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A particle with speed v1 = 2.64 × 10

The answer is 1.9E6 m/s.

I drew a diagram of the scenario. I know that the two final vectors form a right triangle with the initial v

v

v

The first equation broken up into components:

v

0 m/s = v

What do I do? Maybe I'm the world's least intuitive man, but I don't see how this system of equations can be solved. All I need is someone to tell me where to start and I'll probably be good to go.

^{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 v_{1}. The speed of the struck particle after the collision is approximately...The answer is 1.9E6 m/s.

I drew a diagram of the scenario. I know that the two final vectors form a right triangle with the initial v

_{1}. And I have these equations written out:v

_{1i}= v_{1f}+ v_{2f}(since momentum is conserved; the mass can be divided out)v

_{1i}^{2}= v_{1f}^{2}+ v_{2f}^{2}(since KE is conserved and 0.5m can be divided out)The first equation broken up into components:

v

_{1ix}= v_{1fx}+ v_{2fx}0 m/s = v

_{1fy}+ v_{2fy}What do I do? Maybe I'm the world's least intuitive man, but I don't see how this system of equations can be solved. All I need is someone to tell me where to start and I'll probably be good to go.

*[Edit] Am I supposed to put the components in terms of sine and cosine and go from there? [Edit]*
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