What is the Angle Between the Target and Incident Ball Post-Collision?

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In a perfectly elastic collision between two identical steel marbles, one marble collides obliquely with the other, continuing at 55 degrees to its original direction. The angle between the target ball's direction and the incident ball's original direction is not simply 55 degrees. To determine this angle, one must apply the principles of conservation of momentum and conservation of kinetic energy. The calculations will yield the correct angle based on the velocities of both marbles post-collision. Understanding these principles is crucial for solving the problem accurately.
Fusilli_Jerry89
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Homework Statement


A 45 g steel marble collides obliquely w/ an identical stationary marble, and continues at 55 degrees to its original direction. The collision is perfectly elastic. What is the angle between the direction taken by the target ball and the original direction of the incident ball?

The Attempt at a Solution


is it just 55 degrees?
 
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Fusilli_Jerry89 said:

Homework Statement


A 45 g steel marble collides obliquely w/ an identical stationary marble, and continues at 55 degrees to its original direction. The collision is perfectly elastic. What is the angle between the direction taken by the target ball and the original direction of the incident ball?

The Attempt at a Solution


is it just 55 degrees?

It is not 55 degrees. You need to apply conservation of momentum and conservation of energy (kinetic energy in this case) to find the magnitude and direction of the velocity of the target ball.
 
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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