- #1
Su6had1p
- 18
- 7
- Homework Statement
- A water cannon starts shooting a jet of water horizontally, at t = 0, into a heavy trolley
of mass M placed on a horizontal ground. The nozzle diameter of the water cannon is d,
the density of water is ρ, and the speed of water coming out of the nozzle is u. Find the
speed of the trolley as a function of time. Assume that all the water from the jet is
collected in the trolley. Neglect all frictional losses.
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt} = \rho A u$$
$$\therefore M \frac{dv}{dt} + (v - u) \frac{dm}{dt} = \rho A u^2$$
$$M \frac{dv}{dt} + (v - u) \rho A u = \rho A u^2$$
$$\frac{dv}{dt} = \frac{\rho A u [2u - v]}{M}$$
$$\int \frac{dv}{[2u - v]} = \int \frac{\rho A u}{M} \, dt$$
$$-\ln (2u - v) = \frac{\rho A u}{M} t + c$$ \\
given that at $t = 0, v = 0$
$$\therefore C = -\ln (2u)$$
$$ \frac{dM}{dt} = \rho A u \Rightarrow M(t) = M_C + \rho A u t$$
$$\ln (2u) - \ln (2u - v) = \frac{\rho A u t}{M}$$
$$\ln \left( \frac{2u}{2u - v} \right) = \frac{M - M_C}{M}$$
$$\exp \left[ \ln \left( \frac{2u}{2u - v} \right) \right] = \exp \left[ \frac{M - M_C}{M} \right]$$
$$\frac{2u}{2u - v} = \exp \left[ \frac{M - M_C}{M} \right]$$
$$2u \exp \left[ \frac{M_C - M}{M} \right] = 2u - v$$
$$v = 2u \left[ 1 - \exp \left( \frac{M_C - M}{M} \right) \right]$$
this doesnt match with the answer, am i doing anything wrong ? please help
