Elastic glancing collision problem

[kjean]

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?

 

[SammyS]

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?

 

[kjean]

The final velocities for v2 have a x direction and a y direction.

 

[kjean]

https://www.physicsforums.com/showthread.php?t=649385

same question I just don't understand how to use the equations given to solve it.

 

[SammyS]

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 .