Rocket motion - conservation of momentum

In summary, the conversation discusses finding the speed of a rocket when its mass is changing due to ejected fuel. The equation of conservation of momentum is used to derive the equation v2-v1 = u ln(m1/m2). However, there is a discrepancy when defining dm as positive or negative, leading to a conflicting solution. Ultimately, using different letters for the rocket's mass and the ejected fuel's mass will resolve the issue.
  • #1
bensondros
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Homework Statement


Find the speed v of the rocket when the mass of the rocket = m. The rocket starts from rest at with mass M. Fuel is ejected at speed u relative to the rocket.

Homework Equations



m1v1 = m2v2

The Attempt at a Solution


In the textbook, it starts off with a moving rocket with mass m and speed v. The fuel is given the mass (-dm) which is positive. So after a short time dt, the mass of the rocket changes to m+dm and speed v+dv. The mass of the ejected fuel is (-dm) and since it ejects with speed u relative to the rocket traveling at speed v, the ejected fuel travels at speed v-u, which can be positive or negative depending on which of v or u is larger.

Writing down the equation of conservation of momentum:

mv = (m+dm)(v+dv) + (-dm)(v-u)

which then leads to

m dv = -u dm

After a few steps, we get:

v2-v1 = u ln(m1/m2)


That all looks fine and understandable. However, it mentions in the book that I am free to define dm to be positive, and then subtract it from the rocket's mass, and have dm get shot out the back. So I have decided to try it.

Writing down the equation of conservation of momentum:

mv = (m-dm)(v+dv) + dm(v-u)

which then leads to

m dv = u dm (note that at this point, the equation is already different from before)

Moving the variables around and integrating v from v1 to v2, m from m1 to m2 as before, I obtained:

v2-v1 = u ln(m2/m1)

which is clearly wrong because m2<m1, so

ln(m2/m1) < 0

but

v2 > v1


So what went wrong there?
 
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  • #2
Well, thanks for an evening's entertainment! I could not wrap my mind around it. Clearly dm has to be negative in order for the value of m to diminish. But it is natural to think of dm as positive when we write (m-dm). Conflict! I found clarification here
http://ed-thelen.org/rocket-eq.html
Some explanations use one m for the rocket and another m for the propellant.
Anyway, it seems one must use different letters for the m in dm and the M for mass of the rocket. Then dM = - dm and it all works out.

How clever of the author of your book to make us really think!
 
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FAQ: Rocket motion - conservation of momentum

1. What is the law of conservation of momentum?

The law of conservation of momentum states that the total momentum of a system of objects remains constant unless acted upon by an external force. This means that in a closed system, the total momentum before and after an event will be the same.

2. How does rocket motion demonstrate conservation of momentum?

In rocket motion, the momentum of the rocket and its exhaust gases must be conserved. As the rocket accelerates, the exhaust gases are expelled at a high velocity in the opposite direction, resulting in a net momentum change of zero for the system. This demonstrates the conservation of momentum.

3. What is the role of mass and velocity in rocket motion?

The mass of the rocket and the velocity at which the exhaust gases are expelled both play a crucial role in rocket motion. According to the law of conservation of momentum, the product of mass and velocity must remain constant. Therefore, as the mass of the rocket decreases due to the expelled gases, the velocity must increase in order to maintain the same total momentum.

4. How does the shape and design of a rocket affect its motion?

The shape and design of a rocket can greatly impact its motion. A streamlined shape reduces drag and allows the rocket to achieve higher velocities. Additionally, the placement of fins and the size and shape of the nozzle can affect the stability and control of the rocket during motion.

5. What are some real-life applications of the conservation of momentum in rocket motion?

The conservation of momentum is crucial in rocket motion and has many real-life applications. It is used in the design and construction of rockets, as well as in space exploration and satellite launches. It is also applicable in other fields such as aerodynamics, ballistics, and sports like skiing and ice skating.

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