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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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Most popular questions from this chapter

A spacecraft travels along a straight line from Earth to the Moon, a distance of \(3.84 \cdot 10^{8} \mathrm{~m}\). Its speed measured on Earth is \(0.50 c\). a) How long does the trip take, according to a clock on Earth? b) How long does the trip take, according to a clock on the spacecraft? c) Determine the distance between Earth and the Moon if it were measured by a person on the spacecraft.

An astronaut on a spaceship traveling at a speed of \(0.50 c\) is holding a meter stick parallel to the direction of motion. a) What is the length of the meter stick as measured by another astronaut on the spaceship? b) If an observer on Earth could observe the meter stick, what would be the length of the meter stick as measured by that observer?

Robert, standing at the rear end of a railroad car of length \(100 . \mathrm{m},\) shoots an arrow toward the front end of the car. He measures the velocity of the arrow as \(0.300 c\). Jenny, who was standing on the platform, saw all of this as the train passed her with a velocity of \(0.750 c .\) Determine the following as observed by Jenny: a) the length of the car b) the velocity of the arrow c) the time taken by arrow to cover the length of the car d) the distance covered by the arrow

In Jules Verne's classic Around the World in Eighty Days, Phileas Fogg travels around the world in, according to his calculation, 81 days. Due to crossing the International Date Line he actually made it only 80 days. How fast would he have to go in order to have time dilation make 80 days to seem like \(81 ?\) (Of course, at this speed, it would take a lot less than even 1 day to get around the world \(\ldots . .)\)

The Relativistic Heavy Ion Collider (RHIC) can produce colliding beams of gold nuclei with beam kinetic energy of \(A \cdot 100 .\) GeV each in the center-of- mass frame, where \(A\) is the number of nucleons in gold (197). You can approximate the mass energy of a nucleon as approximately \(1.00 \mathrm{GeV}\). What is the equivalent fixed-target beam energy in this case?
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