Complicated Puck Problem (consevation of forces)

  • Thread starter Thread starter Chuck 86
  • Start date Start date
  • Tags Tags
    Forces
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
Chuck 86
Messages
27
Reaction score
0

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
 

Attachments

  • Puck Problem.gif
    Puck Problem.gif
    3 KB · Views: 616
Physics news on Phys.org
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