Bernoulli application — pressure difference ahead of a moving body A body moves through still water at 6 m/s. At a point 1 m ahead of the body, the local water speed is measured as 4 m/s. Estimate the pressure difference p_nose − p_ahead between the stagnation point at the nose and the point 1 m ahead (assume incompressible, same elevation, negligible losses; ρ ≈ 1000 kg/m^3).

Introduction / Context:
Bernoulli’s equation along a streamline relates pressure and dynamic head. Around a blunt nose, the stagnation point has essentially zero local velocity relative to the body, while points upstream can have finite approach velocities different from the far-field speed. The pressure difference between a stagnation point and an upstream point is governed by the difference in dynamic pressures when elevation and losses are negligible.

Given Data / Assumptions:

• Speed at nose (stagnation point) ≈ 0 m/s relative to the body.
• Speed 1 m ahead = 4 m/s (given).
• Body speed relative to still water = 6 m/s (gives context/far-field).
• Same elevation; incompressible water, ρ ≈ 1000 kg/m^3; negligible viscous loss between the two points.

Concept / Approach:
Apply Bernoulli between the nose stagnation point and the measurement point 1 m ahead. With equal elevation and negligible loss: p + 0.5ρv^2 = constant. Hence p_nose − p_ahead = 0.5ρ(v_ahead^2 − v_nose^2). Here, v_nose ≈ 0 and v_ahead = 4 m/s relative to the body; however, the nose stagnation pressure corresponds to decelerating from the local undisturbed approach near the body, effectively from the free-stream around the body (~6 m/s) to zero. The problem provides both 6 m/s and 4 m/s; to find the difference between the nose and the specified upstream point, use the velocity difference 6 m/s → 4 m/s across those locations.

Step-by-Step Solution:
Compute Δv^2 = 6^2 − 4^2 = 36 − 16 = 20.Dynamic head difference = 0.5 * ρ * Δv^2 = 0.5 * 1000 * 20 = 10,000 N/m^2.Therefore p_nose − p_ahead = 10,000 N/m^2.

Verification / Alternative check:
Dimensional check: Pa = kg/(m·s^2) × (m^2/s^2) = N/m^2 (OK). Magnitude (10 kPa) is reasonable for modest speed differences in water.

Why Other Options Are Wrong:
2,000 N/m^2 and 1,000 N/m^2: Underestimate dynamic head difference for the given speeds.
19,620 N/m^2: Corresponds to ~2 m water column; not tied to the stated velocities.
98,100 N/m^2: About 1 atm; far too large for a 6 vs 4 m/s difference.

Common Pitfalls:

• Using only the 4 m/s value against zero and ignoring the 6 m/s free-stream context provided by the problem statement.
• Mixing gauge and absolute pressures; Bernoulli deals with differences so gauge is fine.

Final Answer:
10,000 N/m^2