Solve a Billiard Ball Physics Problem with Spin and Momentum | 210 g Mass

In summary, a billiard ball has mass 210 g and it is hit with spin and bounces off the side of a table. The initial speed of ball is 1.6 ms-1 and its final speed is 0.8 ms-1. The change in momentum due to collision is 0.8 ms-1. The angle at which the collision occurs has no relevance.
  • #1
Metropolis
7
0
Hi guys, I am new here and I am having a LOT of trouble with this new momentum stuff, which I failed in :cry:

Here's a question:

A billiard ball has mass 210 g. It's hit with spin and bounces off the side of a table. The initial speed of ball is 1.6 ms-1 (at angle of 60 degrees) and its final speed is 0.8ms-1 (at angle of 30 degrees)

1) Find the change in the momentum of ball due to collision

I had trouble even starting the question. I drew the normal between the collision, and did 90-30=60 to find the inside angle and 90-60=30 for the other. Is anyone able to give me a hint on what formula to use? I'm not familiar with this because I usually do the problems with same initial and final speed.
 
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  • #2
By definition, the change in the linear momentum is
[tex]m(\vec{v}_f - \vec{v}_i)[/tex]

Hint: Try expressing the velocity vectors in terms of components.
 
  • #3
Thanks for the tip, I tried using it, and I made up a vector component thing, but I think I am totally wrong

http://img95.imageshack.us/img95/1408/vector4sq.jpg [Broken]

I added 60 and 30, and the angle ended up being 90, so right angle

and I used Pythagoras' theorem, but I think that is not right...like am I supposed to do Vector sum? and find the resultant vector?...it's all so puzzling to me!:tongue2:
 
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  • #4
Close! But instead of adding the vectors you need to subtract them. (Hint: Adding the negative of a vector is equivalent to subtracting it.)
 
  • #5
Does the inside angle have any relevance?

So is the change in velocity just 1.6 + (-0.8) = 0.8 ?
 
  • #6
Metropolis said:
Does the inside angle have any relevance?
Of course the angle matters.

So is the change in velocity just 1.6 + (-0.8) = 0.8 ?
No! Just redraw your diagram to show the final velocity minus the initial velocity. (These are vectors and must be treated as such.) You are almost there.

Hint: To find the negative of a vector, just flip its direction 180 degrees.
 
  • #7
I drew this diagram, is it right or not? Does the resultant vector represent the change in velocity or is there more to it than this?

http://img463.imageshack.us/img463/3143/vectordiagram5hh.jpg [Broken]
 
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  • #8
Looks good to me, except that the resultant (the "vertical" arrow) is pointing in the wrong direction. :smile:

V is the final velocity; U is the initial velocity. You drew (correctly) V + (-U). Make sure you draw the resultant correctly.
 
  • #9
Oh! haha duh! I meant to draw the resultant vector pointing the other way!

Anyways, does this mean that the change in velocity is just -0.8? That can't be right..
 
  • #10
Metropolis said:
Anyways, does this mean that the change in velocity is just -0.8? That can't be right..
It's not. Where do you get that conclusion? Find the resultant (just like you did before).

Hint: If the resultant is the hypotenuse of a right triangle, then the Pythagorean theorem applies.

If all you need is the magnitude of the change in momentum, just find the magnitude of the resultant. But if you need the direction as well, you'll need to use a bit of geometry. (Or use components, like I first suggested.)
 
  • #11
I get confused with this question because I'm referring to other similar problems which don't have a direct right angle

so the change in velocity is 1.789 m.s-1? {from P's theorem}
 
  • #12
Metropolis said:
I get confused with this question because I'm referring to other similar problems which don't have a direct right angle
For those problems, where there's no right angle to make your life easier, use the component method to add/subtract vectors.

so the change in velocity is 1.789 m.s-1? {from P's theorem}
Sounds good to me.
 
  • #13
Thanks a lot for your help Doc Al!

Now I think I can handle the other parts of the question

EDIT: So for the 2nd part of the question, I have a major problem (I think)...it tells me to find the force the side of table applies to ball...question is...F= Change in momentum / change in time...but I was not given any time data...is it because there's another approach to it? Or did this question that my teacher set have something missing? (which sometimes happens)
 
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Q: What is the "billiard ball physics problem"?

The billiard ball physics problem refers to a classic physics problem that involves the collision of two billiard balls on a frictionless surface. It is often used as an example to demonstrate the principles of conservation of momentum and conservation of kinetic energy.

Q: What is the setup of the billiard ball physics problem?

In this problem, two billiard balls are initially at rest on a frictionless surface. One ball is struck by a cue stick, causing it to move and collide with the other ball. The goal is to determine the resulting velocities of both balls after the collision.

Q: What principles of physics are involved in solving the billiard ball physics problem?

The billiard ball physics problem involves the principles of conservation of momentum and conservation of kinetic energy. These principles state that the total momentum and kinetic energy of a system remains constant before and after a collision.

Q: What is the formula for calculating the resulting velocities of the billiard balls in this problem?

The formula for calculating the resulting velocities after a collision is known as the conservation of momentum equation: m1v1 + m2v2 = m1v1' + m2v2', where m represents mass and v represents velocity. This equation takes into consideration the momentum of each ball before and after the collision.

Q: Are there any other factors that need to be taken into account when solving the billiard ball physics problem?

Other factors that may need to be considered when solving this problem include the elasticity of the balls and the angle of collision. These factors can affect the amount of kinetic energy lost during the collision and therefore impact the resulting velocities of the balls.

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