Two identical particles are located at x⃗ and y⃗ with reference to the origin of three-dimension

Name: Problem 3, Chapter 5: Objective Physics : NEET 2020 Examination textbook Gurukul Oswal
Uploaded: 2021-09-30T15:45:21-08:00
Duration: 1 min 1 s
Channel: Narayan Hari
Description: Two identical particles are located at x⃗ and y⃗ with reference to the origin of three-dimensional coordinate system. The position vector of centre of mass of the system is given by: (a) x⃗-y⃗ (b) (x⃗+y⃗)/(2) (c) (x⃗-y⃗) (d) (x⃗-y⃗)/(2)

step 1 In this problem, as we know that, as we know that R is equal to M1R1 plus M2 R2 by M1 plus M2. S

Question

Two identical particles are located at $\vec{x}$ and $\vec{y}$ with reference to the origin of three-dimensional coordinate system. The position vector of centre of mass of the system is given by: $\begin{array}{ll}\text { (a) } \vec{x}-\vec{y} & \text { (b) } \frac{\vec{x}+\vec{y}}{2}\end{array}$ (c) $(\vec{x}-\vec{y})$ (d) $\frac{\vec{x}-\vec{y}}{2}$

   Two identical particles are located at $\vec{x}$ and $\vec{y}$ with reference to the origin of three-dimensional coordinate system. The position vector of centre of mass of
the system is given by:
$\begin{array}{ll}\text { (a) } \vec{x}-\vec{y} & \text { (b) } \frac{\vec{x}+\vec{y}}{2}\end{array}$
(c) $(\vec{x}-\vec{y})$
(d) $\frac{\vec{x}-\vec{y}}{2}$

Objective Physics : NEET 2020 Examination textbook Gurukul Oswal

Gurukul Oswal… 2020 Edition

Chapter 5, Problem 3

Instant Answer

Solved by Expert Narayan Hari

09/30/2021

Step 1

Step 1: The position vector of the center of mass of a system of particles is given by the formula: \[\vec{R} = \frac{m_1\vec{r_1} + m_2\vec{r_2}}{m_1 + m_2}\] where $m_1$ and $m_2$ are the masses of the particles and $\vec{r_1}$ and $\vec{r_2}$ are their Show more…

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Two identical particles are located at $\vec{x}$ and $\vec{y}$ with reference to the origin of three-dimensional coordinate system. The position vector of centre of mass of the system is given by: $\begin{array}{ll}\text { (a) } \vec{x}-\vec{y} & \text { (b) } \frac{\vec{x}+\vec{y}}{2}\end{array}$ (c) $(\vec{x}-\vec{y})$ (d) $\frac{\vec{x}-\vec{y}}{2}$

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Key Concepts

Center of Mass

The center of mass of a system is the point where the total mass of the system can be considered to be concentrated. It is calculated by taking the weighted average of the position vectors of all individual particles, with each particle's mass as the weight. For systems with equal masses, the calculation simplifies to the arithmetic mean of the position vectors.

Identical Particles in Two-Body Systems

When dealing with identical particles, their equal masses allow us to simplify the computation of the center of mass. The expression for the center of mass becomes the average of the position vectors of the particles, since each particle contributes equally. This concept is crucial in many areas of physics, including mechanics and dynamics, as it simplifies the analysis of the system.

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