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Physics · WORK, ENERGY AND POWER
हिन्दीMatch List-I with List-II for collisions involving a moving body of mass \( m_1 \) with a stationary target of mass \( m_2 \): List-I List-II A \( m_1 = m_2 \), 1D elastic collision I Particles move together with common velocity \( \frac{m_1 v_{1i}}{m_1 + m_2} \) B \( m_2 \gg m_1 \), 1D elastic collision II Scattered particles move at right angles to each other C \( m_1 = m_2 \), 2D glancing elastic collision III Incident mass reverses velocity: \( v_{1f} \simeq -v_{1i} \) D Completely inelastic collision IV Incident mass comes to rest: \( v_{1f} = 0 \) Choose the correct answer from the options given below:
Match List-I with List-II for collisions involving a moving body of mass \( m_1 \) with a stationary target of mass \( m_2 \):
| List-I | List-II | ||
|---|---|---|---|
| A | \( m_1 = m_2 \), 1D elastic collision | I | Particles move together with common velocity \( \frac{m_1 v_{1i}}{m_1 + m_2} \) |
| B | \( m_2 \gg m_1 \), 1D elastic collision | II | Scattered particles move at right angles to each other |
| C | \( m_1 = m_2 \), 2D glancing elastic collision | III | Incident mass reverses velocity: \( v_{1f} \simeq -v_{1i} \) |
| D | Completely inelastic collision | IV | Incident mass comes to rest: \( v_{1f} = 0 \) |
Choose the correct answer from the options given below:
View Answer & Solution
Correct answer · Option B
A-(IV), B-(III), C-(II), D-(I)
Explanation
Equal masses in 1D elastic collision exchange velocities (A-IV). Heavy stationary target reflects the light projectile (B-III). Equal masses in 2D elastic collision move at 90 degrees (C-II). Completely inelastic collision results in sticking together (D-I).
