Elastic glancing collision problem

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A particle with an initial speed of 2.64 × 10^6 m/s collides elastically with a stationary particle of the same mass, resulting in the struck particle moving at a 45º angle. The key equations governing the collision include conservation of momentum and kinetic energy. The final speed of the struck particle after the collision is approximately 1.9 × 10^6 m/s. Participants in the discussion express confusion about applying the equations to find the final velocities in both x and y directions. Understanding the relationship between these directional components is crucial for solving the problem correctly.
kjean
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Homework Statement



A particle with speed v1 = 2.64 × 10^6 m/s makes a glancing elastic collision with another particle that is at rest. Both particles have the same mass. After the collision, the struck particle moves off at 45º to v1. The speed of the struck particle after the collision is approximately? The answer is 1.9 x 10^6 m/s

Homework Equations



v1i = v1f + v2f

v1i^2 = v1f^2 + v2f^2

v1ix = v1fx + v2fx
0 m/s = v1fy + v2fy

The Attempt at a Solution



I have these equations but I don't understand how to solve it. Can someone please walk me through this problem?
 
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kjean said:

Homework Statement



A particle with speed v1 = 2.64 × 106 m/s makes a glancing elastic collision with another particle that is at rest. Both particles have the same mass. After the collision, the struck particle moves off at 45º to v1. The speed of the struck particle after the collision is approximately? The answer is 4.5 m/s

Homework Equations



v1i = v1f + v2f

v1i^2 = v1f^2 + v2f^2

v1ix = v1fx + v2fx
0 m/s = v1fy + v2fy

The Attempt at a Solution



I have these equations but I don't understand how to solve it. Can someone please walk me through this problem?
Hello kjean. Welcome to PF !

How are v2fx and v2fy related?
 
The final velocities for v2 have a x direction and a y direction.
 
kjean said:
The final velocities for v2 have a x direction and a y direction.
In what direction does the struck particle (particle 2) move after the collision?

The answer to that should tell you how v2fx and v2fy are related .
 
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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