Elastic collision shuffleboard problem

In summary, the conversation discusses a problem involving two shuffleboard disks, one orange and one yellow, colliding elastically. The initial conditions and equations for solving the final velocities of each disk are provided. It is mentioned that the masses of the disks are equal, making it unnecessary to account for them in the calculations. The method for solving the problem is described, with a reminder to be careful of signs in the equations.
  • #1
KD-jay
7
0

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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  • #2
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
 
  • #3
never mind, I misread the problem (deleted now)
 

1. What is an elastic collision shuffleboard problem?

An elastic collision shuffleboard problem is a physics problem that involves calculating the motion and interactions of objects on a shuffleboard table, assuming that the collisions between the objects are perfectly elastic (meaning no energy is lost during the collision).

2. How is the motion of the objects determined in an elastic collision shuffleboard problem?

In an elastic collision shuffleboard problem, the motion of the objects is determined by applying the laws of conservation of momentum and conservation of energy. These laws state that the total momentum and energy in a system remain constant before and after a collision.

3. What are the key factors to consider in an elastic collision shuffleboard problem?

The key factors to consider in an elastic collision shuffleboard problem include the masses of the objects, their initial velocities, the angle and location of the collisions, and the coefficient of restitution (a measure of how bouncy the objects are). These factors will affect the final velocities and paths of the objects after the collision.

4. How is the coefficient of restitution determined in an elastic collision shuffleboard problem?

The coefficient of restitution (COR) is determined by dividing the final velocity of an object after a collision by its initial velocity. This value can range from 0 (perfectly inelastic collision, where the objects stick together) to 1 (perfectly elastic collision, where no energy is lost).

5. Can an elastic collision shuffleboard problem be solved using real-world scenarios?

Yes, an elastic collision shuffleboard problem can be solved using real-world scenarios as long as the assumptions of a perfectly elastic collision hold true. However, in real-world situations, some energy will typically be lost due to factors such as friction and air resistance, so the solutions will not be completely accurate.

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