# Elastic collision of hockey puck

## 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)
mvai = mvaf + mvbf

conservation of kinetic energy (2)
1/2mvai = 1/2mvaf + 1/2mvbf

## 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

vbxf = vaxi - vaxf

vbyf = -vayf

so

vbf = sqrt [(vaxi - vaxf)2 + (-vayf)2]

vbf = sqrt [vaxi2 - 2vaxf + vaxf2 + vayf2]

then i plugged this into 1/2mvai = 1/2mvaf + 1/2mvbf for v_bf

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

i end up getting vfa = 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

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Andrew Mason
Homework Helper
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