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Chapter 35: Problem 6

A proton with a momentum of \(3.0 \mathrm{GeV} / \mathrm{c}\) is moving with what velocity relative to the observer? a) \(0.31 c\) c) \(0.91 c\) e) \(3.2 c\) b) \(0.33 c\) d) \(0.95 c\)

Short Answer

Expert verified
Answer: The velocity of the proton relative to the observer is approximately 0.955c, which is closest to choice (d) 0.95c.

Step by step solution

01

Recall the relativistic momentum formula

The relativistic momentum formula is given by: \(p = \frac{m_0 v}{\sqrt{1 -\frac{v^2}{c^2}}}\), where \(p\) is the momentum, \(m_0\) is the rest mass of the particle, \(v\) is the velocity, and \(c\) is the speed of light. In this exercise, the value \(p = 3.0 \,\text{GeV}/\text{c}\) is given.
02

Substitute the known values

The mass of a proton is \(m_0 = 1.67 \times 10^{-27} \,\text{kg}\), and \(p = 3.0 \,\text{GeV}/\text{c}\) (which is \(3.0 \times 10^9 \,\text{eV}/\text{c}\)). We can now substitute these values in the relativistic momentum formula: \(\frac{3.0 \times 10^9 \,\text{eV}/\text{c}}{c} = \frac{1.67 \times 10^{-27} \,\text{kg} \cdot v}{\sqrt{1 -\frac{v^2}{c^2}}}\).
03

Convert mass-energy units

To properly solve the equation, let's convert the momentum from eV to kg.m/s. Using the conversion factor \(1 \,\text{eV}/\text{c}^2 = 1.783\times 10^{-36} \,\text{kg}\), we can write the momentum as: \(p = \frac{3.0\times 10^9 \,\text{eV}}{c} \times 1.783\times 10^{-36} \,\text{kg}/(\text{eV}/\text{c}^2) = 5.349\times 10^{-27} \,\text{kg.m/s}\).
04

Solve for the velocity

Now, we will input the momentum using SI units in the equation we derived in Step 2: \(\frac{5.349 \times 10^{-27} \,\text{kg.m/s}}{c} = \frac{1.67 \times 10^{-27} \,\text{kg} \cdot v}{\sqrt{1 -\frac{v^2}{c^2}}}\). The next step is to isolate \(v\) and solve for it using algebra. Multiplying both sides by \(c\) and the denominator, we get: \(5.349 \times 10^{-27} \,\text{kg.m/s} = 1.67 \times 10^{-27} \,\text{kg} \cdot v\sqrt{1-\frac{v^2}{c^2}}\). To simplify, let \(x = \frac{v}{c}\), which gives us: \(5.349 \times 10^{-27} \text{kg.m/s} = 1.67 \times 10^{-27} \,\text{kg} \cdot xc\sqrt{1-x^2}\). Solving for \(x\) (using numerical methods or online calculator) gives \(x \approx 0.955\).
05

Find the velocity

Now, we can find the velocity by multiplying \(x\) by the speed of light \(c\): \(v = x\cdot c = 0.955c\). Therefore, the velocity of the proton relative to the observer is approximately \(0.955c\). The correct answer is closest to choice (d) \(0.95c\).

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