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Chapter 26: Problem 37

A constant force, acting for \(3.6 \times 10^{4} \mathrm{s}(10 \mathrm{h}),\) brings a spaceship of mass \(2200 \mathrm{kg}\) from rest to speed \(0.70 c\) (a) What is the magnitude of the force? (b) What is the initial acceleration of the spaceship? Comment on the magnitude of the answer.

Short Answer

Expert verified
Answer: The initial acceleration of the spaceship is approximately 0.0255c/s².

Step by step solution

01

Identify the given information and relevant equations

We are given the following information: - Total time taken: \(t = 3.6 \times 10^4 s\) - Mass of spaceship: \(m_0 = 2200 kg\) - Final speed: \(v = 0.70c\), where \(c\) is the speed of light We will use the following equations in this problem: 1. Relativistic equation for momentum: \(p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}\) 2. Force: \(F = \frac{dp}{dt}\) 3. Newton's second law: \(F = m_0 a\)
02

Calculate the final momentum of the spaceship

First, we'll calculate the final momentum of the spaceship using the relativistic equation: \(p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}\) Plug in given values: \(p = \frac{2200 \cdot 0.70c}{\sqrt{1-\frac{(0.70c)^2}{c^2}}}\) Solve for \(p\) (note that the c values will cancel out): \(p \approx 2.0186 \times 10^3 \ kg \times c\)
03

Calculate the change in momentum per unit of time (Force)

Now we can calculate the force by finding the change in momentum per unit of time: \(F = \frac{dp}{dt}\) Since the initial momentum is 0 (spaceship starts from rest), the change in momentum is just the final momentum: \(F = \frac{2.0186 \times 10^3 \ kg \times c}{3.6 \times 10^4 s}\) Solve for force: \(F \approx 56.072 \ kg \times c/s\) This is the magnitude of the force acting on the spaceship.
04

Calculate the initial acceleration of the spaceship

Finally, we can find the initial acceleration using Newton's second law: \(F = m_0 a\) Plug the given mass, and the calculated force: \(56.072 = 2200a\) Solve for acceleration \(a\): \(a \approx 0.0255 c/s^2\) The initial acceleration is approximately \(0.0255c/s^2\). This acceleration is very small compared to the speed of light, which means the spaceship's acceleration is relatively low in terms of significant relativistic effects.

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Most popular questions from this chapter

Derivation of the Doppler formula for light. A source and receiver of EM waves move relative to one another at velocity \(v .\) Let \(v\) be positive if the receiver and source are moving apart from one another. The source emits an EM wave at frequency \(f_{\mathrm{s}}\) (in the source frame). The time between wavefronts as measured by the source is \(T_{\mathrm{s}}=1 / f_{\mathrm{s}}\) (a) In the receiver's frame, how much time elapses between the emission of wavefronts by the source? Call this \(T_{\mathrm{T}}^{\prime} .\) (b) \(T_{\mathrm{T}}^{\prime}\) is not the time that the receiver measures between the arrival of successive wavefronts because the wavefronts travel different distances. Say that, according to the receiver, one wavefront is emitted at \(t=0\) and the next at \(t=T_{\mathrm{o}}^{\prime} .\) When the first wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}\) When the second wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}+v T_{\mathrm{r}}^{\prime} .\) Each wavefront travels at speed \(c .\) Calculate the time \(T_{\mathrm{r}}\) between the arrival of these two wavefronts as measured by the receiver. (c) The frequency detected by the receiver is \(f_{\mathrm{r}}=1 / T_{\mathrm{r}} .\) Show that \(f_{\mathrm{r}}\) is given by $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
Find the conversion between the momentum unit MeV/c and the SI unit of momentum.
A futuristic train moving in a straight line with a uniform speed of \(0.80 c\) passes a series of communications towers. The spacing between the towers, according to an observer on the ground, is \(3.0 \mathrm{km} .\) A passenger on the train uses an accurate stopwatch to see how often a tower passes him. (a) What is the time interval the passenger measures between the passing of one tower and the next? (b) What is the time interval an observer on the ground measures for the train to pass from one tower to the next?
A plane trip lasts \(8.0 \mathrm{h} ;\) the average speed of the plane during the flight relative to Earth is \(220 \mathrm{m} / \mathrm{s}\). What is the time difference between an atomic clock on board the plane and one on the ground, assuming they were synchronized before the flight? (Ignore general relativistic complications due to gravity and the acceleration of the plane.)
An extremely relativistic particle is one whose kinetic energy is much larger than its rest energy. Show that for an extremely relativistic particle $E \approx p c$.
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