/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 Two balls of equal masses ( \(m\... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 67

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鈥檚 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

Like time dilation, length contraction cannot be seen directly by a single observer. To explain this claim, imagine a rod of proper length \(l_{\mathrm{o}}\) moving along the \(x\) axis of frame \(\mathcal{S}\) and an observer standing away from the \(x\) axis and to the right of the whole rod. Careful measurements of the rod's length at any one instant in frame \(\mathcal{S}\) would, of course, give the result \(l=l_{\mathrm{o}} / \gamma\). (a) Explain clearly why the light which reaches the observer's eye at any one time must have left the two ends \(A\) and \(B\) of the rod at different times. (b) Show that the observer would see (and a camera would record) a length more than \(l\). [It helps to imagine that the \(x\) axis is marked with a graduated scale.] ( \(\mathbf{c}\) ) Show that if the observer is standing close beside the track, he will see a length that is actually more than \(l_{\mathrm{o}}\); that is, the length contraction is distorted into an expansion.

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.

Prove that for any two matrices \(A\) and \(B\), where \(A\) has as many columns as \(B\) has rows, the transpose of \(A B\) satisfies \((A B)=\tilde{B} \tilde{A}\)

Consider two events that occur simultaneously at \(t=0\) in frame \(\mathcal{S},\) both on the \(x\) axis at \(x=0\) and \(x=a\). (a) Find the times of the two events as measured in a frame \(\mathcal{S}^{\prime}\) traveling in the positive direction along the \(x\) axis with speed \(V\). (b) Do the same for a second frame \(\mathcal{S}^{\prime \prime}\) traveling at speed \(V\) but in the negative direction along the \(x\) axis. Comment on the time ordering of the two events as seen in the three different frames. This startling result is discussed further in Section 15.10 .

A particle of mass 3 MeV/c \(^{2}\) has momentum 4 MeV/c. What are its energy (in MeV) and speed (in units of \(c\) )?
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