The position vector of 1 kg object is r⃗ = ( 3î - ĵ )m and its velocity v⃗ = ( 3ĵ + k̂ )ms⁻¹. The magnitude of its angular momentum is √x Nm where x is ___________.

Solution

Physics sir · 4 steps

Angular momentum is m r⃗ × v⃗. Set up the determinant and the cross product does the work.

1. Set up. With m = 1 kg, L⃗ = m(r⃗ × v⃗) = r⃗ × v⃗. Write the components as r⃗ = (3, -1, 0) and v⃗ = (0, 3, 1).

   L⃗ = |î, ĵ, k̂; 3, -1, 0; 0, 3, 1|

2. Expand. î part: (-1)(1) - (0)(3) = -1. ĵ part: -[(3)(1) - (0)(0)] = -3. k̂ part: (3)(3) - (-1)(0) = 9. So L⃗ = (-î - 3ĵ + 9k̂) kg m² s⁻¹.
3. Magnitude. |L⃗| = √((-1)² + (-3)² + 9²) = √(1 + 9 + 81) = √91.
4. So. Matching with √x gives x = 91.

The common slip: 9.54. Reporting the magnitude itself, √91 ≈ 9.54, gives 9.54. The question asks for x, the number under the root.

The idea to keep. For L⃗ = m r⃗ × v⃗ with vectors in component form, always use the determinant. Watch the minus sign on the ĵ term, then square and add.

Answer from the NTA final answer key. Solution by PrepWiser.