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Chapter 8: Problem 55

Consider a two particle system with particles having masses \(m_{1}\) and \(m_{2} .\) If the first particle is pushed towards the centre of mass through a distance \(d\), by what distance should the second particle be moved, 80 as to keep the centre of mass at the same position? IAIEEE 2006] (a) \(\frac{m_{2}}{m_{1}} d\) (b) \(\frac{m_{1}}{m_{1}+m_{2}} d\) (c) \(\frac{m_{1}}{m_{2}} d\) (d) \(d\)

Short Answer

Expert verified
The second particle should be moved by \( \frac{m_1}{m_2}d \), option (c).

Step by step solution

01

Define the Centre of Mass Equation

The centre of mass of a two-particle system can be given by:\[ X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]where \( x_1 \) and \( x_2 \) are the positions of particles with masses \( m_1 \) and \( m_2 \), respectively.
02

Initial Centre of Mass Condition

Since the centre of mass should remain constant despite movement, initially assume that the centre of mass is positioned at a point \( X_{cm} \):\[ X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
03

Change in Particle Positions

When the first particle is moved a distance \( d \) towards the centre of mass, its new position is \( x_1' = x_1 - d \). We need to find the distance \( y \) such that the second particle has to be moved by this distance to keep the centre of mass at the same position.
04

New Centre of Mass Equation

To maintain the same centre of mass, the new equation is:\[ X_{cm} = \frac{m_1 (x_1 - d) + m_2 (x_2 + y)}{m_1 + m_2} \]
05

Solve for the Distance y

Since the initial and final centre of mass are equal, equate and solve for \( y \):\[ \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m_1 (x_1 - d) + m_2 (x_2 + y)}{m_1 + m_2} \]Simplifying gives:\[ m_1 x_1 + m_2 x_2 = m_1 x_1 - m_1 d + m_2 x_2 + m_2 y \]\[ 0 = -m_1 d + m_2 y \]\[ m_2 y = m_1 d \]\[ y = \frac{m_1}{m_2}d \]
06

Select the Correct Answer

The distance by which the second particle must be moved is \( y = \frac{m_1}{m_2}d \), which corresponds to option (c).

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Key Concepts

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

Two-Particle System
In physics, a two-particle system refers to a simplified model in which the dynamics and interactions are studied between only two particles. These particles can have different masses and positions in space.
This concept is often used to make calculations more manageable and to provide insights into more complex multi-body systems. In a two-particle system, each particle can be characterized by its mass and location. These parameters are crucial for determining dynamical properties like momentum and the centre of mass. By focusing on just two particles, students can understand foundational mechanics concepts before moving on to more complex systems.
Masses and Distances
The importance of masses and distances in a two-particle system cannot be overstated. In problems like the one described, understanding how these properties influence system dynamics is crucial. Mass refers to the quantity of matter in an object, which impacts how much force is required to change the object's state of motion.Distance, in this context, refers to how far a particle is moved in relation to another point, such as the centre of mass. It is essential to calculate shifts in positions to keep properties like the centre of mass unchanged. When particle 1 with mass \( m_1 \) is moved, particle 2 must be adjusted accordingly, depending on their masses, to maintain equilibrium.
JEE Main Physics
The JEE Main Physics exams aim to test a student's grasp of fundamental physics concepts applied to complex problems. Questions like the one provided assess a student's ability to understand and apply formulas for the centre of mass and system dynamics. To succeed in this type of exam, students should be familiar with basic principles such as motion, force, and equilibrium. Applying these principles requires a solid understanding of underlying formulas and the ability to manipulate and solve equations correctly.
System Dynamics
In understanding system dynamics in a two-particle system, it is crucial to grasp how movement affects the system's core properties, like the centre of mass. System dynamics involves studying how different qualities, such as mass and velocity, interact and influence the system as a whole. For a stable system, any change in one part must be balanced by changes elsewhere. For example, moving one particle alters its contribution to the centre of mass. To keep the system's equilibrium, another particle must be moved to counterbalance this shift.
By comparing the masses and determining appropriate distances, as solved in the exercise, students learn how interactions in a simple system can predict behaviors in more complex systems.

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Most popular questions from this chapter

A thick uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its, mass is \(0.16 \mathrm{~kg}\) and length is \(1.7 \mathrm{~m} .\) Two particles each of mass \(0.08 \mathrm{~kg}\) are moving on the same surface and towards the bar in the direction perpendicular to the bar, one with a velocity of \(10 \mathrm{~ms}^{-1}\) and other with velocity \(6 \mathrm{~ms}^{-1}\). If collision between particles and bar is completely inelastic, both particles strike with the bar simultaneously. The velocity of centre of masa after collision is (a) \(2 \mathrm{~ms}^{-1}\) (b) \(4 \mathrm{~ms}^{-1}\) (c) \(10 \mathrm{~ms}^{-1}\) (d) \(167 \mathrm{~ms}^{-1}\)

A smooth steel ball strikes a fixed smooth steel plate at an angle \(\theta\) with the vertical. If the coefficient of restitution is e, the angle at which the rebounce will take place is (a) \(\bar{\theta}\) (b) \(\tan ^{-1}\left(\frac{\tan \theta}{e}\right)\) (c) \(e \tan \theta\) (d) \(\tan ^{-1}\left(\frac{c}{\tan \theta}\right)\)

Assertion-Retson type. Each of these contains two Statements: Statement I (Assertion), Statement \(\Pi\) (Rotson). Fach of these questions also has fomr alternatitue choice, only ane of uhich is correct. You hatee to select the correct choices from the coves (al, (b), (c) and (d) ginvn bolow (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true Assertion Torque is time rate of change of a parameter, called angular momentum. Reason This is because in linear motion, force represents time rate of change of linear momentum.

\((n-1)\) equal point masses each of mass \(m\) are placed at the vertices of a regular \(n\)-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the positive vector of centre of mass [NCERT Exemplat] (a) \(\frac{1}{n-1} a\) (b) \(-\frac{1}{(n-1)} a\) (c) \((n-1) a\) (d) \(-\left(\frac{n-1}{a}\right)\)

The density of a non-uniform rod of length \(1 \mathrm{~m}\) is given by \(\rho(x)=a\left(1+b x^{2}\right)\) where \(a\) and \(b\) are constants and \(0 \leq x \leq 1\). The centre of mass of the rod will be at |NCERT Exemplar| (a) \(\frac{3(2+b)}{4(3+b)}\) (b) \(\frac{4(2+b)}{3(3+b)}\) (c) \(\frac{3(2+b)}{4(2+b)}\) (d) \(\frac{4(3+b)}{3(2+b)}\)
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