How does a woodpecker survive the impact?

Question

The head of a woodpecker is moving forward at a speed of 7.49 m/s when the beak makes first contact witha tree limb. The beak stops after penetrating the limb by 1.87 mm. Assuming the acceleration to be constant, find the acceleration magnitued in terms of g.

Note: A rollercoaster is around 3 G's.

Answer 1

V²=2as+U²

V is final velocity = 0
U is initial velocity = 7.49
a is acceleration = a
s = 1.87 mm = 0.00187 m

(0)²=2a(0.00187)+(7.49)²
a=-15000.02674 m/s²


g = 9.8 m/s²
x = 15000.02674

acceleration = - 1530.614973g

Answer 2

this is a good use of the equation:
vf^2=v0^2+2as, where vf is final velocity (here, zero), v0 is initial velocity (here, 7.49 m/s), a is the accel to be determined and s is the distance over which the accel occurs, here (1.87x10^(-3)m)

plugging away:

0=56.1+3.74x10^-3 a=>
a=-1.5x10^4m/s/s or -1500g, a tad much for our human physiques

I read an article some time back in American Scientist (don't remember the reference or author) that talked about research done on Woody to see how he handles the impact. Turns out their brains are enmeshed in a type of cage with fluid that cushions the blow, so that the brain is not experiencing nearly this impact.

This research was used, according to the article, by companies that design and build helmets for American football.

Answer 3

By using this equation, V²=2as+U²

V (final velocity) = 0.00 m/s
U (initial velocity) = 7.49 m/s
s = 1.87 mm
= 1.87*10^(-2)m

Substituting,
(0.00)²=2a(1.87*10^(-2))+(7.49)²
a=15000.02674 m/s²
a=(15000.02674/g)*g m/s²
a=1529.577046g m/s²
Consider the significant figure, it should be
a=1.53*10^(3) .g m/s²

Answer 4

Its too late at night to crunch those kind of numbers, but as it happens, I asked the same question of an ornithology buff. Apparently the woodpecker has a spongy mass in the front of its skull in front of the brain, so it doesn't decelerate as fast as it appears. Besides, all the weaklings died of starvation centuries ago.