Question

Given below are two statements:
Statement I: For a mechanical system of many particles, total kinetic energy is the sum of kinetic energies of all the particles.
Statement II: The total kinetic energy can be the sum of kinetic energy of the center of mass with respect to the origin and the kinetic energy of all the particles with respect to the center of mass as reference.
In the light of the above statements, choose the correct answer from the options given below:

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is false but Statement II is true
• D. Statement I is true but Statement II is false

Hint:

The total kinetic energy of a system can always be decomposed into the kinetic energy of the center of mass motion and the kinetic energy of motion relative to the center of mass.

Answer 1

The Correct Option is A

The question involves understanding the distribution of kinetic energy in a system of particles, which is a fundamental concept in mechanics.

Statement I: For a mechanical system of many particles, total kinetic energy is the sum of kinetic energies of all the particles.

Explanation: 

• The kinetic energy of a particle with mass \(m_i\) and velocity \(\vec{v_i}\) is given by \(\frac{1}{2} m_i v_i^2\).
• For a system of \(N\) particles, the total kinetic energy is indeed the sum of the kinetic energies of all individual particles, i.e., \(T = \sum_{i=1}^{N} \frac{1}{2} m_i v_i^2\).

This confirms that Statement I is true.

Statement II: The total kinetic energy can be the sum of kinetic energy of the center of mass with respect to the origin and the kinetic energy of all the particles with respect to the center of mass as reference.

Explanation:

• The velocity of the center of mass \(\vec{V_{cm}}\) is given by \(\vec{V_{cm}} = \frac{1}{M} \sum_{i=1}^{N} m_i \vec{v_i}\), where \(M = \sum_{i=1}^{N} m_i\) is the total mass of the system.
• The total kinetic energy can be split as follows: 
  \(T = \frac{1}{2} M V_{cm}^2 + \sum_{i=1}^{N} \frac{1}{2} m_i \left(\vec{v_i} - \vec{V_{cm}}\right)^2\).
• This division shows that the total kinetic energy is indeed the sum of the kinetic energy of the center of mass (first term) and the kinetic energy due to motion relative to the center of mass (second term).

This confirms that Statement II is true.

Thus, the correct answer is: Both Statement I and Statement II are true.