Question 5.14: A cart is propelled by a liquid jet issuing horizontally fro...
A cart is propelled by a liquid jet issuing horizontally from a tank as shown in Fig. 5.19. The initial mass of the block is M_{0}. The tank is pressurised so that the jet speed may be considered to be a constant. Obtain an expression for the speed of the block as it accelerates from rest. Assume that the jet issues with a velocity V in the leftward direction, relative to the cart. Also assume the track to be frictionless.
We attach the reference frame xy to the control volume moving with cart (Fig. 5.19 (a)), whereas the reference frame X Y is inertial.
From the conservation of mass for the linearly accelerating control volume, we get
0=\frac{\partial}{\partial t} \int_{C V} d m+\int_{C S}\left(\rho \vec{V}_{x y z} \cdot \hat{n}\right) d A=\frac{d M_{C V}}{d t}+\dot{m}_{e}
\int_{M_{0}}^{M} d M_{C V}=-\int_{0}^{t} \dot{m}_{e} d t
Or M=M_{0}-\dot{m}_{e} t
Where M=M_{C V}
Writing the x component of the linear momentum equation for the linearly accelerating CV, we get
-M \frac{d U}{d t}=-\dot{m}_{e} V=-(\rho A V) V=-\rho A V^{2}Or \frac{d U}{d t}=\frac{\rho A V^{2}}{M}=\frac{\rho A V^{2}}{M_{0}-\dot{m}_{e} t} (5.46)
Separating the variables and integrating,
\int_{0}^{U} d U=\int_{0}^{t} \frac{\rho A V^{2}}{M_{0}-\dot{m}_{e} t} d tU=-\frac{\rho A V^{2}}{\dot{m}_{e}} \ln \left[M_{0}-\dot{m}_{e} t\right]_{0}^{t}
Or U=V \ln \left(\frac{M_{0}}{M_{0}-\dot{m}_{e} t}\right) (5.47)
