Momentum and Collisions in 2 dimensions

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SUMMARY

The discussion focuses on the analysis of an elastic collision between two particles with masses 2m and 3m, moving toward each other along the x-axis with equal speeds v. After the collision, the 2m particle moves downward at a right angle, while the 3m particle's final velocity and angle relative to the horizontal are denoted as v2 and angle w, respectively. The conservation of momentum in both x and y dimensions, along with the conservation of kinetic energy, leads to a system of three equations that can be solved for the unknowns. Participants suggest simplifying the equations and utilizing the identity sin²w + cos²w = 1 to facilitate the solution process.

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Homework Statement


Two particles with masses 2m and 3m are moving toward each other along the x-axis with the same initial speeds v. Particle 2m is traveling to the left, while particle 3m is traveling to the right. They undergo an elastic, glancing collision such that particle 2m is moving downward after the collision at a right angle from its initial direction.

Homework Equations


v2 = final velocity of the 3m object
v3 = final velocity of the 2m object
angle w = the angle of 3m to the horizontal after colliding with the 2m

momentum conserved in x-dimension:
3mv - 2mv = 3m * cos(w) * v2

momentum conversed in y-dimension:
0 = (3m * sin(w) * v2) - (2m * v3)

kinetic energy is also conserved:
.5(3m)(v^2) + .5(2m)(v^2) = .5(3m)(v2^2) + .5(2m)(v3^2)

The Attempt at a Solution


pages of work of which i am too tired now to transcribe into type
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If god is real, then people will help me.
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Your working reads OK.
You have three equations and three unknowns.
How about start by simplifying the equations and doing some substitutions.
Then maybe you can use.
sin²w + cos²w = 1
It's long winded but it works...
 

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