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Physics · JEE | WORK, ENERGY AND POWER

हिन्दी

In a one-dimensional elastic collision, if mass \( m_1 \) collides with stationary mass \( m_2 \), the final velocity of the first mass \( v_{1f} \) is given by:

  1. A
    \( v_{1f} = \frac{2m_1}{m_1 + m_2} v_{1i} \)
  2. B
    \( v_{1f} = \frac{2m_2}{m_1 + m_2} v_{1i} \)
  3. C
    \( v_{1f} = \frac{m_2 - m_1}{m_1 + m_2} v_{1i} \)
  4. D
    \( v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} \)
View Answer & Solution

Correct answer · Option D

\( v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} \)

Explanation

From momentum and kinetic energy conservation equations for 1D elastic collision, \( v_{1f} = (\frac{m_1 - m_2}{m_1 + m_2}) v_{1i} \).
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