Jet impact on a moving plate (normal incidence) A water jet of velocity V and area a strikes a flat plate normally. The plate moves in the jet direction with constant velocity v. The force exerted by the jet on the plate equals:

Introduction / Context:
Force due to a jet on a surface derives from momentum change. When the surface moves in the same direction as the jet, the effective mass flow striking the plate changes because the relative velocity is reduced. This problem is a standard result used in jet propulsion and impulse turbine analyses.

Given Data / Assumptions:

• Jet cross-sectional area = a, jet velocity relative to ground = V.
• Flat plate moves colinearly with the jet at speed v (0 ≤ v < V).
• Jet sticks momentarily and leaves tangentially with zero normal component (idealized), so normal velocity component is brought to rest.

Concept / Approach:
The mass flow actually intercepted by the moving plate is ρ * a * (V − v) because fluid parcels approach with relative speed (V − v). The normal momentum lost per unit time equals mass flow times change in normal velocity, which is (V − v) to 0 in the plate frame. Transforming back to the ground frame gives the same force magnitude in steady conditions.

Step-by-Step Solution:
Relative speed at impact = V_rel = V − v.Mass flow striking plate = ρ * a * V_rel.Normal velocity reduction = V_rel to 0 ⇒ ΔV = V_rel.Force = mass flow * ΔV = (ρ * a * V_rel) * V_rel = ρ * a * (V − v)^2.

Verification / Alternative check:
Limiting cases: v = 0 ⇒ F = ρ a V^2 (stationary plate). v → V ⇒ F → 0 (plate runs with the jet). Both match physical intuition.

Why Other Options Are Wrong:
Options (b) and (c) ignore the reduced intercepted mass flow or double-count terms. Option (d) is unphysical for colinear motion. Option (e) inappropriately halves the correct result.

Common Pitfalls:
Using V instead of V − v for intercepted flow; forgetting that momentum change uses the relative velocity squared in this setup.

Final Answer:
ρ * a * (V - v)^2