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The position vector of the centre of mass of a system of particles in any inertial frame is defined by rCM=Σmr/Σm. If the particles suffer only collision forces, prove that r˙CM=uCM(d/dt); i.e. the centre of mass moves with the velocity of the CM frame. [Hint: Σm, \Sigmami are constant; \Sigmamir is zero between collisions, and at any collision we can factor out the r of the participating particles: rΣm˙=0.]

Short Answer

Expert verified
Question: Prove that the velocity of the center of mass of a system of particles suffering only collision forces is equal to the velocity of the center of mass frame. Answer: Differentiating the given position vector of the center of mass with respect to time and using the hint provided, we find that the velocity of the center of mass (r˙CM) is equal to the velocity of the center of mass frame (uCM). This demonstrates the equality between the two velocities.

Step by step solution

01

Understand the position vector of the center of mass formula

We are given the position vector for the center of mass (rCM) as rCM=Σmr/Σm. This equation means that the position vector of the center of mass is the sum of the product of mass (m) and position vector (r) of each particle divided by the total mass of the system.
02

Differentiating the position vector of the center of mass

To find r˙CM, we need to differentiate the position vector of the center of mass with respect to time. The formula is rCM=Σmr/Σm, so differentiating both sides with respect to time, we get: r˙CM=ddt(ΣmrΣm)
03

Apply the product rule for differentiation

We will use the product rule to differentiate the term inside the brackets: ddt(ΣmrΣm)=(Σmr˙+rΣm˙)(Σm)2
04

Simplify the expression using the hint

Using the hint, we know that Σm is constant and Σm˙r=0 between collisions. Thus, the differentiation term simplifies to: Σmr˙+rΣm˙(Σm)2=Σmr˙(Σm)2
05

Determine the velocity of the center of mass

The velocity of the center of mass (r˙CM) can now be found from the simplified expression: r˙CM=Σmr˙(Σm)2 Since, Σm is constant, it can be factored out from both the numerator and denominator, which results in: r˙CM=uCM This shows that the center of mass moves with the velocity of the center of mass frame.

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Key Concepts

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

Position Vector
A position vector is a fundamental concept in physics that describes the location of a point in space relative to an origin. For a system of particles, the position vector of the center of mass is essentially the average position of all the particles weighted by their masses. The formula rCM=ΣmrΣm conveys that the center of mass position is calculated by summing up the products of each particle's mass and its position vector, then dividing by the total mass of the entire system.

This concept is pivotal because it allows us to simplify the study of a system of multiple particles by reducing the complex system to a single point that behaves as if all the mass and motion of the system are concentrated at that point. The center of mass vector helps physicists and engineers analyze translational motion efficiently.
Inertial Frame
An inertial frame is a reference frame in which an object is either at rest or moving at a constant velocity unless acted upon by a force. Within this frame, the laws of physics, especially the Newtonian mechanics, operate in their standard form such as Newton's First Law.

When considering systems of particles, using an inertial frame allows us to apply consistent and reliable principles to predict how the system evolves over time. For the center of mass, analyzing its motion in an inertial frame ensures that any changes in velocity are due to external forces, and any internal forces (like collisions) do not affect its calculation, as these internal forces cancel each other out due to Newton's Third Law.
System of Particles
A system of particles consists of multiple particles that interact with each other either directly or via forces such as gravity. This exercise looks at how collision forces affect particles within such a system.

In dealing with a system of particles, it is crucial to understand that each particle might possess different properties, such as mass and position, impacting how the system behaves as a whole. By using the center of mass, we can treat this complex collection of particles as a single entity for analysis purposes, which significantly simplifies calculations and interpretations of motion.
Velocity of the Center of Mass Frame
The velocity of the center of mass frame is a vector quantity that represents the velocity at which the center of mass of a system of particles moves through space. It is represented mathematically as r˙CM.

Calculating this velocity involves differentiating the center of mass position vector with respect to time. The significant takeaway from this process is understanding that external forces affect the center of mass velocity, but internal collision forces do not alter it. When we say that the center of mass moves with this velocity, we imply that any motion observed from this frame of reference treats the center of mass as stationary, simplifying many analyses, especially those involving collisions or interactions.

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Most popular questions from this chapter

The mass of a hydrogen atom is 1.00814amu, that of a neutron is 1.00898 amu, and that of a helium atom (two hydrogen atoms and two neutrons) is 4.00388 amu. Find the binding energy as a fraction of the total energy of a helium atom.

If Δx=(cΔt,Δr) is the four-vector join of two events on the worldline of a uniformly moving particle (or photon), prove that the frequency vector of its de Broglie wave is given by Nμ=vcΔtΔxμ, whence v/Δt is invariant. Compare with Exercise III (7).

An excited atom, of total mass m, is at rest in a given frame. It emits a photon and thereby loses internal (i.e. rest) energy ΔE. Calculate the exact frequency of the photon, making due allowance for the recoil of the atom. [Answer: (ΔE/h)(112ΔE/mc2). Hint: use a four-vector argument.]

In an inertial frame S, two photons of frequencies v1 and v2 travel in the positive and negative x-directions respectively. Find the velocity of the CM frame of these photons. [Answer: v/c =(v1v2)/(v1+v2).]

Consider a head-on elastic collision of a "bullet' of rest mass M with a stationary 'target' of rest mass m. Prove that the post-collision γ-factor of the bullet cannot exceed (m2+M2)/2mM. This means that for large bullet energies (with γ-factors much larger than this critical value), almost the entire energy of the bullet is transferred to the target. [Hint: if P,P are the pre-and post-collision four-momenta of the bullet, and Q,Q those of the target, show, by going to the CM frame, that (PQ)20; in fact, in the CM frame PQ has no spatial components.] The situation is radically different in Newtonian mechanics, where the pre- and post-collision velocities of the bullet are related by u/u=(M+m)/(Mm). Prove this.

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