Re: What are the physics involving Ice Hockey

Some previous answers about science and hockey exist on our site
MSN search using the keyword hockey or skating. Physics is everywhere. I will only hit a few examples as an overview of what to consider with hockey, and leave much of it up to you to follow my example for other applications.

Sports in general have a lot of great examples for general mechanics. Pictures and basic physics principles are essential. A Key to a good picture is to define an axis system for use with that problem. I usually start from the basic principles:

1) Conservation of Energy

2) Conservation of Momentum

3) Balance of Forces

Let's take a slap shot as an example. The contact of the stick with the puck is very similar to a golf ball - club or ball - bat collision. We can try to come to a simplified solution using conservation of energy and momentum to figure out how far and how fast the puck will travel after impact. This can be a simple 1 dimension collision to start.

Conservation of Momentum:

Change in Stick Momentum + Change in Puck Momentum = 0 (vector)

or

M_stick * V_{stick initial} + M_puck * V_{puck initial} = M_stick * V_{stick final} + M_puck * V_{puck final}

Conservation of Energy:

Change in Puck Energy = - Change in Stick Energy - Energy Loss

or

1/2 M_stick * V²_{stick initial} + 1/2 M_puck * V²_{puck initial} = 1/2 M_stick * V²_{stick final} + 1/2 M_puck * V²_{puck final} + Energy Lost.

Assume, we know the mass of the puck and the puck is at rest before the shot so its initial velocity (vector) is zero. For a simple approximation, we can estimate the speed of the stick face before and after the collision.

V_{puck final}= (M_stick * V_{stick initial} + M_puck * V_{puck initial} = M_stick * V_{stick final} - M_stick * V_{stick final} )/M_puck

Let's plug in some numbers.

Suppose that the stick face speed is 25 m/sec (90 km/hr or 55.8 MPH) before the collision and 23 m/sec (82.8 km/hr or 51.3 MPH) after the collision. This is the case where the shooter really winds up and connects well with the puck. The puck has a mass of about 2 kg and the stick mass for the collision is effectively 45 kg (this is difficult to say because should we include the whole person in the mass of the stick or not?). Using these numbers, I come up with 45 m/sec, or 162 km/hr or 100.4 MPH. This is All-Star caliber speed (Al MacInnis has broken 100 MPH during All-Star Game skills competition).

Now we can calculate the energy (note that before the collision the only energy is in the kinetic energy of the stick and player) :

KE_{stick/player initial} - KE_{stick/player final} - KE_{puck final}= Energy Loss (positive value)

Plugging in the numbers above:

14062.5 Joules - 11902.5 Joules - 2025 Joules = 135 Joules.

This is a pretty efficient and nearly elastic collision. 135 / 14062.5 = 0.96 % energy loss.

Similar collision analyze could be performed on player to player contact during a check. You could make it more complicated by considering torque and rotation in the collision, but I don't want to get into this overview.

How about considering the puck bouncing off the boards. This is also a conservation of momentum, but unless the puck bounces straight back, the problem must be considered in two dimensions. Here are the things to consider for the basic calculation.

The picture might look like this:

 
 -----------------------------------------------------
                        *|*
   Puck toward wall  *   |   *        Puck away from wall
                  *      |      *
               *         |         *
            *            |            *
                         |                        
                         

The axis is such that a positive momentum or velocity is up (puck toward wall is positive) and negative momentum or velocity is down.

This picture isn't great, but it will do for this discussion.

1) Momentum parallel to the wall will not change because the wall can only exert force perpendicularly.

2) The wall and puck collision is not perfectly elastic, so the puck will lose speed determined by the coefficient of restitution (COR).

3) We will ignore the effects of puck spin.

Setting up the momentum conservation:

Momentum of puck _parallel = M_puck * speed_{parallel before hitting wall} = M_puck * speed_{parallel after hitting wall}

Momentum of puck_{perpendicular} = M_puck * speed_{perpendicular before hitting wall} + Impulse from Wall = M_puck * speed_{parallel after hitting wall} .

From these equations derived from the principles applied, we can see that

speed_{parallel before hitting wall} = speed_{parallel after hitting wall}

and using speed_{parallel after hitting wall} = - COR * speed_{perpendicular before hitting wall} (direction is reversed so it is negative).

we have

Impulse from Wall = M_puck * -1 * (COR + 1) * speed_{perpendicular before hitting wall} .

COR is a number less than 1 (equal for perfectly elastic). With this we can look at the effects of dead spots in the wall versus bouncy parts of the wall. A dead spot would be a COR of 0.0 to 0.2 and a lively wall would be COR (0.8 to 0.95).

I hope you can come up with more examples. I didn't even consider flying pucks(projectile motion), skating/stopping (momentum/forces), the effects of hard ice versus soft ice and puddles, or goal tending.

Sincerely,

Tom "Between the Pipes" Cull