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Chapter 5: Problem 53

Find the momentum of a helium nucleus having a mass of \(6.68 \times 10^{-27} \mathrm{kg}\) that is moving at \(0.200 c\)

Short Answer

Expert verified
The momentum of the helium nucleus is approximately \(4.3 \times 10^{-19} kg \cdot m/s\).

Step by step solution

01

Write down the given values

We are given the mass of the helium nucleus (m) and its speed as a fraction of the speed of light (0.200 c). The mass of the helium nucleus is \(6.68 \times 10^{-27} kg\), and the speed is \(0.200 \times c\), where c is the speed of light (\(3 \times 10^8 m/s\)).
02

Calculate the actual speed in m/s

To find the actual speed of the helium nucleus, multiply the given fraction by the speed of light: \(v = 0.200 \times c = 0.200 \times (3 \times 10^8 m/s) = 6 \times 10^7 m/s\)
03

Apply the relativistic momentum formula

Now, we can use the relativistic momentum formula: \(p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}\) Plug in the given values and solve for p.
04

Plug in the values and solve for p

Using the mass and calculated speed, we can now solve for the momentum: \(p = \frac{(6.68 \times 10^{-27} kg)(6 \times 10^7 m/s)}{\sqrt{1 - \frac{(6 \times 10^7 m/s)^2}{(3 \times 10^8 m/s)^2}}}\) First, let's calculate the value inside the square root: \(\frac{(6 \times 10^7 m/s)^2}{(3 \times 10^8 m/s)^2} = \frac{36 \times 10^{14}m^2/s^2}{9 \times 10^{16}m^2/s^2} = 0.04\) Now, calculate the value inside the square root itself: \(\sqrt{1 - 0.04} = \sqrt{0.96}\) Finally, calculate the relativistic momentum, p: \(p = \frac{(6.68 \times 10^{-27} kg)(6 \times 10^7 m/s)}{\sqrt{0.96}} \approx 4.3 \times 10^{-19} kg \cdot m/s\)
05

State the final answer

The momentum of the helium nucleus is approximately \(4.3 \times 10^{-19} kg \cdot m/s\).

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Most popular questions from this chapter

(a) A nuclear power plant converts energy from nuclear fission into electricity with an efficiency of \(35.0 \%\) How much mass is destroyed in one year to produce a continuous 1000 MW of electric power? (b) Do you think it would be possible to observe this mass loss if the total mass of the fuel is \(10^{4} \mathrm{kg}\) ?

When observed from the sun at a particular instant, Earth and Mars appear to move in opposite directions with speeds \(108,000 \mathrm{km} / \mathrm{h}\) and \(86,871 \mathrm{km} / \mathrm{h}\), respectively. What is the speed of Mars at this instant when observed from Earth?

(a) How fast would an athlete need to be running for a \(100-\mathrm{m}\) race to look 100 yd long? (b) Is the answer consistent with the fact that relativistic effects are difficult to observe in ordinary circumstances? Explain.

Suppose an astronaut is moving relative to Earth at a significant fraction of the speed of light. (a) Does he observe the rate of his clocks to have slowed? (b) What change in the rate of earthbound clocks does he see? (c) Does his ship seem to him to shorten? (d) What about the distance between two stars that lie in the direction of his motion? (e) Do he and an earthbound observer agree on his velocity relative to Earth?

(a) Suppose the speed of light were only \(3000 \mathrm{m} / \mathrm{s}\). A jet fighter moving toward a target on the ground at 800 \(\mathrm{m} / \mathrm{s}\) shoots bullets, each having a muzzle velocity of 1000 \(\mathrm{m} / \mathrm{s} .\) What are the bullets' velocity relative to the target? (b) If the speed of light was this small, would you observe relativistic effects in everyday life? Discuss.
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