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Chapter 37: Problem 39

Compute the kinetic energy of a proton (mass 1.67 \(\times\) 10\(^{-27}\) kg) using both the nonrelativistic and relativistic expressions, and compute the ratio of the two results (relativistic divided by nonrelativistic) for speeds of (a) 8.00 \(\times\) 107 m/s and (b) 2.85 \(\times\) 108 m/s.

Short Answer

Expert verified
Ratios are 1.04 and 2.12 for speeds 8.00脳10^7 m/s and 2.85脳10^8 m/s, respectively.

Step by step solution

01

Nonrelativistic Kinetic Energy Formula

The nonrelativistic kinetic energy (KE) is calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the particle.
02

Nonrelativistic Kinetic Energy for 8.00 脳 10^7 m/s

Substitute the given values into the formula: \( KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (8.00 \times 10^7)^2 \). Calculate to find \( KE \approx 5.34 \times 10^{-13} \) J.
03

Nonrelativistic Kinetic Energy for 2.85 脳 10^8 m/s

Similarly, substitute into the formula: \( KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (2.85 \times 10^8)^2 \). Calculate to find \( KE \approx 6.78 \times 10^{-12} \) J.
04

Relativistic Kinetic Energy Formula

Relativistic kinetic energy is calculated by \( KE_{rel} = ( rac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1)mc^2 \), where \( c \) is the speed of light, \( 3.00 \times 10^8 \) m/s.
05

Relativistic Kinetic Energy for 8.00 脳 10^7 m/s

Substitute the values into the relativistic formula: \( KE_{rel} = (\frac{1}{\sqrt{1 - \frac{(8.00 \times 10^7)^2}{(3.00 \times 10^8)^2}}} - 1) \times 1.67 \times 10^{-27} \times (3.00 \times 10^8)^2 \). Calculate \( KE_{rel} \approx 5.54 \times 10^{-13} \) J.
06

Relativistic Kinetic Energy for 2.85 脳 10^8 m/s

Substitute into the relativistic formula: \( KE_{rel} = (\frac{1}{\sqrt{1 - \frac{(2.85 \times 10^8)^2}{(3.00 \times 10^8)^2}}} - 1) \times 1.67 \times 10^{-27} \times (3.00 \times 10^8)^2 \). Calculate \( KE_{rel} \approx 1.44 \times 10^{-11} \) J.
07

Compute Ratio for 8.00 脳 10^7 m/s

The ratio of relativistic to nonrelativistic kinetic energy is \( \frac{KE_{rel}}{KE_{nonrel}} = \frac{5.54 \times 10^{-13}}{5.34 \times 10^{-13}} \approx 1.04 \).
08

Compute Ratio for 2.85 脳 10^8 m/s

The ratio for this speed \( \frac{KE_{rel}}{KE_{nonrel}} = \frac{1.44 \times 10^{-11}}{6.78 \times 10^{-12}} \approx 2.12 \).

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Key Concepts

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

Nonrelativistic Kinetic Energy
Kinetic energy helps us understand how objects in motion carry energy due to their speed. The nonrelativistic kinetic energy applies when speeds are significantly lower than the speed of light. The formula for this is straightforward: \( KE = \frac{1}{2}mv^2 \). In this formula:
  • \( m \) represents the mass of the object.
  • \( v \) stands for the velocity of the object.
  • \( KE \) is the kinetic energy.
For example, using the speed of 8.00 \( \times \) 10\(^7\) m/s and a mass of 1.67 \( \times \) 10\(^{-27}\) kg, we simply substitute the values into the equation to find the kinetic energy: \( KE \approx 5.34 \times 10^{-13} \) J. This method is based on the classical Newtonian physics trail, which works perfectly for speeds much less than the speed of light. But, keep in mind, as speeds increase, especially nearing the speed of light, this formula becomes less accurate.
Relativistic Kinetic Energy
When velocities approach the speed of light, the relativistic kinetic energy becomes important as it accounts for the effects of Einstein's theory of relativity. The equation is more complex but essential for high-velocity objects:\[ KE_{rel} = \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right)mc^2 \]In this equation:
  • \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \) m/s.
  • \( v \) is the speed of the object.
  • \( m \) is the mass.
For example, when the speed is 8.00 \( \times \) 10\(^7\) m/s, the relativistic kinetic energy computes to around \( 5.54 \times 10^{-13} \) J. Notice that this value is just slightly higher than the nonrelativistic value for the same speed. The relativistic formula accounts for the increase in mass a particle undergoes as it accelerates towards the speed of light, ensuring energy calculations remain accurate.
Energy Ratios
Understanding the difference between nonrelativistic and relativistic kinetic energy is crucial, particularly when comparing their results. The energy ratio is the comparison of these two forms of kinetic energy. It's calculated by dividing the relativistic kinetic energy by the nonrelativistic kinetic energy.Let's look at the example with a speed of 8.00 \( \times \) 10\(^7\) m/s:- Relativistic KE: \( 5.54 \times 10^{-13} \) J.- Nonrelativistic KE: \( 5.34 \times 10^{-13} \) J.Thus, the ratio \( \frac{KE_{rel}}{KE_{nonrel}} \approx 1.04 \). This ratio tells us the relativistic kinetic energy is slightly higher at this speed.When this speed increases to 2.85 \( \times \) 10\(^8\) m/s, the difference becomes more pronounced:- Relativistic KE: \( 1.44 \times 10^{-11} \) J.- Nonrelativistic KE: \( 6.78 \times 10^{-12} \) J.Here, the ratio is \( \approx 2.12 \), indicating a more significant difference due to the object's increased speed. These ratios highlight how essential relativistic physics becomes as speeds increase, showing that classical equations eventually fall short at accurately describing reality.

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Most popular questions from this chapter

Everyday Time Dilation. Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 h. By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (\(Hint\): Since \(u \ll c\), you can simplify \(\sqrt{1 - u^2/c^2}\) by a binomial expansion.)

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light in the laboratory reference frame, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 \(\mu\)s. They are produced when cosmic rays bombard the upper atmosphere about 10 km above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2 - \(\mu\)s lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the 2.2- \(\mu\)s lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999c, what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only 2.2 \(\mu\)s, so how does it make it to the ground? What is the thickness of the 10 km of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

Two protons (each with rest mass \(M\) = 1.67 \(\times\) 10\(^{-27}\) kg) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces an \(\eta_0\) particle (see Chapter 44). The rest mass of the \(\eta_0\) is m = 9.75 \(\times\) 10\(^{-28}\) kg. (a) If the two protons and the \(\eta_0\) are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in MeV. (c) What is the rest energy of the \(\eta_0\), expressed in MeV? (d) Discuss the relationship between the answers to parts (b) and (c). 37.39 .

An enemy spaceship is moving toward your starfighter with a speed, as measured in your frame, of 0.400c. The enemy ship fires a missile toward you at a speed of 0.700c relative to the enemy ship (Fig. E37.18). (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure that the enemy ship is 8.00 * 106 km away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?
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