Elastic collision of hockey puck

[mybrohshi5]

Homework Statement

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially traveling at 15.2 m/s and is deflected 30.0^\circ from its initial direction. The pucks are made of superball-like material, so you may assume that the collision is perfectly elastic.

Find the final speed of the puck A after the collision

Homework Equations

conservation of momentum for one object at rest (1)
mv_ai = mv_af + mv_bf

conservation of kinetic energy (2)
1/2mv_ai = 1/2mv_af + 1/2mv_bf

The Attempt at a Solution

I have tried this about 10 times and cannot get it right :(

i solved equation 1 for the x and y components of the velocity final for puck b

v_bxf = v_axi - v_axf

v_byf = -v_ayf

so

v_bf = sqrt [(v_axi - v_axf)² + (-v_ayf)²]

v_bf = sqrt [v_axi² - 2v_axf + v_axf² + v_ayf²]

then i plugged this into 1/2mv_ai = 1/2mv_af + 1/2mv_bf for v_bf

i would show all my steps of reducing but it would take forever.

i end up getting v_fa = cos(30) = .866 but this is wrong

am i right up to this point of what i have done?

any help would be great. Thank you

 

[Andrew Mason]

The directional components of momentum before and after are the same. So:

$$mv_{ai} = mv_{af}cos(\alpha) + mv_{bf}cos(\theta)$$ and

$$mv_{af}sin(\alpha) + mv_{bf}sin(\theta) = 0$$

Now the trick is to realize that the conservation of energy equation gives you the angle between the two pucks after collision:

$$v_{ai}^2 = v_{af}^2 + v_{bf}^2$$

so with that angle you can find $\alpha$

AM