Complicated Puck Problem (consevation of forces)

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SUMMARY

The discussion focuses on calculating the speed of a blue puck after a collision with a green puck, where the blue puck's mass is 24.6% greater than the green puck's mass. The green puck has an initial speed of 12.0 m/s and the collision results in a loss of half the kinetic energy. Key equations include conservation of momentum and energy, specifically EFx and EFy equations, which relate the masses and velocities of both pucks before and after the collision.

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  • Ability to solve equations involving trigonometric functions
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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:

[tex]\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[/tex]

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:

[tex]M_{blue}\vec{V}_{blue} + M_{green}\vec{V}_{green} = M_{blue}\vec{V}'_{blue} + M_{green}\vec{V}'_{green}[/tex]

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

AM
 

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