Two particles of equal mass m have respective
initial velocities uî and u( (î + ĵ)/2 ).
They collide completely inelastically. The energy lost in the process is :

Solution

Physics sir · 4 steps

Momentum fixes the final velocity, so compare the kinetic energy before and after; the difference is the loss.

1. Initial kinetic energy. The second particle has velocity (u/2, u/2), so its speed squared is u²/4 + u²/4 = u²/2. Total initial KE is 1/2mu² + 1/2m·u²/2 = 3/4mu².
2. Common final velocity. In a completely inelastic collision the two masses stick together, and momentum is conserved. So 2mv⃗ = m uî + mu/2(î + ĵ), which gives v⃗ = 3u/4î + u/4ĵ.
3. Final kinetic energy. v² = 9u²/16 + u²/16 = 5u²/8. The combined mass is 2m, so the final KE is 1/2(2m)·5u²/8 = 5/8mu².

   Δ K = 3/4mu² - 5/8mu² = 1/8mu²

4. So. The energy lost is 1/8mu², which is option (b). As a check, Δ K = 1/2μ |u⃗₁ - u⃗₂|² with μ = m/2 and |u⃗₁ - u⃗₂|² = u²/2 gives the same 1/8mu².

The idea to keep. In a perfectly inelastic collision, the energy lost is 1/2μ vᵣₑₗ², where μ is the reduced mass. Only the relative velocity matters, so work with vectors.

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