Q.A particle slides down a frictionless parabolic track shaped like $y = x^2$. It starts from rest at point A on the upper-left arm of the parabola, slides down through point B at the vertex (the lowest point of the parabola), and continues up the right arm to point C, which is at a height less than that of A. At C the particle leaves the track and moves freely through the air as a projectile, launched at some speed at an angle above the horizontal, and reaches the highest point of its flight at P. Given this, which of the following are correct? (Note: more than one option may be correct.)

Concept

The parabolic track is frictionless and, after C, the particle is in free projectile motion. In both phases the only forces are the (normal, does no work) track reaction and gravity (conservative). Therefore total mechanical energy $E = KE + PE$ is conserved throughout. Take B (the vertex) as the reference level, and let $h_A$ and $h_C$ be the heights of A and C above B, with $h_A > h_C$.

Working through each option

(C) total energy at P = total energy at A — TRUE. With no energy dissipation, $E$ is constant, so $E_P = E_A$. This is the safe consequence of energy conservation.

(A) KE at P = KE at B — FALSE. At B all of A's potential energy has become kinetic: $KE_B = mgh_A$. At the apex P the vertical velocity is zero but the horizontal velocity $v_x = v_0\cos\theta$ survives, so $KE_P = \tfrac12 m v_0^2\cos^2\theta$. Since $\tfrac12 m v_0^2 = mg(h_A-h_C)$, we get $KE_P = mg(h_A-h_C)\cos^2\theta < mgh_A = KE_B$.

(B) height at P = height at A — FALSE. By energy conservation $mgh_A = KE_P + mgh_P$, so

$h_P = h_A - \frac{KE_P}{mg} = h_A - (h_A-h_C)\cos^2\theta < h_A,$ …

Concept: Settle Each Claim by Contradiction, Using Only "Horizontal Velocity Never Changes"

Method: Proof by Contradiction for (A), (B), (D); Direct Energy Conservation for (C)

Rather than computing $h_P$, $KE_P$ explicitly in terms of the launch angle at C, this method assumes each false statement is true and shows it forces an impossible conclusion -- using only one physical fact that's true throughout the entire flight, on the track and in the air: gravity and the (frictionless) normal force never touch the particle's horizontal momentum, so the horizontal component of velocity, once the particle leaves the track at C, is fixed for the rest of the flight and equal to whatever it was at C. Since C is on the curved, rising part of the track (not its flat vertex), that horizontal speed at C is strictly positive.

Setting up the one fact everything rests on

Once airborne, the only force is gravity (vertical) -- the standard projectile setup -- so $v_x(t) = v_{Cx} > 0$ for the entire flight after C, including at the apex P. ($v_{Cx}>0$ because C sits on the sloped right arm of the parabola, where the track's own tangent has a nonzero horizontal component -- only at the vertex B would the tangent be perfectly vertical, which C is not.)

(C): total energy at P = total energy at A -- proved directly, not by contradiction

The track (frictionless) and the air (gravity only) both do work only via a conservative force, so mechanical energy $E=KE+PE$ is conserved across the entire journey A$\to$B$\to$C$\to$P with no exceptions. Hence trivially $E_P=E_A$. (C) is true.

(A): KE at P = KE at B -- refuted by contradiction

Assume $KE_P = KE_B$. Since $E$ is conserved (from (C)) and $E = KE+mgh$, equal $KE$ at two points forces equal height too: $h_P=h_B=0$ (taking B as the reference level). But P is the highest point of a projectile launched with $v_{Cx}>0$ -- for that to have zero height above B, C itself would already have to be at height $0$ (i.e. $C=B$), contradicting that C is a distinct point higher up the right arm. Contradiction $\Rightarrow$ (A) is false.

(B): height at P = height at A -- refuted by contradiction

Assume $h_P=h_A$. By conservation of energy, this forces $KE_P=KE_A=0$ (the particle started from rest at A). But $KE_P \geq \tfrac12 m v_{Cx}^2 > 0$, since the horizontal velocity $v_{Cx}$ never vanishes during the flight (Step 1) -- the particle cannot have zero kinetic energy at P while still carrying nonzero horizontal speed. Contradiction $\Rightarrow$ (B) is false.

(D): time A$\to$B equals time B$\to$P -- refuted by a structural mismatch, not a computation …