Pool Ball Grade 12 - Issue Resolved Thanks to Trajan22

  • Thread starter Thread starter hassan123456
  • Start date Start date
  • Tags Tags
    Balls Grade 12
AI Thread Summary
The discussion revolves around resolving a Grade 12 physics problem related to pool balls and momentum. The initial confusion stemmed from a lack of suitable examples in the textbook, particularly regarding angles. Trajan22 provided guidance on breaking the problem into x and y components using momentum equations, emphasizing the importance of sine and cosine for determining these components. The conversation clarified how to apply these concepts, especially in relation to angles of deflection. The user expressed gratitude for the assistance and noted the value of the forum for future questions.
hassan123456
Messages
8
Reaction score
0
pool balls grade 12 please help

issue resolved tyvm to Trajan22
 
Last edited:
Physics news on Phys.org
i know this is asking quite a bit but its very apreciated
 
What have you done so far?
 
gotten confused beyond beileif and reileized the book doesn't include a suitable example, this is because the example totally exludes angles.
 
Well I am going to assume two things so tell me if I am wrong. The mass of both balls are the same and that you know the equation for momentum.

The way you start this problem and all problems like it is to break this up into components, find the sum of the momentums in the x direction and then find another equation for the sum of the momentums in the y direction.
 
Last edited:
yes the mass is the same and would i be using the right formula (numbers are supposed to be subscript) if I am using: M1V1+M2V2=M1V'1+M2V'2




edit: one problem i have no idea how to discern the x and y direction am i right in saying that i should try using pascals triangle or something?
 
Last edited:
The equation you are using is right but you will end up with two equations.
The mass is the same no matter which direction so all you are concerned with is the velocity which moves with different x and y components depending on the angle. So just use sin's and cos's to find these components.
So since the first ball is traveling horizontally initially its y component is zero. However it deflects at an angle to where the velocity has both an x and y component to it.

One more question is the 41.5% in the answer supposed to be 41.5 degrees
 
yes it is sorry about that


ok now you lost me how do i use the sin and cos
ps: I am really thankfull that your spending your time
 
Last edited:
Well here is a general form of the equation.
M_{1}V_{1x}+M_{2}V_{2x}=M_{1}V'_{1x}+M_{2}V'_{2x}
and another equation for the y components
M_{1}V_{1y}+M_{2}V_{2y}=M_{1}V'_{1y}+M_{2}V'_{2y}
but in the problem they give you an angle that the first ball deflects at being 29.7 degrees below the x-axis so in order to find the x component you would simply have a Vcos(theta) to be the x component of the ball. Where V is the velocity the ball is traveling and theta is the angle. Its easiest to think about it like a triangle with the hypoteneuse going in the direction the ball is traveling and the x and y components being the opposite and adjacent sides.
 
  • #10
thank you so much lol if only you were my teacher instead. you have no idea how gratefull i am right now. again thank you so much
 
  • #11
No problem, if you have any other questions just ask.
 
  • #12
will do i even bookmarked this site and sent emails to my friends about how they should us it if need be
 
Back
Top