- #1
Samuelriesterer
- 110
- 0
Problem Statement:
Moe Mentum and Ken Ettik are playing pool. Moe hits the cue ball, sending it towards the 6 ball at 2 m/s. It strikes the stationary 6 ball off-center, moving off at a 60 degree angle from the original direction after a perfectly elastic collision. Both balls have a mass of 160 grams.
Relative Equations:
m1v1i + m2v2i = m1v1f + m2v2f
½ m1v1i + ½ m2v2i = ½ m1v1f + ½ m2v2f
v1i = v1f + v2f
v1i^2 = v1f^2 + v2f^2
Work done so far:
(1) Use conservation of kinetic energy to determine the angle at which the six ball moves after collision, then draw a diagram of the collision.
The angle will be 30 degrees to form a right angle triangle because:
v1i^2 = v1f^2 + v2f^2
(2) Determine the direction of the forces of impact on the balls during collision and set up a pair of axes.
Only an applied force to the 6 ball because we are assuming no friction or air resistance and mg and N cancel. We will put the x-axis through the centers of the balls and the y-axis along the center tangent to the balls
(3) Resolve the initial and final velocities into components along the axes you have chosen. What happens to the components perpendicular to the forces of impact?
v1i_x = v1i
v1i_y = 0
v2i_x = 0
v2i_y = 0
v1f_x = v1f cos 60
v1f_y = v1f sin 60
v2f_x = v2f cos (30)
v2f_y = v2f sin (-30)
Components perpendicular to the forces of impact remain unchanged while the x components changes sign
(4) For the axis in the direction of the forces, calculate the velocity of the center of mass before collision.
Here is where things seem complicated. I know how to calculate the CM for equal masses, simply average the initial velocities:
CM_x = (v1i – 0) / 2
CM_y = (0 – 0) / 2 = 0
But since the CM_y = 0 and the y components = 0 and v2i = 0, I can’t see how the balls will break at an angle if the cue ball is traveling along the x-axis into a collision with the 6 ball which is at rest. That is, are we assuming the 6 ball’s center is along the axis? Because it can’t be, otherwise the two balls will continue in the x direction with no vertical component. When I work the CM above, it comes out with no y component. And this is understandable.
Now if we know that the 6 ball is off center, then how do I factor that into the CM?
(5) Transform the velocities along the axis in (4) to the center of mass frame.
Once I have the CM I can subtract if from the initial velocities to convert it to the CM frame.
(6) Now determine the final velocities along the axis in the CM frame and then transform them back to the pool table frame.
The final velocities in the CM frame can be found by changing the sign of the initial velocities in the CM frame. Then to get the final velocities in the lab frame I add the CM to the final velocities in the CM frame.
(7) Recombine the final velocity components to get the final velocities for the two pool balls.
Again, I need a working CM first.
(8) The six ball encounters a pencil on the table and flies off the table, bouncing off the cushion to fly upward, 60° above horizontal moving at the same speed it had on the table. What is the vertical component of the launch velocity and the maximum height it reaches?
Moe Mentum and Ken Ettik are playing pool. Moe hits the cue ball, sending it towards the 6 ball at 2 m/s. It strikes the stationary 6 ball off-center, moving off at a 60 degree angle from the original direction after a perfectly elastic collision. Both balls have a mass of 160 grams.
Relative Equations:
m1v1i + m2v2i = m1v1f + m2v2f
½ m1v1i + ½ m2v2i = ½ m1v1f + ½ m2v2f
v1i = v1f + v2f
v1i^2 = v1f^2 + v2f^2
Work done so far:
(1) Use conservation of kinetic energy to determine the angle at which the six ball moves after collision, then draw a diagram of the collision.
The angle will be 30 degrees to form a right angle triangle because:
v1i^2 = v1f^2 + v2f^2
(2) Determine the direction of the forces of impact on the balls during collision and set up a pair of axes.
Only an applied force to the 6 ball because we are assuming no friction or air resistance and mg and N cancel. We will put the x-axis through the centers of the balls and the y-axis along the center tangent to the balls
(3) Resolve the initial and final velocities into components along the axes you have chosen. What happens to the components perpendicular to the forces of impact?
v1i_x = v1i
v1i_y = 0
v2i_x = 0
v2i_y = 0
v1f_x = v1f cos 60
v1f_y = v1f sin 60
v2f_x = v2f cos (30)
v2f_y = v2f sin (-30)
Components perpendicular to the forces of impact remain unchanged while the x components changes sign
(4) For the axis in the direction of the forces, calculate the velocity of the center of mass before collision.
Here is where things seem complicated. I know how to calculate the CM for equal masses, simply average the initial velocities:
CM_x = (v1i – 0) / 2
CM_y = (0 – 0) / 2 = 0
But since the CM_y = 0 and the y components = 0 and v2i = 0, I can’t see how the balls will break at an angle if the cue ball is traveling along the x-axis into a collision with the 6 ball which is at rest. That is, are we assuming the 6 ball’s center is along the axis? Because it can’t be, otherwise the two balls will continue in the x direction with no vertical component. When I work the CM above, it comes out with no y component. And this is understandable.
Now if we know that the 6 ball is off center, then how do I factor that into the CM?
(5) Transform the velocities along the axis in (4) to the center of mass frame.
Once I have the CM I can subtract if from the initial velocities to convert it to the CM frame.
(6) Now determine the final velocities along the axis in the CM frame and then transform them back to the pool table frame.
The final velocities in the CM frame can be found by changing the sign of the initial velocities in the CM frame. Then to get the final velocities in the lab frame I add the CM to the final velocities in the CM frame.
(7) Recombine the final velocity components to get the final velocities for the two pool balls.
Again, I need a working CM first.
(8) The six ball encounters a pencil on the table and flies off the table, bouncing off the cushion to fly upward, 60° above horizontal moving at the same speed it had on the table. What is the vertical component of the launch velocity and the maximum height it reaches?
