# Rocket sled

## 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.

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 in to 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?

## Answers and Replies

kuruman
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Your starting equation does not look right. dp/dt is the generalized form of ma that takes into account the changing mass. The right side should be the sum of all the forces in the horizontal direction. Also, what is v0 doing there and where did the negative sign come from?

I do have the sum of the horizontal forces in the horizontal direction. That is what ##(M_0 + \gamma t)\frac{dv}{dt} - v_0 \gamma = - \mu (m_0 + \gamma t) g##, where the LHS is the rate at which the momentum changes and the RHS is the sum of the forces, which is only the friction force acting to the left since the rocket is moving to the right. Also, ##v_0## is the relative velocity of the expelled mass to the rocket sled (I'm not sure why it is called ##v_0##, but that is how is stated in my book. You could just pretend that it is ##u## or something, as it has nothing to do with the initial velocity of the rocket sled. Also, the negative sign where? The negative sign in the beginning equation comes from the rocket equation.

kuruman
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I see, v0 confused me a bit. The way you use γ is also confusing. I would use a positive γ and then write M(t) = M0 - γt. The final equation would be easier to read and interpret then. Also, you are looking for the maximum speed. You calculated the final speed. Is that the maximum? It would be if v(t) were monotonically increasing. Is it? It's hard to tell with γ being implicitly negative.

I see, v0 confused me a bit. The way you use γ is also confusing. I would use a positive γ and then write M(t) = M0 - γt. The final equation would be easier to read and interpret then. Also, you are looking for the maximum speed. You calculated the final speed. Is that the maximum? It would be if v(t) were monotonically increasing. Is it? It's hard to tell with γ being implicitly negative.
When I use negatives I get ##\displaystyle v = v_0 \log (\frac{M_0}{M_0 - \gamma t}) - \mu g t##. This doesn't seem like a monotonically increasing function, but it seems like it would have to be because until the engines are cut off the force from the sled is greater than the friction, which is why it is accelerating in the first place

kuruman
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The function is what it is and does not have to be anything else. How can you tell if it has an extremum, a maximum or a minimum?

Well if the function is monotonically increasing, then the maximum would have to be when the engines turn off, because at that point friction kicks in immediately. If the function is not monotonically increasing, then we could do the first derivative test to determine if there is an extremum.

kuruman
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Right, do the first derivative rule and see what you get. If there is an extremum, you need to determine if it's a maximum or a minimum.
Well if the function is monotonically increasing, then the maximum would have to be when the engines turn off, because at that point friction kicks in immediately.
Friction "kicked in" as soon as the sled started to move. What you want to say is that if the function is monotonically increasing, the maximum value is reached at the moment when the rocket engine shuts down because after that the velocity will be monotonically decreasing.

Right, do the first derivative rule and see what you get. If there is an extremum, you need to determine if it's a maximum or a minimum.
Friction "kicked in" as soon as the sled started to move. What you want to say is that if the function is monotonically increasing, the maximum value is reached at the moment when the rocket engine shuts down because after that the velocity will be monotonically decreasing.
So I got ##t = \frac{M_0}{\gamma} + \frac{v_0}{\mu g}## as an extremum. When I plug this in to the derivative of the acceleration, I get ##v_0 (\frac{\gamma}{M_0 - \gamma t})^2##, which is negative because ##v_0 < 0## (right?).

kuruman
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... because v0<0v_0 < 0 (right?
It depends. Look at the derivation of the rocket equation that you used. If the relative velocity of the expelled gases is written as v + v0, it is negative; if it is written as v - v0, it is positive. I have seen both forms used.

It depends. Look at the derivation of the rocket equation that you used. If the relative velocity of the expelled gases is written as v + v0, it is negative; if it is written as v - v0, it is positive. I have seen both forms used.
In the derivation, it is written as ##v + v_0##, so it would be negative

kuruman
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OK, I agree.
When I plug this in to the derivative of the acceleration, I get
You mean the derivative of the velocity, not acceleration. Anyway, do you see what you need to do now?

OK, I agree.
You mean the derivative of the velocity, not acceleration. Anyway, do you see what you need to do now?
Do I plug ##t = \frac{M_0}{\gamma} + \frac{v_0}{\mu g}## into ##\displaystyle v = v_0 \log (\frac{M_0}{M_0 - \gamma t}) - \mu g t## to find the minimum velocity?

kuruman
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You could do that, but how will you be able to tell whether the velocity is a minimum or a maximum at that time? Also, how do you know that the rocket fuel will last until that time?

