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What is the percent difference between the classical kinetic energy, \(K_{\mathrm{cl}}=\frac{1}{2} m_{0} v^{2},\) and the correct relativistic kinetic energy, \(K=m_{0} c^{2} / \sqrt{1-v^{2} / c^{2}}-m_{0} c^{2},\) at a speed of (a) \(0.10 c\) and (b) \(0.90 c ?\)

Short Answer

Expert verified
At 0.10c, the percent difference is about 0.5%; at 0.90c, it's approximately 219.3%.

Step by step solution

01

Define the Percent Difference Formula

To find the percent difference between two values, use the formula \( \text{Percent Difference} = \left( \frac{K_{\text{relativistic}} - K_{\text{classical}}}{K_{\text{classical}}} \right) \times 100\% \). This tells us how much the values differ relative to the classical value.
02

Calculate Classical Kinetic Energy at 0.10c

The speed given is \( v = 0.10c \). Substitute into the formula for classical kinetic energy, \( K_{\mathrm{cl}} = \frac{1}{2} m_{0} v^{2} \). Thus, \( K_{\mathrm{cl}} = \frac{1}{2} m_{0} (0.10c)^{2} = \frac{1}{2} m_{0} \times 0.01c^{2} = 0.005m_{0}c^{2} \).
03

Calculate Relativistic Kinetic Energy at 0.10c

Using the relativistic formula, \( K = \frac{m_{0} c^{2}}{\sqrt{1-(0.10c)^{2}/c^{2}}} - m_{0} c^{2} \). This simplifies to \( K = \frac{m_{0} c^{2}}{\sqrt{1-0.01}} - m_{0} c^{2} = \frac{m_{0} c^{2}}{\sqrt{0.99}} - m_{0} c^{2} \). Calculate this to be approximately \( K \approx 0.005025 m_{0} c^{2} \).
04

Compute Percent Difference at 0.10c

Substitute the values obtained into the percent difference formula: \( \text{Percent Difference} = \left( \frac{0.005025m_{0}c^{2} - 0.005m_{0}c^{2}}{0.005m_{0}c^{2}} \right) \times 100\% \), resulting in approximately \( 0.5\% \).
05

Calculate Classical Kinetic Energy at 0.90c

Here, \( v = 0.90c \). Substitute into the classical kinetic energy formula: \( K_{\mathrm{cl}} = \frac{1}{2} m_{0} (0.90c)^{2} = \frac{1}{2} m_{0} \times 0.81c^{2} = 0.405m_{0}c^{2} \).
06

Calculate Relativistic Kinetic Energy at 0.90c

Using the relativistic kinetic energy formula at \( v=0.90c \), we have \( K = \frac{m_{0} c^{2}}{\sqrt{1-(0.90c)^{2}/c^{2}}} - m_{0} c^{2} = \frac{m_{0} c^{2}}{\sqrt{0.19}} - m_{0} c^{2} \). Calculate this to be approximately \( K \approx 1.2942 m_{0} c^{2} \).
07

Compute Percent Difference at 0.90c

Substitute these values into the percent difference formula: \( \text{Percent Difference} = \left( \frac{1.2942m_{0}c^{2} - 0.405m_{0}c^{2}}{0.405m_{0}c^{2}} \right) \times 100\% \), which calculates to approximately \( 219.3\% \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Percent Difference
Understanding percent difference is key to comparing two quantities, such as classical and relativistic kinetic energy. It measures how much the values deviate from each other relative to a reference point, usually the classical value in physics. Define percent difference using:
  • Formula: \( \text{Percent Difference} = \left( \frac{K_{\text{relativistic}} - K_{\text{classical}}}{K_{\text{classical}}} \right) \times 100\% \)
This formula expresses the difference as a percentage of the classical kinetic energy.
This approach is crucial, especially when dealing with relativistic effects, which can cause significant deviations at high speeds. When the speed reaches a large fraction of the speed of light \((c)\), the percent difference becomes more prominent, as seen when evaluating at \(0.90c\).
Using percent difference helps highlight the limitations of classical mechanics and the necessity of considering relativistic effects found in Einstein's theory of Special Relativity.
Classical Kinetic Energy
Classical Kinetic Energy has been a fundamental concept in physics for centuries, represented by the formula:
  • \( K_{\text{cl}} = \frac{1}{2} m_{0} v^{2} \)
This elementary formula calculates the energy based on mass \(m_{0}\) and velocity \(v\). It presumes that mass and velocity do not change with speed, and is suitable for everyday speeds.
At lower speeds, such as \(0.10c\), classical kinetic energy provides a good approximation. However, as objects move faster, nearing a significant fraction of \(c\), this model becomes inadequate due to relativistic effects.
John's bicycle, for example, doesn't need relativistic considerations. But for particles in accelerators or speeds approaching that of light, Einstein's relativistic model is more accurate, where energy significantly differs from classical predictions.
Special Relativity
Special Relativity, introduced by Albert Einstein, reforms our understanding of high-velocity physics. It accounts for how time and space are interconnected, affecting kinetic energy calculations substantially.
In Einstein's framework, the relativistic kinetic energy for a moving object is given by:
  • \( K = \frac{m_{0} c^{2}}{\sqrt{1-v^{2} / c^{2}}} - m_{0} c^{2} \)
This formula accounts for mass increases at higher velocities, approaching the speed of light \(c\). As objects reach relativistic speeds, such as \(0.90c\), the relativistic kinetic energy grows much larger than its classical counterpart.
Understanding this concept reveals why at speeds close to \(c\), the computations yield a high percent difference, underscoring the inadequacies of classical perspectives. Thus, Special Relativity not only amends classical equations but forms the backbone for modern physics, amplifying the nuances of energy at high velocities.

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