Centre of Mass of Multiple Masses on Axis – Rankers Physics

Calculation of Center of Mass: Practice Problem & Solution

Three masses are placed on the $x$-axis : $300\text{ g}$ at origin, $500\text{ g}$ at $x = 40\text{ cm}$ and $400\text{ g}$ at $x = 70\text{ cm}$. The distance of the center of mass from the origin is : (2012 Mains)
$40\text{ cm}$
$45\text{ cm}$
$50\text{ cm}$
$30\text{ cm}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Calculation of Center of Mass. Understanding the underlying formula is key to arriving at the correct answer below:

Use the centre of mass formula $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$. Substituting the given values: $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400} = \frac{48000}{1200} = 40\text{ cm}$. Option (a) is correct.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The centre of mass of a system of particles depends on
2. Two objects of mass $10\text{ kg}$ and $20\text{ kg}$ respectively are connected to the two ends of a rigid rod of length $10\text{ m}$ with negligible mass. The distance of the centre of mass of the system from the $10\text{ kg}$ mass is : (2022)
3. Assertion (A): Maximum energy loss occurs when the particles get stuck together as a result of collision. Reason (R): A point particle of mass (m\) moving with speed (v\) collides with stationary point particle of mass (M\). Then the maximum energy loss possible is given \( \frac{m}{(m+M)}\left(\frac{1}{2}mv^2\right)\).  
4. Center of mass is closer to
5. Two persons of masses $55\text{ kg}$ and $65\text{ kg}$ respectively, are at the opposite ends of a boat. The length of the boat is $3.0\text{ m}$ and weighs $100\text{ kg}$. The $55\text{ kg}$ man walks up to the $65\text{ kg}$ man and sits with him. If the boat is in still water the center of mass of the system shifts by: (2012 Pre)