Elastic collision shuffleboard problem

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The discussion revolves around solving an elastic collision problem involving two shuffleboard disks, one orange and one yellow, where the orange disk strikes the yellow disk at rest. The orange disk moves at 5.40 m/s and makes a 40° angle with its initial direction after the collision, while the disks' velocities are perpendicular post-collision. Participants confirm the use of conservation of momentum and energy equations to find the final speeds of both disks. There is a suggestion to simplify calculations by eliminating mass from the equations since both disks have equal mass. Overall, the approach is correct, and attention to detail in signs and calculations is emphasized.
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)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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