Elastic collision - Trigonometry equations

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In an elastic collision between two identical billiard balls, one moving at 4.00 m/s and the other stationary, the stationary ball moves away at 1.69 m/s after the collision. The discussion focuses on determining the final speed and direction of the incoming ball, as well as the direction of the stationary ball. Participants emphasize using conservation of energy and momentum equations, along with trigonometric identities, to solve for the unknown angles. A suggested approach involves forming a triangle with the momentum vectors and applying the cosine rule for simplification. This method aims to resolve the complexity of the trigonometric equations involved in the problem.
Dandi Froind
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Homework Statement


A billiard ball moves at a speed of 4.00m/s and collides elastically with an identical stationary ball. As a result, the stationary ball flies away at a speed of 1.69m/s, as shown in Figure A2.12. Determine:
  1. the final speed and direction of the incoming ball after the collision.
  2. the direction of the stationary ball after the collision.

Homework Equations


Conservation of Energy and momentum:
E before = E after
Σpx=0
Σpy=0

The Attempt at a Solution


As you can see, I have 2 equations with 2 angles that I don't know.
However, I couldn't find the right way to solve it due to trigonometry complexity.
I am sure that taking into account the fact that (sin(theta))^2 +(cos(theta))^2 = 1 would help in some way.
https://image.ibb.co/gv6YVw/Capture.jpg
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Hi again,

As long as I can see, you use the right method. You are right, the equation sin2θ1+cos2θ1=1 can help you.

You must think how you can form this (sin2θ1+cos2θ1). Try to "play" with these two equations. (Hint: You will end up solving an equation with only one of the two angles firstly).
 
Dandi Froind said:

Homework Statement


A billiard ball moves at a speed of 4.00m/s and collides elastically with an identical stationary ball. As a result, the stationary ball flies away at a speed of 1.69m/s, as shown in Figure A2.12. Determine:
  1. the final speed and direction of the incoming ball after the collision.
  2. the direction of the stationary ball after the collision.

Homework Equations


Conservation of Energy and momentum:
E before = E after
Σpx=0
Σpy=0

The Attempt at a Solution


As you can see, I have 2 equations with 2 angles that I don't know.
However, I couldn't find the right way to solve it due to trigonometry complexity.
I am sure that taking into account the fact that (sin(theta))^2 +(cos(theta))^2 = 1 would help in some way.
Yes, it would, Arrange the momentum equations in the form sin(θ2)=... cos(θ2)=...
Square them and add.
 
Thank you very much.
It was very helpful.
 
Dandi Froind said:

Homework Statement


A billiard ball moves at a speed of 4.00m/s and collides elastically with an identical stationary ball. As a result, the stationary ball flies away at a speed of 1.69m/s, as shown in Figure A2.12. Determine:
  1. the final speed and direction of the incoming ball after the collision.
  2. the direction of the stationary ball after the collision.

Homework Equations


Conservation of Energy and momentum:
E before = E after
Σpx=0
Σpy=0

The Attempt at a Solution


As you can see, I have 2 equations with 2 angles that I don't know.
However, I couldn't find the right way to solve it due to trigonometry complexity.
I am sure that taking into account the fact that (sin(theta))^2 +(cos(theta))^2 = 1 would help in some way.
https://image.ibb.co/gv6YVw/Capture.jpg
View attachment 214226

If you take the two resultant momentum vectors after the collision and add them (i.e. put them tip to tail with each other), then, by conservation of momentum, you have the initial momentum vector.

This gives you a triangle where you know the lengths of all three sides. You could, therefore, use the cosine rule to get a simpler equation for each angle separately.

You could try this technique, as I find it a simpler method for collision problems.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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