/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 The energy to accelerate a stars... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 44

The energy to accelerate a starship comes from combining matter and antimatter. When this is done the total rest energy of the matter and antimatter is converted to other forms of energy. Suppose a starship with a mass of \(2.0 \times 10^{5} \mathrm{kg}\) accelerates to \(0.3500 c\) from rest. How much matter and antimatter must be converted to kinetic energy for this to occur?

Short Answer

Expert verified
Answer: To calculate the amount of matter and antimatter required, follow these steps: 1. Calculate the Lorentz factor, 纬, using the formula: 纬 = 1 / sqrt(1 - (v^2 / c^2)). For a speed of 0.3500c, 纬 is approximately 1.1843. 2. Calculate the final kinetic energy, K, using the formula: K = (纬 - 1) * mc^2. With the mass of the starship given as 2.0 x 10^5 kg, K is approximately 4.667 x 10^18 Joules. 3. Find the amount of matter and antimatter, 螖m, using the equation: 螖m = K / c^2. The required amount of matter and antimatter is approximately 5.185 x 10^10 kg.

Step by step solution

01

Find the final kinetic energy of the starship

First, we need to calculate the final kinetic energy of the starship using the relativistic kinetic energy formula, which is given by: \(K = ( \gamma - 1)mc^2\) where \(K\) is the kinetic energy, \(\gamma\) is the Lorentz factor, \(m\) is the mass of the object, and \(c\) is the speed of light. The Lorentz factor is given by: \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) Given that \(m = 2.0 \times 10^5 \mathrm{kg},\ v = 0.3500c\). We will first calculate the Lorentz factor and then find the kinetic energy.
02

Calculate the Lorentz factor

We have the formula for the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) Substituting the values, we get: \(\gamma = \frac{1}{\sqrt{1-(0.3500)^2}}\) Solve for \(\gamma\).
03

Calculate the kinetic energy

Now that we have the Lorentz factor, we can find the kinetic energy using the relativistic kinetic energy formula: \(K = ( \gamma - 1)mc^2\) Substitute the values: \(K = (\gamma - 1)(2.0 \times 10^5 \mathrm{kg})(3.0 \times 10^8 \mathrm{m/s})^2\) Solve for \(K\).
04

Find the amount of matter and antimatter

The total amount of matter and antimatter converted is equal to the total kinetic energy of the starship. We can find this using the rest energy equation: \(E = mc^2\) We already know the value of \(K\) and \(c\). Let's implement the equation like this: \(K = (\Delta m) c^2\) Now we need to solve for \(\Delta m\): \(\Delta m = \frac{K}{c^2}\) Substitute the values of \(K\) and \(c\) to find the required amount of matter and antimatter.

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Most popular questions from this chapter

Refer to Example \(26.2 .\) One million muons are moving toward the ground at speed \(0.9950 c\) from an altitude of \(4500 \mathrm{m} .\) In the frame of reference of an observer on the ground, what are (a) the distance traveled by the muons; (b) the time of flight of the muons; (c) the time interval during which half of the muons decay; and (d) the number of muons that survive to reach sea level. [Hint: The answers to (a) to (c) are not the same as the corresponding quantities in the muons' reference frame. Is the answer to (d) the same?]
An astronaut in a rocket moving at \(0.50 c\) toward the Sun finds himself halfway between Earth and the Sun. According to the astronaut, how far is he from Earth? In the frame of the Sun, the distance from Earth to the Sun is \(1.50 \times 10^{11} \mathrm{m}\).
Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.
A rectangular plate of glass, measured at rest, has sides \(30.0 \mathrm{cm}\) and \(60.0 \mathrm{cm} .\) (a) As measured in a reference frame moving parallel to the 60.0 -cm edge at speed \(0.25 c\) with respect to the glass, what are the lengths of the sides? (b) How fast would a reference frame have to move in the same direction so that the plate of glass viewed in that frame is square?
An observer in the laboratory finds that an electron's total energy is $5.0 \mathrm{mc}^{2} .$ What is the magnitude of the electron's momentum (as a multiple of \(m c\) ), as observed in the laboratory?
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