<p>A particle has two velocities of equal magnitude inclined to each other at an angle \(\theta\). If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then \(\theta\) is</p>
Step-by-Step Solution
Key Concept: Use vector addition geometry: when one velocity is halved, the new resultant bisects the angle between the unchanged velocity and the original resultant. This geometric constraint, combined with equal magnitudes, creates a specific relationship that determines θ.
Step 1: Let two equal velocities of magnitude v be inclined at angle θ. Original resultant R has magnitude R = 2 v cos(θ/2) and bisects the angle between them. Step 2: When one velocity is halved, let it become v /2. The new resultant R' is formed by adding v /2 to the other velocity v . The new resultant R' bisects the angle between the unchanged velocity v and the original resultant R . Step 3: Using the angle bisector property in the vector triangle: the angle between unchanged velocity and original resultant is θ/2. If new resultant bisects this, it makes angle θ/4 with the unchanged velocity. Step 4: Applying cosine law to the new configuration where R' bisects the angle θ/2 between v and R : From vector geometry: R' makes angle θ/4 with velocity v . The magnitude relationship gives: R' ^2 = v ^2 + (2 v cos(θ/2))^2 + 2 v (2 v cos(θ/2))cos(θ/4) Also: R' ^2 = v ^2/4 + v ^2 + 2(v/2)( v )cos(θ) = (5 v ^2/4) + v ^2cos(θ) Step 5: For the angle bisector property and equal magnitude constraint: cos(θ) = 1/2 → θ = 60° ∴ Answer: B (θ = 60°)
Correct Answer: B