Momentum and Collisions in 2 dimensions

[cheese825]

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.
If god is not real, then people will help me.
Everything is appreciated!

 

[Carid]

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...

 

[thomate1]

I can provide a response to this content by first acknowledging that the problem statement is describing a situation involving conservation of momentum and energy in a two-dimensional collision. The given equations are correct and can be used to solve for the final velocities and angle of particle 2m after the collision.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant in the absence of external forces. In this case, the initial momentum in the x-direction is 3mv - 2mv = mv, and since the collision is elastic, the final momentum in the x-direction must also be mv. This leads to the equation 3mv - 2mv = 3m * cos(w) * v2, where v2 is the final velocity of particle 2m.

Similarly, in the y-direction, the initial momentum is 0 since both particles are moving in opposite directions with the same speed. After the collision, the final momentum in the y-direction must also be 0, which leads to the equation (3m * sin(w) * v2) - (2m * v3) = 0, where v3 is the final velocity of particle 3m.

Using these equations, we can solve for the final velocities v2 and v3 as well as the angle w. Additionally, we can check the conservation of kinetic energy by using the equation .5(3m)(v^2) + .5(2m)(v^2) = .5(3m)(v2^2) + .5(2m)(v3^2). If this equation holds true, then we can be confident that our solution is correct.

In conclusion, the problem statement presents a scenario involving conservation of momentum and energy in a two-dimensional collision. By applying the principles of conservation and using the given equations, we can solve for the final velocities and angle of particle 2m after the collision. This type of problem is commonly encountered in physics and can be solved using mathematical principles and equations.