Elastic collision shuffleboard problem

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SUMMARY

The discussion focuses on solving an elastic collision problem involving two shuffleboard disks, one orange and one yellow, with equal mass. The orange disk, initially moving at 5.40 m/s, strikes the stationary yellow disk, resulting in a collision where the disks move at right angles post-collision. The conservation of momentum and kinetic energy equations are applied to determine the final speeds of both disks, with the orange disk moving at an angle of 40.0° after the collision.

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KD-jay
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Homework Statement


Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.40 m/s. After the collision, the orange disk moves along a direction that makes an angle of 40.0° with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.

Homework Equations


pi = pf
KEi=KEf

The Attempt at a Solution


X-Direction

Orange Shuffleboard
Initial Momentum = 5.4m
Final Momentum = Vofcos(40)m
Initial KE = (1/2)(m)(5.4)2
Final KE = (1/2)(m)(Vof)2

Yellow Shuffleboard
Initial Momentum = 0
Final Momentum = Vyfcos(90-40)m
Initial KE = 0
Final KE = (1/2)(m)(Vyf)2

Conservation of Momentum
5.4m + 0 = Vofcos(40)m + Vyfcos(50)m

Conservation of Energy
(1/2)(m)(5.4)2 + 0 = (1/2)(m)(Vof)2 + (1/2)(m)(Vyf)2

Y-Direction

Orange Shuffleboard
Initial Momentum = 0
Final Momentum = Vofsin(40)m
Initial KE = (1/2)(m)(5.4)2
Final KE = (1/2)(m)(Vof)2

Yellow Shuffleboard
Initial Momentum = 0
Final Momentum = Vyfsin(50)m
Initial KE = 0
Final KE = (1/2)(m)(Vfy)2

Conservation of Momentum
0 + 0 = Vofsin(40)m + Vyfsin(50)m

Conservation of Energy
(1/2)(m)(5.4)2 + 0 = (1/2)(m)(Vof)2 + (1/2)(m)(Vfy)2

I know I'm supposed to use all these equations to solve for the unknowns but I just want to make sure I'm on the right track. Are all my variables correct? Also, it seems extremely tedious to solve this system of equations, would it be safe to just cross out the masses on the momentum conservation and the (1/2)m on the energy conservations? Thanks.
 
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The masses are all equal so no need to account for that.

Basically it looks like you have the right method. Careful of your signs in solving
 
never mind, I misread the problem (deleted now)
 

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