A particle of mass m is projected with a speed u from the ground at an angle θ = π/3 w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity uî . The horizontal distance covered by the combined mass before reaching the ground is:

Solution

Physics sir · 4 steps

At the top only horizontal speed is left: conserve momentum along x, then let the pair fall from rest.

1. Velocity at the top. At maximum height the vertical velocity is zero. The particle moves only horizontally, with u cos 60° = u/2.
2. Collision. The other particle has the same mass and moves with u along x. In a completely inelastic collision they stick together, and momentum along x is conserved.

   m(u/2) + m u = 2m v ⇒ v = 3u/4

3. Time to fall. The combined mass starts with zero vertical velocity from H = (u²sin² 60°)/2g = 3u²/8g. The fall time is t = √(2H/g) = (√3 u)/2g, the same as the time taken to rise.
4. So. Horizontal distance = v t = 3u/4 × (√3 u)/2g = 3√3/8u²/g, which is option (b).

The idea to keep. A collision at the top of a projectile changes only the horizontal speed. The fall time stays equal to the rise time, so the distance is the new horizontal speed times that time.

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