College Physics 2.32 – Acceleration of a woodpecker’s head


A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?


Solution:

Part A

The acceleration is computed based from the formula, \displaystyle \text{v}^2=\text{v}_0^2+2\text{a}\Delta \text{x}. If we solve for acceleration, we have

\displaystyle \text{a}=\frac{\text{v}^2-\text{v}_0^2}{2\Delta \text{x}}

\displaystyle \text{a}=\frac{\left(0\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(2.00\times 10^{-3}\right)\text{m}}

\text{a}=-90.0\:\text{m/s}^2

In terms of g.

\displaystyle \frac{\left|a\right|}{\left|g\right|}=\frac{90.0\:\text{m/s}^2}{9.80\:\text{m/s}^2}

\displaystyle \frac{\left|a\right|}{\left|g\right|}=9.18

Therefore,

a=9.18\text{g}

Part B

From the formula \Delta \text{x}=\frac{1}{2}\left(\text{v}_0+\text{v}\right)\text{t}, we have

\displaystyle \text{t}=\frac{2\Delta \text{x}}{\text{v}_0+\text{v}}

\displaystyle \text{t}=\frac{2\left(2.00\times 10^{-3}\:\text{m}\right)}{0.600\:\text{m/s}+0\:\text{m/s}}

\text{t}=6.67\times 10^{-3}\:\text{s}

Part C

\displaystyle a=\frac{v^2-v_0^2}{2\Delta x}

\displaystyle a=\frac{\left(0\:m/s\right)^2-\left(0.600\:m/s\right)^2}{2\left(4.50\times 10^{-3}\:m\right)}

\displaystyle a=-40.0\:m/s^2

In terms of g.

\displaystyle \frac{\left|a\right|}{\left|g\right|}=\frac{40.0\:m/s^2}{9.80\:m/s^2}

\displaystyle \frac{\left|a\right|}{\left|g\right|}=4.08

Therefore,

\displaystyle a=4.08g