Balls elastic collision off center

In summary: Finally, ...In summary, 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: 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. 2) Determine the direction of the forces of impact on the balls during collision and set up a pair of
  • #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?
 
Physics news on Phys.org
  • #2
Now if we know that the 6 ball is off center, then how do I factor that into the CM?
... once you have describe the vectors in your coordinate system, you have already factored it in. Look at your diagram and see which way the forces act.
 
  • #3
Here is how I calculate this:

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)

V_cm_x = (V1i)/2
V_cm_y = 0

cmV1i_x = V1i_x - V_cm_x = V1i/2
cmV1i_y = 0
cmV2i_x = V2i_x - V_cm= -V1i/2
cmv2i_y = 0

cmV1f_x = -cmV1i_x = -V1i/2
cmV1f_y = 0
cmv2f_x = -cmV2i_x = V1i/2
cmv1f_y = 0

labV1f_x = 0
labV1f_y = 0
labV2f_x = V1i
labV2f_y = 0

As you can see the outcome treats this as a head on collision along the x axis.

What am I doing wrong?
 
  • #4
You have defined the x-axis as the line joining the two centers at the time of contact.
In the lab frame, his will put the initial momenta off the axes. For com frame, ... the trick is to sketch the diagrams for initial and final states. To convert back you need the motion of the com.
 
  • #5
Im confused. Can you elaborate on that last post?
 
  • #6
The way I read it, you were wondering that the resulting motion was parallel the axes you set up ... but you had set up the axes so that this would be the case. Using that coordinate system, the velocity of the center of mass will point at an angle to the x-axis.
You were concerned that the off-center hit was not taken into account ... take a closer look at the position of the center of mass compared with the head-on case; then see how that affects the velocity.
 
  • #7
I'm sorry I'm just not getting this. I can find no help online or in the 3 textbooks that I have on Physics about this particular problem. I don't get the rotation of the coordinate system. I thought that was taken into account when I converted the coordinates from the lab to the CM and then back again.
 
  • #8
I don't know what you are talking about either.
The way you have set up the coordinate system, it does not rotate.
But if you define one axis as through the centers of the two balls, then the coordinate system will rotate against the usual one.
To see how this happens, try sketching the balls at different separations and draw the axis in.
 

Related to Balls elastic collision off center

1. What is an off-center elastic collision?

An off-center elastic collision is a type of collision between two objects where the point of impact is not directly in the center of the objects. This means that the objects do not collide head-on but rather at an angle.

2. How does an off-center elastic collision differ from a center elastic collision?

In a center elastic collision, the objects collide directly in the center, resulting in a transfer of energy and momentum that causes the objects to bounce off each other. In an off-center elastic collision, the objects collide at an angle, causing a transfer of energy and momentum in a different direction.

3. What factors affect the outcome of an off-center elastic collision?

The outcome of an off-center elastic collision is affected by the mass, velocity, and angle of the objects involved. The elasticity of the objects and any external forces also play a role in the collision.

4. What is the conservation of energy and momentum in an off-center elastic collision?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In an off-center elastic collision, the total energy before the collision is equal to the total energy after the collision. The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

5. How is the velocity of the objects after an off-center elastic collision calculated?

The velocity of the objects after an off-center elastic collision can be calculated using the equations of conservation of energy and momentum. These equations take into account the masses, velocities, and angles of the objects before and after the collision, as well as the elasticity of the objects and any external forces.

Similar threads

  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
933
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top