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

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Homework Help Overview

The discussion revolves around a 2D elastic collision problem involving two pucks on a frictionless air-hockey table. The original poster attempts to find the final speeds and angles of the pucks after the collision, given their masses and initial conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conservation of momentum in both x and y directions, leading to equations involving the unknown angles θ and φ. There is an attempt to apply the conservation of kinetic energy as well.

Discussion Status

Some participants provide hints regarding trigonometric identities and suggest re-evaluating the algebraic manipulations. There is ongoing exploration of how to correctly apply these identities to solve for the angles involved.

Contextual Notes

Participants note potential errors in the algebraic steps taken, particularly regarding the squaring of terms and the application of the trigonometric identity. There is also mention of the challenge posed by having multiple unknowns to solve for simultaneously.

amanda.ka
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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.
 

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