Welcome Guest!
For persistent fullscreen mode, use F11 function key.

# Problem/CM-02001

For page specific messages
For page specific messages

Question

A sprinkler wagon wets asphalt on a hot summer day. the power of the motor is barely great enough to over the combined friction of the groundand the wheels, of the air, and the axle bearings. The vehicle therefore behaves as if under no forces. Let $m$ be the mass of the water in the water tank  at any instant plus the constant mass of the empty vehicle. Let the amount
of water squirted out per unit time $\mu = -\dot{m}$ , be its exit velocity towards rear, $q$, a seen from the wagon, or $v-q$ as seen from the street, $v$ be the speed of the vehicle.

Then using Newton's second law the rate of change of momentum of the

water + wagon = Force on the wagon.

or,
\begin{eqnarray}
\dot{p} = 0 \\
\frac{d}{dt}(m v) =0\\
\dot{m} v + m \frac{dv}{dt} =0 \\
or, m \frac{dv}{dt}= \mu v
\end{eqnarray}
It would then appear that the acceleration of the wagon is independent of the exit velociy $q$. This is paradoxical.