Solve for 2D Elastic Collision: Find θ & φ Angles

AI Thread Summary
In a 2D elastic collision problem involving two pucks on a frictionless air-hockey table, puck 1 (0.5 kg) moves at 4 m/s and collides with puck 2 (0.3 kg) initially at rest. The goal is to find the final speed of puck 2 and the angles θ and φ after the collision. The conservation of momentum equations for both x and y directions are set up, leading to two equations involving cosθ and sinθ. Participants discuss the need to correctly manipulate these equations, emphasizing the importance of using the trigonometric identity cos²θ + sin²θ = 1 to solve for the unknown angles. Clarifications are made regarding algebraic errors and the proper approach to isolate variables for solving the angles.
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Homework Statement


A 2D elastic collision:
Two pucks (masses m1 = 0.5 kg and m2 = 0.3 kg) collide on a frictionless air-hockey table. Puck 1 has an initial velocity of 4 m/s in the positive x direction and a final velocity of 2 m/s in an unknown direction, θ. Puck 2 is initially at rest. Find the final speed of puck 2 and the angles θ and φ.

I get stuck at the end of this problem when I have to use the two equations to solve for 2 unknown angles. If someone could show me how to do that last step that would be great. Thanks in advance!

Homework Equations

The Attempt at a Solution


Since the collision is elastic I found the final kinetic energy using Ei = Ef and it equals 4.47 m/s

Then conservation of the x and y components of total momentum:

X DIRECTION: Pi = Pf
m1v1 + m2v2i = m1v1 + m2v2f
(0.5)(4) = (0.5)(2cosθ) + (0.3)(4.47cosφ)
2 = cosθ + 1.341cosφ

Y DIRECTION: Pi = Pf
m1v1 + m2v2i = m1v1 + m2v2f
0 = (O.5)(2sinθ) -(0.3)(4.47sinφ)
0 = sinθ - 1.341sinφ
 
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Hint: cos2θ + sin2θ = 1

(Your work looks good so far.)
 
TSny said:
Hint: cos2θ + sin2θ = 1

(Your work looks good so far.)
So I solve the first equation for cosφ and the second equation for sinφ?
1.341cosφ = 2-cosθ
and
-1.341sinφ = sinθ

then square each term:
1.3412cos2φ = 22-cos2θ
and
-1.3412sin2φ = sin2θ

then I added them but I'm still a bit lost/don't know the next step :/

1.3412cos2φ + (-1.341)2sin2φ = 22-cos2θ + sin2θ
 
Take another look at (2-cosθ)2. It doesn't look right. Also, I thought it might have been a little simpler to solve for sinθ and cosθ, then square, then add the equations and then implement TSny's hint.
 
amanda.ka said:
1.341cosφ = 2-cosθ
and
-1.341sinφ = sinθ

then square each term:
1.3412cos2φ = 22-cos2θ
and
-1.3412sin2φ = sin2θ

Note that (2 - cosθ)2 ≠ 22-cos2θ. In general (a + b)2 ≠ a2 + b2.

In order to use the trig identity cos2θ + sin2θ = 1, you could solve the first equation for cosθ. You already have an expression for sinθ, except I believe you have a sign error in -1.341sinφ = sinθ.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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