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kicyunya [14]
3 years ago
11

Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro

n Collider (LHC) at the CERN Laboratory in Geneva, Switzerland. The machine accelerates protons to high kinetic energies in an underground ring 27 km in circumference.
a. What speed v of proton in the LHC if the proton's kinetic energy is 6.7 TeV? (Because v is very close to c, write v=(1−Δ)c and give your answer in terms of Δ).
b. Find the relativistic mass, mrel, of the accelerated protons in terms of their rest mass.
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0

Solution :

Energy of photon, E = 6.7 eV

                              E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

   $=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r  = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

   = 0.00000000933

Relative mass, $m_{rel}=r.m$

                                $=7131 \times 1.6728 \times 10^{-27}$

                               $=1.1927 \times 10^{-23}$ kg

                                 

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Answer:

2.583 s, 29.77 ft and 1.219 s

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Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

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To Maximum height reached can be calculated with the formula

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The total time it will be in air can be calculated with the formula below

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0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

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