We now examine further applications of Newton's 3rd law, focusing on the rocket principle, which is one of the most dynamic forms of it. Early in the last century it was firmly believed that no heavier than air craft would ever work in space because there was "nothing to push against". It couldn't be imagined how something like a rocket could work in a vacuum.
What wasn't figured in is that a kind of "automatic momentum" change is built into the rocket system (see, e.g. attached images). Thus, its motion changes when a fraction of its mass (delta m) is released in the form of ejected gases. Since these ejected masses acquire their own momentum, the rocket itself receives a compensating momentum in the opposite direction. Therefore, the rocket is accelerated as a result of a push or thrust from the gases. In free space, or a vacuum, the entire system moves independently of the propulsion process.
Using the lower diagram as a guide, assume at somt time t, the momentum of the rocket shown plus fuel is: (M + delta m)v. Then at some later time (t + delta t), tjhe rocket ejects some fraction of mass, delta m, so the rocket's speed increases to (v + delta v). By using Newton's 3rd law in an application of conservation of momentum, we can write: total initial momentum of the rocket system = total final momentum of the system and we get (writing 'd' instead of delta to save space):
(M + dm)v = M(v + dv) + dm(v - v')
where v' is the velocity with which the fuel is ejected relative to the rocket. The equation can then be simplified to give:
M dv = v' dm
Which can easily be integrated. However, since this is a blog on basic mechanics and not advanced, calculus-based mechanics, I will simply give the final result after some further working. We than obtain the "rocket equation":
v(f) - v(i) = v' ln (M(i)/M(f))
where the velocities on the left side denote the final and initial values, respectively, and M(i) refers to the initial rocket mass, while M(f) refers the final rocket mass (with fuel expended). In general, for most rockets, M(i) >> M(i). Let's look at how it is applied, say for a model rocket launch such as shown in the photo.
A group in central Colorado fires a model rocket of mass 1 kg, which uses a solid fuel ZnS formula and the exhaust gases attain a velocity of 100 m/s relative to the rocket, and for 3 seconds. After this time the rocket mass has decreased to 0.05 kg.Find the rocket's acceleration and estimate the altitude assuming zero air drag and a perfectly vertical launch.
We have for the model rocket: M(i) = 1 kg, M(f) = 0.05 kg
Therefore
:
v(f) - v(i) = v' ln (M(i)/M(f)) = [100 m/s] ln (1/0.05) = (100 ms) ln(20)
v(f) - v(i) = 300 m/s
The acceleration is then: a = [v(f) - v(i)]/ delta t
a = 300 m/s/ 3 s = 100 m/s/s
The altitude can then be estimated using the kinematic equation:
s = ½ a t^2 where s is the vertical displacement.
s = ½ (300 m/s^2) (3s)^2 = 450 m = 1485 ft.
Note that Newton's 3rd law applies in many other forms as well, including to any devices which use something similar to the rocket principle such as balloons. In addition, the application of conservation of linear momentum can be used in such varied settings as finding the recoil velocity of a rifle firing a bullet at some defined speed, as well as the fission of a radioactive nucleus.
Problems:
1) A liquid fuel rocket traveling in space has a speed of 3, 000 m/s. Its engines are turned on and fuel is ejected in a direction opposite to the rocket's motion at a speed of 5,000 m/s relative to the rocket. Find the speed of the rocket once its mass is reduced to one half its mass before ignition.
2) A toy balloon is found to move at a speed of 10 m/s after all its air is expelled. Assuming its initial speed = 0.001 m/s estimate the mass of the balloon alone if its initial mass is 0.1g and the velocity of ejected air relative to a stationary frame is found to be 4 m/s.
3)A rifle of mass 5 kg fires a bullet of mass 0.002 kg which attains a velocity of 200 m/s. Find the velocity of recoil of the rifle.
4)A toy trolley of 1 kg mass has a wooden support affixed to one end and rests on a frictionless track. An arrow of mass 100 g is fired at the wood support with a velocity of 20 m/s. Find the velocity with which the trolley moves off with the arrow embedded in it after impact.
5) A radionuclide of mass 210u (u denotes atomic mass units) undergoes fission with the release of an alpha-particle of mass 4u. Find the kinetic energy of the residual nucleus.
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