Problem:
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:
We are given the following values: v_0=0.600 \ \text{m/s}; v_f=0.000\:\text{m/s}; and \Delta x=0.002\:\text{m}.
Part A
The acceleration is computed based on the formula,
\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x
Solving for acceleration a in terms of the other variables, we have
a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}Substituting the given values,
\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\
a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}In terms of g taking absolute values of the acceleration , we have
\begin{align*}
a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\
a & = \frac{90.0}{9.80}g \\
a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
We shall use the formula
\Delta x=v_{ave}twhere v_{ave} is the average velocity computed as
v_{ave}=\frac{v_0+v_f}{2}Solving for time t in terms of the other variables, we have
t=\frac{2\Delta x}{v_0+v_f}Substituting the given values, we have
\begin{align*}
t & =\frac{2\Delta x}{v_0+v_f} \\
t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\
t & =6.67\times 10^{-3}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part C
Employing the same formula we used in Part A, we have
\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\
a & =-40.0\:\text{m/s}^2
\end{align*}In terms of g, taking absolute values of the acceleration
\begin{align*}
a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\
a & =\frac{40.0}{9.80}g \\
a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}College Physics Chapter 2 Problems
College Physics 2nd Edition Solutions Table of Contents
Chapter 1: Introduction: The Nature of Science and Physics
Chapter 3: Two-Dimensional Kinematics
Chapter 4: Dynamics: Force and Newton’s Law of Motion
Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity
Chapter 6: Uniform Circular Motion and Gravitation
Chapter 7: Work, Energy, and Energy Resources
Chapter 8: Linear Momentum and Collisions
Chapter 10: Rotational Motion and Angular Momentum
Chapter 12: Fluid Dynamics and Its Biological and Medical Applications
Chapter 13: Temperature, Kinetic Theory, and the Gas Laws
Chapter 14: Heat and Heat Transfer Methods
Chapter 15: Thermodynamics
Chapter 16: Oscillatory Motion and Waves
Chapter 17: Physics of Hearing
Chapter 18: Electric Charge and Electric Field
Chapter 19: Electric Potential and Electric Field
Chapter 20:
Electric Current, Resistance, and Ohm’s Law
Chapter 21: Circuits and DC Instruments
Chapter 22: Magnetism
Chapter 23: Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 24: Electromagnetic Waves
Chapter 25: Geometric Optics
Chapter 26: Vision and Optical Instrument
Chapter 27: Wave Optics
Chapter 28: Special Relativity
Chapter 29: Introduction to Quantum Physics
Chapter 30: Atomic Physics
Chapter 31: Radioactivity and Nuclear Physics
Chapter 32: Medical Applications of Nuclear Physics
Chapter 33: Particle Physics
Chapter 34: Frontiers of Physics
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