Conservation of mechanical energy

Steps:

• Frictional forces are negligible, so total mechanical energy (KE + PE) is conserved between X and Y.
• Set PE = 0 at Y: at X, KE_X = ½ m (4)² = 8 m, PE_X = m g (0.4).
• At Y, KE_Y = ½ m v², PE_Y = 0, so 8 m + m g (0.4) = ½ m v².
• v = √[ (4)² + 2 g (0.4) ]; with g = 10 m/s², v = √(16 + 8) = √24 ≈ 5 m/s.

Why C is correct:

• Conservation of energy gives v_Y = √(v_X² + 2gh) ≈ 5 m/s exactly matching option C.

Why the others are wrong:

• A: Speed unchanged only if height difference h = 0 (no PE conversion to KE).
• B: Matches if v_X = 0 (ignores initial KE at X).
• D: Far too high; perhaps from erroneous v = gh or unrelated formula.

Final answer: C