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Physics · WORK, ENERGY AND POWER
हिन्दीA billiard ball moving with speed \( v \) undergoes an elastic collision with an identical stationary billiard ball. After collision, the first ball moves at an angle of \( 30^\circ \) to its initial line of motion. The speed of the second ball after the collision is:
Correct answer · Option 4
\( \frac{v}{2} \)
Explanation
For equal mass glancing elastic collision, \( \theta_1 + \theta_2 = 90^\circ \implies \theta_2 = 60^\circ \). Using momentum conservation perpendicular to initial motion: \( v_{1f} \sin 30^\circ = v_{2f} \sin 60^\circ \implies v_{1f} = \sqrt{3}v_{2f} \). Conservation of kinetic energy: \( v_{1f}^2 + v_{2f}^2 = v^2 \implies 4v_{2f}^2 = v^2 \implies v_{2f} = \frac{v}{2} \).
