[ans: 0.25 kg] 2. A proton moves with a speed of 0 .800c through a particle accelerator In the accelerator's frame of reference,calculate (a) the total and (b) the kinetic energies of the proton in megaelectron-volts TA A [ans: (a) 1.57times 10^3 MeV; (b) 6.26times 10^2MeV]

Here's the solution:<br /><br />**Understanding the Problem**<br /><br />We're dealing with relativistic energies here because the proton's speed is a significant fraction of the speed of light (c). We'll need to use the relativistic formulas for total energy and kinetic energy.<br /><br />**Given:**<br /><br />* v = 0.800c (proton's speed)<br />* m₀ (rest mass of a proton) = 938.3 MeV/c² (This is a standard value you should know or have access to)<br /><br />**Formulas:**<br /><br />* **Total Relativistic Energy (E):** E = γm₀c²<br />* **Relativistic Kinetic Energy (K):** K = (γ - 1)m₀c²<br />* **Lorentz Factor (γ):** γ = 1 / sqrt(1 - (v²/c²))<br /><br />**Calculations:**<br /><br />**(a) Total Energy (E):**<br /><br />1. **Calculate γ:**<br /> γ = 1 / sqrt(1 - (0.800c)²/c²)<br /> γ = 1 / sqrt(1 - 0.64)<br /> γ = 1 / sqrt(0.36)<br /> γ = 1 / 0.6<br /> γ = 1.667 (approximately)<br /><br />2. **Calculate E:**<br /> E = γm₀c²<br /> E = (1.667) * (938.3 MeV/c²) * c²<br /> E = 1564.5 MeV ≈ 1.57 x 10³ MeV<br /><br />**(b) Kinetic Energy (K):**<br /><br />1. **Use the calculated γ:**<br /> K = (γ - 1)m₀c²<br /> K = (1.667 - 1) * (938.3 MeV/c²) * c²<br /> K = (0.667) * 938.3 MeV<br /> K = 625.9 MeV ≈ 6.26 x 10² MeV<br /><br /><br />**Final Answers:**<br /><br />* (a) Total Energy: 1.57 x 10³ MeV<br />* (b) Kinetic Energy: 6.26 x 10² MeV<br /><br />Therefore, the provided answers in the original question are correct.<br />