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QExample - 1: HC Verma: Concepts of Physics - Volume 1 Physics and Mathematics Example - 1 Solution: Two vectors having equal magnitudes A make an…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often fail to apply the half-angle identities $1 + \cos\theta = 2\cos^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ simultaneously, leading to algebraic errors in simplifying the resultant magnitude and direction. Ensure both numerator and denominator are converted to half-angles to correctly cancel terms and derive $R = 2A\cos(\theta/2)$.

Two vectors having equal magnitudes A make an angle θ with each other. Find the magnitude and direction of the resultant.

Step-by-Step Verified Solution

Solution:

1. Magnitude of the Resultant (R):

  • We use the law of cosines to find the magnitude of the resultant vector:
  • R = √(A² + A² + 2A * A * cos(θ))
  • R = √(2A² + 2A²cos(θ))
  • R = √(2A²(1 + cos(θ)))
  • Using the trigonometric identity 1 + cos(θ) = 2cos²(θ/2), we get:
  • R = √(2A² * 2cos²(θ/2))
  • R = √(4A²cos²(θ/2))
  • R = 2Acos(θ/2)

2. Direction of the Resultant (θ'):

  • Let θ' be the angle between one of the vectors and the resultant vector.
  • We use the formula for the tangent of the angle between the resultant and one of the vectors:
  • tan(θ') = (Asin(θ)) / (Acos(θ) + A)
  • tan(θ') = sin(θ) / (cos(θ) + 1)
  • Using the trigonometric identities sin(θ) = 2sin(θ/2)cos(θ/2) and 1 + cos(θ) = 2cos²(θ/2), we get:
  • tan(θ') = (2sin(θ/2)cos(θ/2)) / (2cos²(θ/2))
  • tan(θ') = sin(θ/2) / cos(θ/2)
  • tan(θ') = tan(θ/2)
  • Therefore:
  • θ' = θ/2

Conclusion:

  • The magnitude of the resultant vector is 2Acos(θ/2).
  • The resultant vector bisects the angle between the two original vectors, making an angle of θ/2 with each of them.

All Questions in this Chapter

QExample - 1 QExample - 2 QExample - 3 QExample - 4 QExample - 5 QExample - 6 QExample - 7

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