A particle (m = 1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaching its highest point P (height 1 m), the kinetic energy of the particle (in )) is :
(Figure drawn is schematic and not to scale; take g = 10 ms⁻²)

Solution

Physics sir · 4 steps

No friction and no air drag, so the 20 J the particle starts with is still all there at P.

1. Energy at A. The particle starts from rest, so all its energy is potential: mgh_A = 1 × 10 × 2 = 20 J, measured from the level of O.
2. Nothing is lost. The track is frictionless and the flight in air has only gravity acting. So total mechanical energy stays at 20 J along the track and through the projectile path.
3. Energy at P. At P the height is 1 m, so the potential energy is 1 × 10 × 1 = 10 J. The rest must be kinetic energy. P is the top of a projectile, so only the vertical velocity is zero there; the horizontal velocity is still present.

   K_P = mgh_A - mgh_P = 20 J - 10 J = 10 J

4. So. The kinetic energy of the particle at P is 10 J.

The common slip: 0. Treating P as a point where the particle stops gives 0, but at the top of a projectile only the vertical velocity is zero, so the horizontal motion still carries 10 J.

The idea to keep. With no friction or drag, track plus flight is one energy story: K = mg(hₛₜₐᵣₜ - h_(now)). At the top of a projectile, KE is not zero because the horizontal velocity remains.

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