The position vectors of two 1 kg particles, (A) and (B), are given by r⃗_A=(α₁ t² î+α₂ t ĵ+α₃ t k̂) m and r⃗_B=(β₁ t î+β₂ t² ĵ+β₃ t k̂) m , respectively; (α₁=1 m / s², α₂=3 n m / s, α₃=2 m / s, β₁=2 m / s, β₂=-1 m / s², β₃=4 pm / s), where t is time, n and p are constants. At t=1 s,|V_A⃗|=|V_B⃗| and velocities V⃗_A and V⃗_B of the particles are orthogonal to each other. At t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is √L kgm² s⁻¹. The value of L is _________.

Solution

Physics sir · 4 steps

Fix n and p from the two velocity conditions, then cross A's position from B with A's velocity.

1. Find the velocities. Differentiate each position vector (read the β₁ term as β₁ t î). At t = 1 s this gives V⃗_A = 2î + 3nĵ + 2k̂ and V⃗_B = 2î - 2ĵ + 4pk̂.
2. Fix n and p. Equal speeds give 8 + 9n² = 8 + 16p², so 3n = ± 4p. Orthogonal velocities give 4 - 6n + 8p = 0. The choice 3n = 4p leaves 4 = 0, so 3n = -4p, which gives p = -1/4 and n = 1/3.
3. Position from B. At t = 1 s, r⃗_A = î + ĵ + 2k̂ and r⃗_B = 2î - ĵ - k̂. So r⃗_(AB) = r⃗_A - r⃗_B = -î + 2ĵ + 3k̂, and V⃗_A = 2î + ĵ + 2k̂.

   L⃗ = m r⃗_(AB) × V⃗_A = |î, ĵ, k̂; -1, 2, 3; 2, 1, 2| = î + 8ĵ - 5k̂

4. So. With m = 1 kg, |L⃗| = √(1 + 64 + 25) = √90 kg m² s⁻¹, so L = 90.

The common slip: 5. Taking the angular momentum about the origin, r⃗_A × V⃗_A = 2ĵ - k̂, gives 5, not 90. The reference point here is particle B.

The idea to keep. Angular momentum about a point needs the position vector measured from that point: use r⃗_A - r⃗_B, not r⃗_A. When a sign choice appears, test both branches against every condition.

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