- #1

Mr Davis 97

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## Homework Statement

A rocket sled moves along a horizontal plane, and is retarded by a friction force ##f_{friction} = \mu W##, where ##\mu## is the kinetic friction constant and ##W## is the weight of the sled.

The sled's initial mass is ##M_0##, and its rocket engine expels mass at a constant rate ##\displaystyle \frac{dM}{dt} = \gamma##; the expelled mass has constant speed ##v_0## relative to the rocket.

The rocket sled starts from rest and the engine stops when half the sled's total mass is gone. Find an expression for the maximum speed.

## Homework Equations

Rocket equation

## The Attempt at a Solution

We start by using the rocket equation, and defining that movement to the right is positive and movement to the left is negative, and that the sled is moving to the right:

##\displaystyle \frac{dP}{dt} = M\frac{dv}{dt} - v_0 \frac{dM}{dt}##

then since ##M = M_0 + \gamma t## and ##\displaystyle \frac{dM}{dt} = \gamma##

##\displaystyle \frac{dP}{dt} = (M_0 + \gamma t)\frac{dv}{dt} - v_0 \gamma##

Then since there is external friction force acting on the system, we have that ##\displaystyle \frac{dP}{dt} = (M_0 + \gamma t)\frac{dv}{dt} - v_0 \gamma = - \mu (m_0 + \gamma t) g##

then

##\displaystyle \frac{dv}{dt} = \frac{v_0 \gamma - \mu (M_0 + \gamma t) g}{M_) + \gamma t} = \frac{v_0 \gamma}{M_0 + \gamma t} - \mu g##

Solving for velocity, we find that

##\displaystyle v = v_0 \log (1 + \frac{\gamma t}{M_0}) - \mu g t##

Now the final velocity of the rocket sled is when the mass expelled is half of the initial mass: ##M_0 + \gamma t_f = \frac{1}{2} M_0 \implies t_f = \frac{-m_0}{2 \gamma}##. Plugging this into our formula for velocity, we get ##\displaystyle v_f = \mu g \frac{M_0}{2 \gamma} - v_0 \log (2)##

Is this the correct expression for the maximum velocity?