# Detailed Solution in English ## **Problem Statement** A cart is propelled by a jet of liquid coming out of a tank as shown in the figure. The horizontal resistance (friction) is negligible. The tank is pressurized so that the **jet velocity** is **constant**. Obtain a general expression for the velocity of the cart when it starts from rest. --- ## **Given Data and Variables** - **Initial mass of the cart + tank + liquid:** \( M_ \) - **Jet velocity (relative to the cart):** \( V \) (constant) - **Liquid density:** \( \rho \) - **Area of jet orifice:** \( A \) - **Velocity of the cart (to be found):** \( U \) - **Initial velocity of the cart:** \( \) (starts from rest) - **No friction** --- ## **Approach** This is a **variable mass system** (rocket-like problem). We'll use the **conservation of momentum**. --- ### **Step 1: Mass Flow Rate** The **mass flow rate** of the liquid exiting: \[ \dot{m} = \rho A V \] where: - \( \rho \) = density of liquid, - \( A \) = area of nozzle, - \( V \) = jet velocity (relative to cart). --- ### **Step 2: Conservation of Momentum** Let: - \( M \) = mass of cart + remaining liquid at any instant, - \( U \) = velocity of cart at that instant. At a small time \( dt \): - Mass loss: \( dM = -\dot{m} dt \) - The ejected mass in time \( dt \) is \( dm = -dM \) - The ejected mass leaves at velocity \( V \) **relative to the cart**. #### **Total momentum before:** \[ (M)U \] #### **Total momentum after:** - Cart + tank: \( (M + dM)(U + dU) \) - Ejected mass: \( dm \) at velocity \( U - V \) (absolute velocity is \( U - V \), since jet leaves backward). So: \[ \text{Total after} = (M + dM)(U + dU) + dm \cdot (U - V) \] --- ### **Step 3: Apply Conservation of Momentum** \[ (M)U = (M + dM)(U + dU) + dm (U - V) \] Neglecting second order terms and substituting \( dM = -dm \): \[ MU = MU + MdU + UdM + dM dU + dm (U - V) \] Since \( dM dU \) is second order, neglect it: \[ = M dU + U dM + dm (U - V) \] But \( dM = -dm \): \[ = M dU + U dM + dM (U - V) \] \[ = M dU + U dM + dM U - dM V \] \[ = M dU + 2U dM - V dM \] But, since the cart's mass is decreasing (liquid is expelled), and the **ejected mass has already left, so only consider \( dU \) and \( dM \) terms**: Better to use the standard rocket equation form: \[ M \frac{dU}{dt} = V \frac{dM}{dt} \] But since the liquid is expelled **backwards** (relative velocity \( -V \)), the sign is negative: \[ M \frac{dU}{dt} = -V \frac{dM}{dt} \] --- ### **Step 4: Separate and Integrate** \[ M \frac{dU}{dt} = -V \frac{dM}{dt} \] \[ M dU = -V dM \] \[ dU = -V \frac{dM}{M} \] Integrate both sides from \( U = \) (initial) to \( U \) (final), and from \( M = M_ \) (initial mass) to \( M \) (mass at time t): \[ \int_{}^{U} dU = -V \int_{M_}^{M} \frac{dM}{M} \] \[ U = -V [\ln(M) - \ln(M_)] \] \[ U = -V \ln\left(\frac{M}{M_}\right) \] \[ U = V \ln\left(\frac{M_}{M}\right) \] --- ## **Final General Expression** \[ \boxed{ U = V \ln\left(\frac{M_}{M}\right) } \] where: - \( U \) = velocity of the cart at mass \( M \), - \( V \) = jet velocity (relative to cart), - \( M_ \) = initial mass (cart + tank + liquid), - \( M \) = current mass (cart + tank + remaining liquid). --- ### **Notes** - The answer is analogous to the rocket equation (Tsiolkovsky rocket equation). - Assumes **constant jet velocity** and **no friction**. - If you want the velocity when all liquid is expelled, set \( M = \) final mass (cart + empty tank). --- ## **Summary Table** | Symbol | Meaning | |--------|-------------------------------------| | \( U \) | Velocity of cart (solution) | | \( V \) | Jet velocity (relative to cart) | | \( M_ \) | Initial total mass (cart + tank + fluid) | | \( M \) | Mass at time \( t \) (cart + tank + remaining fluid) | | \( \rho \) | Density of liquid | | \( A \) | Area of nozzle | --- **If you need the velocity as a function of time, you can relate \( M \) to time using:** \[ M = M_ - \dot{m} t = M_ - \rho A V t \] Then: \[ U(t) = V \ln \left( \frac{M_}{M_ - \rho A V t} \right) \] Let me know if you need this time-dependent version or further details!