# Complicated Puck Problem (consevation of forces)

1. Oct 16, 2009

### Chuck 86

1. The problem statement, all variables and given/known data
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.

2. Relevant 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)

3. The attempt at a solution
I think these equations are correct but i dont knbow how im going to get the mass of them. That would help a lot

#### Attached Files:

• ###### Puck Problem.gif
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2. Oct 17, 2009

### Andrew Mason

The above says nothing at all. Perhaps you meant to write:

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

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