Complicated Puck Problem (consevation of forces)

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The discussion revolves around calculating the speed of a blue puck after a collision, given that its mass is 24.6% greater than that of a green puck, which has an initial speed of 12.0 m/s. The pucks approach each other with equal and opposite momenta, and it is stated that half of the kinetic energy is lost during the collision. The relevant equations for momentum and kinetic energy conservation are provided, but there is confusion regarding the correct application of these equations. The key focus is on using conservation of momentum and energy principles to determine the post-collision speed of the blue puck. The conversation emphasizes the need for clarity in the equations and their proper formulation to solve the problem effectively.
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Homework Statement


The mass of the blue (dark) puck in the figure below is 24.6% greater than the mass of the green (light) one.

Before colliding, the pucks approach each other with equal and opposite momenta, and the green puck has an initial speed of 12.0 m/s. The angle q = 33.5o. Calculate the speed of the blue puck after the collision if half the kinetic energy is lost during the collision.

Homework Equations


EFx=0=M(Green)V(green)cos(x)-M(blue)V(blue)Cos(x)
EFy=0=M(green)V(green)sin(x)-M(blue)V(blue)Sin(x)

( (1/2M(blue)V(blue)^2+M(green)V(green)^2)/2)=(1/2)M(green)V(green)^2+(1/2)M(blue)V(blue)^2

M(green)V'(green)=M(blue)V'(blue)

M(green)= (M(blue))/(2.46)
M(blue)= (2.46)M(green)


The Attempt at a Solution


I think these equations are correct but i don't knbow how I am going to get the mass of them. That would help a lot
 

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Chuck 86 said:

Homework Statement


The mass of the blue (dark) puck in the figure below is 24.6% greater than the mass of the green (light) one.

Before colliding, the pucks approach each other with equal and opposite momenta, and the green puck has an initial speed of 12.0 m/s. The angle q = 33.5o. Calculate the speed of the blue puck after the collision if half the kinetic energy is lost during the collision.

Homework Equations


EFx=0=M(Green)V(green)cos(x)-M(blue)V(blue)Cos(x)
EFy=0=M(green)V(green)sin(x)-M(blue)V(blue)Sin(x)

( (1/2M(blue)V(blue)^2+M(green)V(green)^2)/2)=(1/2)M(green)V(green)^2+(1/2)M(blue)V(blue)^2
The above says nothing at all. Perhaps you meant to write:

\frac{1}{2}M_{green}V_{green}^{'2} + \frac{1}{2}M_{blue}V_{blue}^{'2} = (\frac{1}{2}M_{blue}V_{blue}^2 + \frac{1}{2}M_{green}V_{green}^2)/2

M(green)V'(green)=M(blue)V'(blue)
What is this?

Conservation of momentum means that the total momentum of the balls before collision is the same as the total momentum of the balls after:

M_{blue}\vec{V}_{blue} + M_{green}\vec{V}_{green} = M_{blue}\vec{V}'_{blue} + M_{green}\vec{V}'_{green}

Now, using those two equations and the given information, try to find the velocities after collision.

AM
 
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