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Two balls of equal masses ( \(m\) each) approach one another head-on with equal but opposite velocities of magnitude 0.8c. Their collision is perfectly inelastic, so they stick together and form a single body of mass \(M\). What is the velocity of the final body and what is its mass \(M ?\)

Short Answer

Expert verified
Final velocity is 0 and mass is \(3.3333m\).

Step by step solution

01

Define the System

We have two balls, each of mass \(m\), with velocities \(v_1 = 0.8c\) and \(v_2 = -0.8c\), colliding perfectly inelastically. Perfectly inelastic collisions imply that the objects stick together, so we must conserve momentum and consider mass-energy equivalence.
02

Use the Relativistic Momentum Formula

In a reference frame moving with constant velocity, the relativistic momentum of an object is given by \( p = \frac{mv}{\sqrt{1 -\frac{v^2}{c^2}}} \). The total initial momentum for the system is \( p_1 + p_2 = \frac{m(0.8c)}{\sqrt{1 - (0.8)^2}} + \frac{m(-0.8c)}{\sqrt{1 - (0.8)^2}} = 0 \). Thus the total momentum is zero.
03

Consider the Conservation of Momentum

Since the total momentum before the collision is zero, the total momentum of the system after the collision must also be zero. Hence the resulting body must be at rest, meaning the velocity \( V = 0 \).
04

Use the Energy Conservation Principle

According to the energy conservation principle and considering relativistic effects, the total initial energy includes the rest mass energy and kinetic energy. Calculate the rest mass energy: \( E_{rest} = 2mc^2 \). The total initial energy is \( 2mc^2 + \text{kinetic energy terms} \).
05

Calculate the Mass of M

Since the velocity of the resulting body is zero, all the kinetic energy converts into the mass of the body. The initial kinetic energy can be calculated as \( 2 \times \gamma mc^2 \), with \( \gamma = \frac{1}{\sqrt{1 - (0.8)^2}} = 1.6667 \). This gives an equivalent mass \( M = 2\gamma m = 3.3333 m \).

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Key Concepts

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

Conservation of Momentum
In physics, the conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act upon it. Momentum, the quantity of motion an object possesses, is the product of its mass and velocity. In the context of the original exercise, two balls collide head-on with equal yet opposite velocities.
  • The formula for momentum in this relativistic context becomes adjusted to incorporate the effects of high velocity, using the equation for relativistic momentum: \[ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \]
  • Since the balls approach each other with equal speeds but in opposite directions, their initial momenta sum to zero.
  • Consequently, the momentum after collision remains zero, illustrating the conservation principle.
This principle is essential in solving the problem, as it determines that the combined mass of the two balls, post-collision, comes to rest because their momenta cancel out. Understanding this can greatly simplify the analysis of collisions where external forces are negligible.
Inelastic Collision
An inelastic collision is one in which the colliding bodies stick together and kinetic energy is not conserved, although momentum is. In the given exercise, the balls undergo a perfectly inelastic collision. This means:
  • After collision, the objects combine to become a single body.
  • Kinetic energy is partly transferred into other forms of energy, such as heat or internal energy.
  • While kinetic energy is not conserved, momentum is conserved.
In this setup, two balls with velocities of magnitude \(0.8c\) collide and merge into one body. Their kinetic energy before collision converts into other energy forms leading to an increase in mass, as well as being vital to solving the discussed problem. **Perfectly inelastic** implies maximal energy loss to move to internal states. This new mass body, despite having all the previous system's energy, shows zero kinetic energy since it remains at rest.
Mass-Energy Equivalence
Mass-energy equivalence is one of the fundamental principles of Einstein’s Theory of Relativity, expressed through the famous equation \(E = mc^2\). According to this principle, mass can be converted into energy and vice versa, which is a cornerstone of understanding relativistic collisions.
  • The total energy of an object includes both its rest mass energy and its kinetic energy, especially notable at high speeds approaching the speed of light, \(c\).
  • For the given exercise, this equivalence implies that any kinetic energy lost in the collision manifests as an increase in mass.
  • The relativistic kinetic energy mus transform into mass, leaving us with a new mass greater than the sum of the original bodies' rest masses.
This equivalence is the reason the mass \( M \) after the collision is calculated as \( M = 2\gamma m = 3.3333 m \), with \( \gamma \) reflecting the Lorentz factor due to relativistic speeds. These conversions between energy forms highlight the profound interrelation between mass and energy in relativistic physics.

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Most popular questions from this chapter

The pion \(\left(\pi^{+} \text {or } \pi^{-}\right)\) is an unstable particle that decays with a proper half-life of \(1.8 \times 10^{-8}\) s. (This is the half-life measured in the pion's rest frame.) (a) What is the pion's half-life measured in a frame \(\mathcal{S}\) where it is traveling at \(0.8 c ?\) (b). If 32,000 pions are created at the same place, all traveling at this same speed, how many will remain after they have traveled down an evacuated pipe of length \(d=36 \mathrm{m} ?\) Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive. (c) What would the answer have been if you had ignored time dilation? (Naturally it is the answer (b) that agrees with experiment.)

Consider two events that occur at positions \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) and times \(t_{1}\) and \(t_{2} .\) Let \(\Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}\) and \(\Delta t=t_{2}-t_{1} .\) Write down the Lorentz transformation for \(\mathbf{r}_{1}\) and \(t_{1}\), and likewise for \(\mathbf{r}_{2}\) and \(t_{2},\) and deduce the transformation for \(\Delta \mathbf{r}\) and \(\Delta t\). Notice that differences \(\Delta \mathbf{r}\) and \(\Delta t\) transform in exactly the same way as \(\mathbf{r}\) and \(t\). This important property follows from the linearity of the Lorentz transformation.

As a meter stick rushes past me (with velocity v parallel to the stick), I measure its length to be \(80 \mathrm{cm} .\) What is \(v ?\)

The muons created by cosmic rays in the upper atmosphere rain down more-or- less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about \(1.5 \mu\) s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of \(2000 \mathrm{m}\), and in the course of an hour detects 650 muons traveling at \(0.99 c\) toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive.) Needless to say, the relativistic answer agrees with experiment.

Since the four-velocity \(u=\gamma(\mathbf{v}, c)\) is a four-vector its transformation properties are simple. Write down the standard Lorentz boost for all four components of \(u\). Use these to deduce the relativistic velocity- addition formula for v.

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