Particle A of mass m₁ moving with velocity ( √3î + ĵ )ms⁻¹ collides with another particle B of mass m₂ which is at rest initially. Let V₁⃗ and V₂⃗ be the velocities of particles A and B after collision respectively. If m₁ = 2m₂ and after
collision V₁⃗ =( î + √3 ĵ ) , the angle between V₁⃗ and V₂⃗ is :

Solution

Physics sir · 4 steps

Momentum conservation alone gives you V₂⃗, then compare its direction with V₁⃗.

1. Conserve momentum. No external force acts, so total momentum before the collision equals total momentum after it. B starts at rest, so only A carries momentum at the start.

   m₁(√3 î + ĵ) = m₁(î + √3 ĵ) + m₂V₂⃗

2. Find V2. Put m₁ = 2m₂ and cancel m₂: V₂⃗ = 2(√3-1)î + 2(1-√3)ĵ = 2(√3-1)(î - ĵ). The x and y parts are equal in size and opposite in sign, so V₂⃗ points at -45° to the x-axis.
3. Direction of V1. V₁⃗ = î + √3 ĵ has tan θ = √3, so it points at +60° to the x-axis.
4. So. The angle between them is 60° - (-45°) = 105°. A dot-product check agrees: cos θ = -(√3-1)/2√2 ≈ -0.259 = cos 105°. The answer is (a).

The idea to keep. Momentum is conserved as a vector in every collision, so one equation gives the unknown velocity. Then measure each vector's angle from the x-axis and subtract, minding the signs.

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