Colenso said:
The answer I think he was looking for was a speed of Sin 45 degrees x 5mph = sqrt2/2 x 5 = approx 3.5mph

The cue ball angle would be 45 degrees from the original path.

Assuming frictionless surface on tables and balls.
Click to expand...
Correct Calc man...you sure live up to your name.

Follow up pop quiz...

The CB shoots off to the right 45 degrees at 3.5mph, and the OB shoots off to the left 45 degrees also at 3.5mph. How can these two velocities sum to equal 7.0mph, if the original CB velocity was only 5.0mph? Where did the extra 2.0mph come from? Whatever happened to conservation of momentum? :D
 
jsp said:
Follow up pop quiz...

The CB shoots off to the right 45 degrees at 3.5mph, and the OB shoots off to the left 45 degrees also at 3.5mph. How can these two velocities sum to equal 7.0mph, if the original CB velocity was only 5.0mph? Where did the extra 2.0mph come from? Whatever happened to conservation of momentum? :D
Click to expand...

I'll bite...

The "momentum" that is conserved is the momentum of the system, i.e. the velocity of the center of mass of the two balls multiplied by the mass of the two balls. After contact, the two balls are each moving away from each other and down-table (I'm picturing this as a spot-shot), and their center of mass is moving directly downtable, at a velocity equal to the downtable component of the velocity of each of the balls. This velocity happens to be exactly 2.5 mph. You've still got exactly 5 mph-balls (funny units in this problem) of momentum.

The other thing that's conserved, since this problem is assuming an elestic collision, is kinetic energy, which involves the square of the velocity, not velocity itself. Turns out (non-coincidentally) that the number colin approximated as 3.5 is exactly the square root of 12.5. The sum of the squares of the velocities of the two balls is equal to the square of the CB's initial velocity. If that sounds like the pythagorean theorem, that's because it is. The two ball paths are at right angles to each other. Treating them as vectors and resolving them yields their hypotenuse, which is exactly the vector representing the CB's initial velocity.

Physics are phun.

-Andrew
 
jsp said:
Correct Calc man...you sure live up to your name.

Follow up pop quiz...

The CB shoots off to the right 45 degrees at 3.5mph, and the OB shoots off to the left 45 degrees also at 3.5mph. How can these two velocities sum to equal 7.0mph, if the original CB velocity was only 5.0mph? Where did the extra 2.0mph come from? Whatever happened to conservation of momentum? :D
Click to expand...

The velocity of the balls are vectors and not a scalors. Vectors don't add like scalors, thus it appears that the total speed (which is a scalor measurement) has increased.
 
Andrew Manning said:
I'll bite...

The "momentum" that is conserved is the momentum of the system, i.e. the velocity of the center of mass of the two balls multiplied by the mass of the two balls. After contact, the two balls are each moving away from each other and down-table (I'm picturing this as a spot-shot), and their center of mass is moving directly downtable, at a velocity equal to the downtable component of the velocity of each of the balls. This velocity happens to be exactly 2.5 mph. You've still got exactly 5 mph-balls (funny units in this problem) of momentum.

The other thing that's conserved, since this problem is assuming an elestic collision, is kinetic energy, which involves the square of the velocity, not velocity itself. Turns out (non-coincidentally) that the number colin approximated as 3.5 is exactly the square root of 12.5. The sum of the squares of the velocities of the two balls is equal to the square of the CB's initial velocity. If that sounds like the pythagorean theorem, that's because it is. The two ball paths are at right angles to each other. Treating them as vectors and resolving them yields their hypotenuse, which is exactly the vector representing the CB's initial velocity.

Physics are phun.

-Andrew
Click to expand...
Couldn't have said it better myself!
You sir are an honorary Calculator:D

Funily enough, my old road partner was known as Pythagoras. Mainly because he'd study every possible line and angle of a shot whenever he got into a strategical fix. I've never seen anyone who could win more games from seemingly hopeless situations.
 
Colin Colenso said:
Couldn't have said it better myself!
You sir are an honorary Calculator:D
Click to expand...

This is the greatest honor of my young life.

-Andrew

(It's not, really. I was just using extreme exaggeration for humor value. It's actually even funnier now that I've explained it paranthetically. I follow up all my jokes with paranthetical explanations. (I don't, really. That was just another joke.))
 
Andrew Manning said:
This is the greatest honor of my young life.

-Andrew

(It's not, really. I was just using extreme exaggeration for humor value. It's actually even funnier now that I've explained it paranthetically. I follow up all my jokes with paranthetical explanations. (I don't, really. That was just another joke.))
Click to expand...
uh ha...a comedy expert!
So why is it that Clowns talk funny? :rolleyes: :p
 
Very well said Andrew and Zeeder. :)

To put it yet another way...

Momentum (mass*velocity) is a vector quanity, which has both magnitude and direction. Momentum is still conserved, if you look at the vertical and horizontal components of the CB and OB momentum vectors after impact (horizontal components cancel out, vertical components add to 5mph-balls).

Kinetic energy, however, is a scalar quanity, and they should add. But as Andrew pointed out, kinetic energy is equal to the square of the velocity. So the velocity of the CB squared plus the velocity of the OB squared should equal the initial velocity of the CB squared...which it does.

Okay, class is dismissed. Recess! Yay. :p
 
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