| MadSci Network: Physics |
Dear Bryan,
Let the mass of the puck be m = 6 ounces (actually a weight) = 0.170 kg,
and let the initial speed of the puck be v = 85 miles/hour = 38 m/s.
The final speed of the puck is zero if the goalie brings it to a stop.
If the goalie does not recoil and if the puck does not bounce off the
goalie's equipment, then the change in momentum (or the impulse)
imparted to the goalie is mv.
To calculate the average force experienced by the goalie, one must know
the time t that it took the goalie to decelerate the puck from its
initial speed to zero.
average force = impulse / time = mv/t
If the stopping time is t = 0.1 second, then the average force is
F = (0.170 kg)*(38 m/s)/(0.1 s) = 64.6 newtons = 14 pounds.
If the stopping time is ten times smaller (t = 0.01 s), then the average
force will be ten times larger (F = 140 pounds).
Now you need to do a little research to find out how long it typically
takes a goalie to stop a puck.
By the way, if the hard rubber puck rebounds elastically from, say, the
goalie's helmet then the momentum of the incoming puck will still be mv,
but now the final momentum will be -mv instead of zero. The impulse or
change in momentum will be 2mv imparted to the goalie, so the average
force exerted on the goalie will be twice as large as in the completely
inelastic collision described above.
--Randall J. Scalise http://www.phys.psu.edu/~scalise/
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