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In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]

Short Answer

Expert verified
If momentum is conserved, the forces between particles must satisfy Newton's third law: \( \mathbf{F}_{12} = -\mathbf{F}_{21} \).

Step by step solution

01

Understand the Problem Statement

We need to demonstrate that if the conservation of momentum holds for any group of particles, the forces between these particles follow Newton's third law. Specifically, if momentum is conserved for two particles without external forces, then the forces they exert on each other must be equal in magnitude and opposite in direction: \( \mathbf{F}_{12} = -\mathbf{F}_{21} \).
02

Apply Conservation of Momentum

Consider two particles 1 and 2 in a system with no external forces. According to the conservation of momentum, the total momentum of this system remains constant over time. That means \( \frac{d}{dt}(\mathbf{p}_1 + \mathbf{p}_2) = 0 \) where \( \mathbf{p}_1 \) and \( \mathbf{p}_2 \) are the momenta of particles 1 and 2, respectively.
03

Express Momentum in Terms of Forces

Write the momentum \( \mathbf{p} \) as mass times velocity: \( \mathbf{p}_1 = m_1 \mathbf{v}_1 \) and \( \mathbf{p}_2 = m_2 \mathbf{v}_2 \). Taking the derivative with respect to time, we have \( \frac{d\mathbf{p}_1}{dt} + \frac{d\mathbf{p}_2}{dt} = 0 \). By Newton's second law, this is equivalent to \( m_1 \mathbf{a}_1 + m_2 \mathbf{a}_2 = \mathbf{F}_{12} + \mathbf{F}_{21} = 0 \).
04

Infer the Forces Between Particles

Since \( \mathbf{F}_{12} + \mathbf{F}_{21} = 0 \), we can conclude that \( \mathbf{F}_{12} = -\mathbf{F}_{21} \). This shows that the force exerted by particle 1 on particle 2 is equal in magnitude and opposite in direction to the force exerted by particle 2 on particle 1. This is precisely Newton's third law.

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Key Concepts

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

Newton's third law
Newton's third law is a fundamental principle in physics. It states that for every action, there is an equal and opposite reaction. In simpler terms, if one object exerts a force on another object, the second object exerts a force of equal strength in the opposite direction on the first object. This principle explains the reciprocal interactions between objects.

For example, when you push against a wall, the wall pushes back with an equal force on you. It might not be obvious because the wall doesn't move, but this is because its mass is much larger than yours, causing a negligible acceleration.

Newton's third law is crucial in understanding how forces work in pairs. It is important to note that these action-reaction force pairs act on different objects and don't cancel each other out because they are not acting on the same object. This law is the reason why rockets propel themselves forward, as the gases they expel backward push on the rocket with equal force in the opposite direction.
Interparticle forces
Interparticle forces refer to the forces between particles or objects in a physical system. These forces can be contact forces, such as friction or tension, or of a fundamental nature, such as gravitational, electromagnetic, nuclear, or weak forces. Understanding these forces helps us explain how particles interact and bind together to form stable systems.

In the context of momentum conservation, when two particles interact with no external force acting on them, the interparticle forces they exert on each other must follow certain rules. As suggested by momentum conservation, and proven by Newton's laws, any force one particle exerts on another is met with an equal and opposite force.

This equality and opposition ensure momentum conservation within the system, showing that in every interaction, the momentum lost by one particle is gained by the other. By focusing on just two particles, without external forces, we can prove the forces between them obey Newton's third law, allowing us to model complex interactions like collisions or bonding processes in chemistry.
Momentum conservation proof
The conservation of momentum is a foundational concept in physics, stating that in a closed system, the total momentum remains constant unless acted upon by an external force. This principle makes analyzing collisions and other interactions simpler, as it allows us to predict the motion of particles before and after events.In proving momentum conservation implies Newton’s third law, we begin by considering two particles with masses and velocities. According to the conservation of momentum:\[\frac{d}{dt}(\mathbf{p}_1 + \mathbf{p}_2) = 0\]where \(\mathbf{p}_1\) and \(\mathbf{p}_2\) are the momenta of the particles. Expressing momentum as mass times velocity, and differentiating, leads us to:\[m_1 \mathbf{a}_1 + m_2 \mathbf{a}_2 = \mathbf{F}_{12} + \mathbf{F}_{21}\]Here, \(\mathbf{a}_1\) and \(\mathbf{a}_2\) are accelerations, and forces are derived from Newton’s second law. Knowing that the derivative of a constant zero means no change in momentum, it also tells us that the sum of forces is zero:\[\mathbf{F}_{12} + \mathbf{F}_{21} = 0\]This equation clarifies that \(\mathbf{F}_{12} = -\mathbf{F}_{21}\), proving that forces are equal and opposite, thus showing momentum conservation implies Newton's third law, completing the proof with elegance.

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Most popular questions from this chapter

Imagine two concentric cylinders, centered on the vertical \(z\) axis, with radii \(R \pm \epsilon,\) where \(\epsilon\) is very small. A small frictionless puck of thickness \(2 \epsilon\) is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates \((\rho, \phi, z)\) for its position (Problem 1.47 ), then \(\rho\) is fixed at \(\rho=R,\) while \(\phi\) and \(z\) can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion.

If you have some experience in electromagnetism and with vector calculus, prove that the magnetic forces, \(\mathbf{F}_{12}\) and \(\mathbf{F}_{21}\), between two steady current loops obey Newton's third law. [Hints: Let the two currents be \(I_{1}\) and \(I_{2}\) and let typical points on the two loops be \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\). If \(d \mathbf{r}_{1}\) and \(d \mathbf{r}_{2}\) are short segments of the loops, then according to the Biot- Savart law, the force on \(d \mathbf{r}_{1}\) due to \(d \mathbf{r}_{2}\) is $$\frac{\mu_{0}}{4 \pi} \frac{I_{1} I_{2}}{s^{2}} d \mathbf{r}_{1} \times\left(d \mathbf{r}_{2} \times \hat{\mathbf{s}}\right)$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2} .\) The force \(\mathbf{F}_{12}\) is found by integrating this around both loops. You will need to use the " \(B A C-C A B\) " rule to simplify the triple product.]

One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors \(\mathbf{b}=(1,2,4)\) and \(\mathbf{c}=(4,2,1)\) by evaluating their scalar product.

By applying Pythagoras's theorem (the usual two-dimensional version) twice over, prove that the length \(r\) of a three-dimensional vector \(\mathbf{r}=(x, y, z)\) satisfies \(r^{2}=x^{2}+y^{2}+z^{2}\)

A particle moves in a circle (center \(O\) and radius \(R\) ) with constant angular velocity \(\omega\) counter-clockwise. The circle lies in the \(x y\) plane and the particle is on the \(x\) axis at time \(t=0 .\) Show that the particle's position is given by $$\mathbf{r}(t)=\hat{\mathbf{x}} R \cos (\omega t)+\hat{\mathbf{y}} R \sin (\omega t)$$ Find the particle's velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion.

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