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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A ball AA moving with certain velocity along positive xx-axis collides with a stationary ball BB. After collision their directions of motion make angles α\alpha and β\beta with the xx-axis. The possible values of α\alpha and β\beta are
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Correct option: A
α=30∘\alpha=30^{\circ} and β=−45∘\beta=-45^{\circ}
Option A (α=30∘\alpha=30^{\circ} and β=−45∘\beta=-45^{\circ}) is correct because the y-component of momentum must be conserved (zero) and the x-component must be non-zero.

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