Question

Four particles A, B, C, D of mass $\frac{m}{2}$, $m$, $2m$, $4m$ have the same momentum, respectively. The particle with maximum kinetic energy is:

• A. D
• B. C
• C. A
• D. B

Answer 1

The Correct Option is C

The kinetic energy is calculated as:

\[ KE = \frac{p^2}{2m} \]

Given that all particles possess identical momentum, their kinetic energy is inversely related to their mass:

\[ KE \propto \frac{1}{m} \]

Consequently, the particle exhibiting the smallest mass will possess the highest kinetic energy.

Considering the provided particles:

\[ m_A = \frac{m}{2}, \quad m_B = m, \quad m_C = 2m, \quad m_D = 4m \]

Therefore, \(\frac{m}{2}\) (corresponding to particle A) exhibits the maximum kinetic energy.