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johnhuntsman
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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 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 v1. And I have these equations written out:
v1i = v1f + v2f (since momentum is conserved; the mass can be divided out)
v1i2 = v1f2 + v2f2 (since KE is conserved and 0.5m can be divided out)
The first equation broken up into components:
v1ix = v1fx + v2fx
0 m/s = v1fy + v2fy
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]
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 v1. And I have these equations written out:
v1i = v1f + v2f (since momentum is conserved; the mass can be divided out)
v1i2 = v1f2 + v2f2 (since KE is conserved and 0.5m can be divided out)
The first equation broken up into components:
v1ix = v1fx + v2fx
0 m/s = v1fy + v2fy
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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