Calculating Angle Theta in a Two-Puck Elastic Collision

[NotMrX]

Suppose there are two puck of equal mass. That the first puck moves to the east at a velocity of 3.6 m/s while the second one is at rest. Then there is an elastic collision. The first puck moves at an angle theta north of east with a velocity of 1.3 m/s. The second puck moves south of east at angle phi. What is the angle theta?

I have attempted the problem but haven't gotten it to work out yet. Here is what I did.

Momentum in the x direction:
$$v_{1i}=v_{1f}*cos(\theta )+v_{2f}cos(\phi )$$

Momentum in the y direction:
$$0=v_{1f}*sin(\theta )+v_{2f}sin(\phi )$$

Consevation of Energy:
$$v_{1i}^2=v_{1f}^2+v_{2f}^2$$

 

[berkeman]

It looks like you can get v2 from the conservation of energy equation, and then you have 2 equations and the two unknown angles...

 

[NotMrX]

I have a hard time solving the equations once I get them.

From energy we can say:
$$v_{2f}=\sqrt{v_{1i}^2-v_{1f}^2$$

From y momentum we get:
$$sin(\phi )=\frac{-v_{1f}*sin(\theta )}{v_{2f}}$$

From the x momentum we get:
$$cos(\phi )=\frac{v_{1i}-v_{1f}*cos(\theta )}{v_{2f}}$$

Here is the trick. I draw the triangles then I say that the adjacent legs are the same value since both have the same hypotnuse v2f and the same angle phi.
$$v_{1i}-v_{1f}*cos(\theta )=\sqrt{v_{2f}^2-(v_{1f}*sin(\phi )^2}$$

From here it is possible to solve I just wondered if anyone knew any easier way.

 

[NotMrX]

Is the angle between the final velocity of the pucks supposed to be 90 degrees because when I worked it out I got something like 88.6 degrees?

 

[NotMrX]

I read that it is always 90 degrees so I guess I can say:

90= theta - phi

since phi is going in the negative direction.

Then I could use the double angle thereom which might be a little easier. The final expression I ended up with was:

$$cos(\theta )=\frac{V_{1f}}{V_{1i}}$$

Which seems so simple, yet I had to go through so many hoops to get it. I feel like there must be some short cut.

 

[NotMrX]

yep it is much easier to say that:
sin(Phi)=sin(pheta-90)=-cos(pheta)