Didn't I just find at that time the velocity is at a minimum? Also, could I just take the difference between the time when the engines shut off and the time at which be have an extemum to determine which one is after the other?

kuruman
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Didn't I just find at that time the velocity is at a minimum?
You found the time when the velocity is an extremum, this means either a maximum or a minimum.
Also, could I just take the difference between the time when the engines shut off and the time at which be have an extemum to determine which one is after the other?
You can try that, but without numbers for the variables (except g), how can you tell which is the shorter time? All you can hope for is a condition involving an inequality.

Think this through.
1. If the fuel runs out first, no extremum is reached, so it doesn't matter if the extremum is a maximum or a minimum.
2. If the extremum is reached, and it is a maximum, then that maximum is the maximum speed.
3. If the extremum is reached, and it is a minimum, then the final speed is the maximum speed.

You found the time when the velocity is an extremum, this means either a maximum or a minimum.

You can try that, but without numbers for the variables (except g), how can you tell which is the shorter time? All you can hope for is a condition involving an inequality.

Think this through.
1. If the fuel runs out first, no extremum is reached, so it doesn't matter if the extremum is a maximum or a minimum.
2. If the extremum is reached, and it is a maximum, then that maximum is the maximum speed.
3. If the extremum is reached, and it is a minimum, then the final speed is the maximum speed.
Isn't the extremum a minimum since we we plug it into the the derivative of the acceleration we get a value that is always negative, namely
##v_0 (\frac{\gamma}{M_0 - \gamma t})^2##?

• haruspex
kuruman
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f(x) = x2 has a minimum at x = 0. Its second derivative at x = 0 is positive.

But I thought that I was using the second derivative test to test the sign of the extremum...

One more thing. Why would the author give me the information that the rocket's engines turn off when it is half its initial mass if this is not the exact maximum time?

kuruman
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But I thought that I was using the second derivative test to test the sign of the extremum...
You are, except you remember the test incorrectly. That's what I tried to tell you with the f(x) = x2 example.
One more thing. Why would the author give me the information that the rocket's engines turn off when it is half its initial mass if this is not the exact maximum time?
I can only guess what the author could have been thinking. It is easy to figure out the time when the fuel is half the mass. You can see what's going on here, the forward thrust v0(dM/dt) competes with the frictional force μMg to accelerate whatever mass M the sled has at that moment. Depending on the value of M(t), γ and v0, it is conceivable that γv0 = μM(t)g, in which case the sled moves (instantaneously) at constant velocity. That's when the extremum occurs.

I am also concerned with your last expression for the velocity (post #13). If v0 is negative, then both terms in v(t) are negative since the argument of the log is greater than 1. Stated slightly differently, in the absence of friction (μ = 0), the velocity v(t) should be positive and your expression does not show that.

haruspex
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Why would the author give me the information that the rocket's engines turn off when it is half its initial mass if this is not the exact maximum time
Are you sure the rocket starts moving immediately? What if M0>γv0/(gμ)?

Last edited:
So let us just assume that ##\displaystyle v = v_0 \log (\frac{M_0}{M_0 - \gamma t}) - \mu g t## is correct for the velocity (which I think it is). To find an extremum, we take the derivative and set it to zero. We know that ##\frac{dv}{dt} = \frac{- v_0 \gamma}{M_0 - \gamma t} - \mu g## So if we set this to zero we get ##t = \frac{M_0}{\gamma} + \frac{v_0}{\mu g}## We can tell if this is maximum or minimum if put this time in to the second derivative of velocity, which is ##v_0 (\frac{\gamma}{M_0 - \gamma t})^2##. This is a minimum because ##v_0## carries its sign with it. This means that the t is a minimum.

So what am I doing wrong?

haruspex
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So let us just assume that ##\displaystyle v = v_0 \log (\frac{M_0}{M_0 - \gamma t}) - \mu g t## is correct for the velocity (which I think it is). To find an extremum, we take the derivative and set it to zero. We know that ##\frac{dv}{dt} = \frac{- v_0 \gamma}{M_0 - \gamma t} - \mu g## So if we set this to zero we get ##t = \frac{M_0}{\gamma} + \frac{v_0}{\mu g}## We can tell if this is maximum or minimum if put this time in to the second derivative of velocity, which is ##v_0 (\frac{\gamma}{M_0 - \gamma t})^2##. This is a minimum because ##v_0## carries its sign with it. This means that the t is a minimum.

So what am I doing wrong?
You don't need to differentiate to find the extremum, in fact it won't work. As you wrote in your original post, the greatest velocity will be just as the fuel is exhausted. Once it is moving, the net force increases monotonically as the fuel is consumed.
Did you read and understand my post #22?