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In the collision shown, energy is conserved. because both objects have the same speed and mass after as before the collision. Since the collision merely reverses the velocities, the final (total) momentum is opposite the initial. Thus, momentum can be conserved ooly if it is zero. (a) Using the relativistically correct expression for momentum, show that the total momentum is zero that momentum is conserved. (Masses are in arbitrary units.) (b) Using the relativistic velocity transformation, find the four velocities in a frame moving to the right at \(0.6 c\) (c) Verify that momentum is conserved in che new frame.

Short Answer

Expert verified
By using the relativistically correct expression for momentum, it is shown that the total momentum is indeed zero, hence conserved, before and after the collision. Then, using the relativistic velocity transformation formula, the velocities of the colliding objects in a frame moving to the right at \(0.6 c\) are calculated. Lastly, it is verified that momentum is conserved in this new frame by showing that the initial and final total momenta are zero.

Step by step solution

01

Expression for Relativistic Momentum

The relativistically correct expression for momentum is given by \(p = \frac{mv}{\sqrt{1 - (v/c)^2}}\), where \(m\) is the mass, \(v\) is the velocity, and \(c\) is the speed of light. For the two objects, the initial momentum is \(p_i = m(v_1-v_2)/\sqrt{1 - (v/c)^2}\). After the collision, the velocities are reversed, so the final momentum is \(p_f = m(-v_1+v_2)/\sqrt{1 - (v/c)^2} = -p_i\). Since \(p_i + p_f = 0\), the total momentum is zero and momentum is conserved.
02

Relativistic Velocity Transformation

The relativistic velocity transformation formula is given by \(v' = (v - u) / (1 - uv/c^2)\), where \(v'\) is the velocity in the new frame, \(v\) is the original velocity, \(u\) is the velocity of the new frame relative to the original frame, and \(c\) is the speed of light. Using the given value of \(u = 0.6c\), the four velocities in the new frame are given by applying this formula to each of \(v_1\), \(-v_1\), \(v_2\), and \(-v_2\). This provides the new velocities \(v'_1\), \(-v'_1\), \(v'_2\), and \(-v'_2\).
03

Verification of Momentum Conservation in New Frame

With the transformed velocities from Step 2, the momentum in the new frame can be calculated using the formula for relativistic momentum. The initial momentum in the new frame is \(p'_i = m(v'_1-v'_2)/\sqrt{1 - (v'_i/c)^2}\) and the final momentum in the new frame is \(p'_f = m(-v'_1+v'_2)/\sqrt{1 - (v'_f/c)^2}\). If \(p'_i + p'_f = 0\), then momentum is conserved in the new frame, proving the initial claim.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum Conservation
In physics, one of the most fundamental concepts is the conservation of momentum. This principle states that within a closed system, the total momentum before and after an event must remain constant, provided no external forces are acting upon it. In simpler terms, momentum is a property of moving objects that remains unchanged unless influenced by an outside force or agent. In the realm of relativistic physics, momentum conservation is applied using the relativistically correct expression for momentum, which accounts for the effects of relativity when velocities approach the speed of light. In this scenario, momentum is determined using \[p = \frac{mv}{\sqrt{1 - (v/c)^2}},\]where:
  • \(p\) is momentum,
  • \(m\) is mass,
  • \(v\) is velocity,
  • \(c\) is the speed of light.
This equation shows how increasing velocity significantly impacts momentum due to the relativistic factor \(\frac{1}{\sqrt{1 - (v/c)^2}}\). Even if mass and speed appear similar pre- and post-collision, relativistic corrections are crucial for accurate calculations in physics.
Relativistic Velocity Transformation
When studying relativistic physics, understanding how velocities transform from one reference frame to another is essential. This is where the relativistic velocity transformation formula comes into play:\[v' = \frac{v - u}{1 - uv/c^2},\]where:
  • \(v'\) is the velocity in the new reference frame,
  • \(v\) is the velocity in the original frame,
  • \(u\) is the relative velocity of the moving frame,
  • \(c\) is the speed of light.
This equation ensures that the sum of the velocities doesn't exceed the speed of light, as predicted by Einstein's theory of relativity. For example, if a frame is moving to the right at \(0.6c\), this formula allows us to calculate velocities in this new frame accurately. By applying this transformation to each velocity, we can understand how relativistic effects alter our observations of speed within varying reference frames.
Collision in Physics
Collisions are events where two or more objects impact each other, exchanging momentum and energy. Understanding collisions is crucial in both classical and relativistic physics because it helps illustrate fundamental conservation laws, like momentum conservation and energy conservation. There are various types of collisions:
  • Elastic, where both momentum and kinetic energy are conserved.
  • Inelastic, where momentum is conserved but kinetic energy is not.
  • Perfectly inelastic, where the colliding bodies stick together.
In this specific relativistic collision scenario, we observe that by reversing velocities post-collision, the system maintains conservation of momentum. This example perfectly demonstrates that, despite the complexities introduced by high velocities, basic physics principles still hold true when appropriately adjusted for relativistic effects.
Relativistic Physics
Relativistic physics extends classical physics to account for the phenomena observed when objects are moving at speeds close to the speed of light. Special relativity, formulated by Albert Einstein, revolutionized our understanding of space, time, and motion.A cornerstone of relativistic physics is that the laws of physics are the same for all observers, regardless of their relative motion. This leads to surprising results when dealing with high velocities, such as:
  • Time dilation: moving clocks run slower relative to a stationary observer.
  • Length contraction: moving objects are measured to be shorter in the direction of motion.
  • Mass-energy equivalence: summarized by the famous equation \(E=mc^2\).
Relativistic corrections are vital in accurately predicting the outcomes of events involving high-speed particles, which are essential in modern technologies like particle accelerators and GPS systems. These principles ensure our calculations remain precise even as our understanding of the universe expands.

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Most popular questions from this chapter

Prove that if \(v\) and \(u^{\prime}\) are less than \(c\), it is impossible for a speed \(u\) greater than \(c\) to result from equation \((2-19 b)\). [Hint: The product (c \(\left.-u^{\prime}\right)(c-v)\) is positive.]

You are in a bus traveling on a straight road ar \(20 \mathrm{~m} / \mathrm{s}\). As you pass a gas station, your clock and a clock in the station read precisely 0. You pass another gas station \(900 \mathrm{~m}\) farther down the road. ( In the frame of reference of the gas stations. all gas station clocks are synchronized.) (a) As you pass the second station. do you find its clock to be ahead of, or behind your own clock, and (b) by how much?

The Lorentz transformation equations have \(x\) and \(t\) and \(x^{\prime}\) and \(t^{\prime}\). Why no \(v\) and \(v^{\prime} ?\)

You are gliding over Earth's surface at a high speed. carrying your high- precision clock. At points \(X\) and \(Y\) on the ground are similar clocks, synchronized in the ground frame of reference. As you pass over clock X, it and your clock both read 0 . (a) According to you, do clocks \(X\) and \(Y\) advance slower or faster than yours? (b) When you pass over clock \(Y\), does it read the same time, an earlier time or a later time than yours? (Make sure your answer aguces with what ground observers should see.) (c) Reconcile any seeming contradictions berween your answers to parts \((a)\) and \((b)\).

Appearing in the time-dilation and length-contraction formulas, \(\gamma_{y}\) is a reasonable measure of the size of relativistic effects. Roughly speaking, at what speed would observations deviate from classical expectations by \(1 \% ?\)

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