Question

Water jet coming out of a stationary horizontal tube at speed $v$ strikes horizontally a massive wall moving in opposite direction with same speed. Water comes to rest relative to wall after striking. Treating $A$ as cross-section of jet and density of water as $\rho$. Select the correct alternative(s)

• A. force exerted on the wall is $2\rho A{v}^2$
• B. force exerted on the wall is $4\rho A{v}^2$
• C. rate of change of kinetic energy of water jet striking the wall is $8\rho A{v}^3$
• D. rate of change of kinetic energy of water jet striking the wall is zero.

Answer 1

Answer: (B), (D)

The correct options are
B force exerted on the wall is $4\rho A{v}^2$
D rate of change of kinetic energy of water jet striking the wall is zero.

As water comes to rest relative to wall after striking, velocity of water after striking is equal to velocity of wall itself.

Change in velocity during strike
$=\Delta v={\overrightarrow{v}}_f-{\overrightarrow{v}}_i=(-v)-\left(v\right)=-2v$
Mass rate of strike $\frac{dm}{dt}=A\left(2v\right)\rho$
Thrust force on jet $=\Delta v\frac{dm}{dt}=\frac{dm}{dt}(-2v)=-4\rho A{v}^2$
Thrust force on wall $=-$ Thrust force on jet $=4\rho A{v}^2$
Rate of change of kinetic energy is
$\frac{d\left(KE\right)}{dt}=\frac{1}{2}\frac{dm}{dt}\left(v_f^2\right)-\frac{1}{2}\frac{dm}{dt}\left(v_i^2\right)$
$=\frac{1}{2}\frac{dm}{dt}[{(-v)}^2-v^2]=0$