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A particle with mass \(m\) has speed \(c / 2\) relative to inertial frame S. The particle collides with an identical particle at rest relative to frame \(S .\) Relative to \(S,\) what is the speed of a frame \(S^{\prime}\) in which the total momentum of these particles is zero? This frame is called the center of momentum frame.

Short Answer

Expert verified
The speed of frame \( S' \) is \( \frac{c}{4} \).

Step by step solution

01

Identify Initial Conditions

We have two particles with identical mass \( m \). The first particle, \( P_1 \), has an initial velocity \( v_1 = \frac{c}{2} \) and the second particle, \( P_2 \), is at rest, so \( v_2 = 0 \), relative to frame \( S \).
02

Understand Total Momentum in Frame S

In frame \( S \), the total momentum of the system is given by the sum of the momenta of both particles. So, the total momentum is \( p_{total} = m v_1 + m v_2 = m \frac{c}{2} + 0 = \frac{mc}{2} \).
03

Define Conditions for the Center of Momentum Frame

In the center of momentum frame \( S' \), the total momentum should be zero. This means that the velocity of the center of momentum frame, \( V_{cm} \), relative to frame \( S \), should offset the momentum of the two particles.
04

Calculate the Velocity of the Center of Momentum Frame

The velocity \( V_{cm} \) should satisfy \( m(v_1 - V_{cm}) + m(v_2 - V_{cm}) = 0 \). Simplifying, \( m \left( \frac{c}{2} - V_{cm} \right) + m(0 - V_{cm}) = 0 \). Factoring out and solving gives\( m \left( \frac{c}{2} - 2V_{cm} \right) = 0 \). Thus, \( V_{cm} = \frac{c}{4} \).

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Key Concepts

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

Momentum Conservation
Momentum conservation is a fundamental principle in physics. It tells us that the total momentum of a closed system remains constant if no external forces act on it. Momentum is the product of an object's mass and velocity. In mathematical terms, it is expressed as: \[ p = mv \] where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
In the exercise, two identical particles are involved. At first glance, it might seem unimportant, but because they have the same mass, calculations become simpler.
Initially, the total momentum of the system in frame \( S \) is calculated by adding the momentum of each particle: \[ p_{total} = mv_1 + mv_2 \]
When dealing with isolated systems, physicists often look for a frame where the total momentum becomes zero. This special frame simplifies many calculations and helps reveal intrinsic properties of the system.
For any collision or interaction, examining the momentum conservation helps us understand how objects interact, how they change speed, and what happens during and after the interaction.
Inertial Frame
An inertial frame is a reference frame where Newton's laws of motion hold true without the need for corrections due to non-inertia, like acceleration. In simpler terms, it is a non-accelerating frame where objects tend to stay at rest or in constant motion unless acted on by a force.
In our problem, two frames are particularly important: frame \( S \) and frame \( S' \). Frame \( S \) is stationary with respect to one particle and is an inertial frame. This allows us to easily apply Newton's laws to determine initial conditions like velocity and momentum.
Frame \( S' \), the center of momentum frame, is also inertial, but its speed relative to \( S \) is chosen such that the total momentum in \( S' \) is zero. This does not mean the particles stop moving in frame \( S' \), but rather their velocities balance out, making calculations more straightforward.
Choosing the right inertial frame is crucial to simplifying complex physics problems, especially involving momentum and collisions.
Particle Collision
A particle collision is an interaction where two or more particles come close enough to exert forces on each other, often leading to a transfer of energy and momentum. Collisions can be elastic, where kinetic energy is conserved, or inelastic, where kinetic energy is not conserved.
In this exercise, the collision is analyzed through momentum considerations. We are not given specifics about elastic or inelastic nature, but rather focus on how momentum is redistributed as a result of the collision.
To understand the unique perspective of the collision, we analyze it in the center of momentum frame \( S' \). Here, the total momentum before and after the collision is zero. With this condition, calculations and predictions about particle paths and speeds post-collision become more evidently clear and manageable.
Collisions are a key topic in physics, allowing deeper insights into particle interactions, energy transformation, and underlying forces affecting interacting bodies.

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Most popular questions from this chapter

What is the speed parameter for the following speeds: (a) a typical rate of continental drift ( 1 in \(/ y\) ); (b) a typical drift speed for electrons in a current-carrying conductor \((0.5 \mathrm{~mm} / \mathrm{s}) ;\) (c) a highway speed limit of \(55 \mathrm{mi} / \mathrm{h} ;\) (d) the root-mean-square speed of a hydrogen molccule at room temperature; (c) a supersonic plane flying at Mach \(2.5(1200 \mathrm{~km} / \mathrm{h}) ;\) (f) the escape speed of a projectile from the Earth's surface; (g) the speed of Earth in its orbit around the Sun; (h) a typical recession speed of a distant quasar due to the cosmological expansion \(\left(3.0 \times 10^{4} \mathrm{~km} / \mathrm{s}\right) ?\)

A pitot tube (Fig. \(14-48\) ) is used to determine the airspeed of an airplane. It consists of an outer tube with a number of small holes \(B\) (four are shown) that allow air into the tube; that tube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole \(A\) at the front end of the device, which points in the direction the plane is headed. At \(A\) the air becomes stagnant so that \(v_{A}=0 .\) At \(B,\) however, the speed of the air presumably equals the airspecd \(v\) of the plane. (a) Use Bernoulli's equation to show that $$ v=\sqrt{\frac{2 \rho g h}{\rho_{\text {ar }}}} $$ where \(\rho\) is the density of the liquid in the \(U\) -tube and \(h\) is the difference in the liquid levels in that tube. (b) Suppose that the tube contains alcohol and the level difference \(h\) is \(26.0 \mathrm{~cm}\). What is the plane's speed relative to the air? The density of the air is \(1.03 \mathrm{~kg} / \mathrm{m}^{3}\) and that of alcohol is \(810 \mathrm{~kg} / \mathrm{m}^{3}\)

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipc (Fig. \(14-50) ;\) the cross-scetional arca \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional arca. Between the cntrance and exit, the fluid flows from the pipe with speed \(V\) and then through a narrow "throat" of cross- scetional area \(a\) with speed \(v\). A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's specd is accompanied by a change \(\Delta p\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manomcter. (Here \(\Delta p\) means pressure in the throat minus pressurc in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. \(14-50,\) show that $$ V=\sqrt{\frac{2 a^{2} \Delta p}{\rho\left(a^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-scctional arcas are \(64 \mathrm{~cm}^{2}\) in the pipe and \(32 \mathrm{~cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and \(41 \mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?

Caught in an avalanche, a skier is fully submerged in flowing snow of density \(96 \mathrm{~kg} / \mathrm{m}^{3}\). Assume that the average density of the skier, clothing, and skiing equipment is \(1020 \mathrm{~kg} / \mathrm{m}^{3}\). What percentage of the gravitational force on the skier is offset by the buoyant force from the snow?

(a) If \(m\) is a particle's mass, \(p\) is its momentum magnitude, and \(K\) is its kinctic energy, show that $$ m=\frac{(p c)^{2}-K^{2}}{2 K c^{2}} $$ (b) For low particle speeds, show that the right side of the equation reduces to \(m\). (c) If a particle has \(K=55.0 \mathrm{MeV}\) when \(p=\) \(121 \mathrm{MeV} i c,\) what is the ratio \(\mathrm{m} / \mathrm{m}_{e}\) of its mass to the clectron mass?

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