When a constant horizontal stream of water of fixed cross-sectional area strikes a flat ve… — JAMB UTME

When a constant horizontal stream of water of fixed cross-sectional area strikes a flat vertical wall normally and comes to rest without rebounding, the magnitude of the force exerted on the wall is directly proportional to the square of the speed of the water stream.

Answer: Answer

Answer

True. The force exerted on the wall is directly proportional to the square of the speed of the water stream (

F \propto v^2

According to Newton's second law, force is the rate of change of linear momentum (

F = \frac{\Delta p}{\Delta t}

). For a continuous fluid stream,

\frac{\Delta m}{\Delta t} = \rho A v

. Since each unit mass loses speed

v

upon impact,

F = \left(\frac{\Delta m}{\Delta t}\right) v = \rho A v^2

. Thus, force is directly proportional to the square of the speed.

Step-by-Step Solution

Determine the mass of water striking the wall in time interval

\Delta t

\Delta m = \rho A v \Delta t

, where

\rho

is fluid density,

A

is cross-sectional area, and

v

is speed.

The volume of water reaching the wall per unit time is given by

A v

Apply Newton's Second Law in terms of rate of change of linear momentum.

F = \frac{\Delta p}{\Delta t} = \frac{\Delta m \cdot v}{\Delta t} = \frac{(\rho A v \Delta t) v}{\Delta t} = \rho A v^2

Since the water comes to rest, its change in velocity per unit mass is

v

Analyze the dependence of force

F

on speed

v

Since density

\rho

and cross-sectional area

A

are constant,

F = (\text{constant}) \times v^2 \implies F \propto v^2

Both mass flow rate and momentum change per unit mass are proportional to

v

Key Concept

Newton's Second Law and continuous mass flow momentum change

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