Ref: Momentum Basics
Momentum is mass times velocity. It typically uses symbol: "p". p = m * v p = (2.0 kg)(4.0 m/s) p = 8.0 kgm/s Thus, we calculate this object's momentum to be 8.0 kgm/s. Momentum's dimensions are kgm/s, pronounced: 'kilogram meter per second'. Momentum is also directly proportional to mass. With constant velocity, momentum varies with the mass. Example: triple the mass of an object, then the momentum also triples. Momentum is a conserved quantity. Within a closed system of interacting objects, total momentum of entire system maintains a constant. When object A collides with object B in an isolated system, the total momentum of the two objects before the collision is the same as the total momentum after the collision. Any momentum lost by object A goes to object B.  Thus, total momentum of a system of objects is "conserved"; total amount of mass times velocity is static. To further understand momentum conservation, consider Newton's Law of Momentum Conservation. When two objects collide, the forces acting between the two objects are equal in magnitude and opposite in direction. In equation form, m*v = M * V. The forces between the two objects are equal in magnitude and opposite in direction, which leads to Newton's Law of Momentum Conservation. In a collision, the momentum change of object A is equal and opposite to the momentum change of object B. That is, the momentum lost by object A is equal to the momentum gained by object B. In a collision between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object A loses 75 units of momentum, then object B gains 75 units of momentum. Yet, the total momentum of the two objects (object A plus object B) is the same before the collision as it is after the collision; the total momentum of the system (the collection of two objects) is conserved. For any collision occurring in an isolated system, momentum is conserved  the total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. Momentum can be readily observed by watching billiard balls collide. 
Similarly, high velocity gas goes in one direction, and rocket goes in opposite direction. Action causes reaction.
Recall momentum exchange: M * V = m * v
Rocket and jets propel themselves due to momentum transfer where high speed exhaust particles move a much larger vehicle in the opposite direction to a much slower but still useful speed. This momentum exchange between fast moving fuel particles and a slow moving spaceship can be expressed:
First, we choose to increase the speed of a large space vessel by 10 m/s in one second. Thus, set V_{ship} be a constant 10 m/s. This relates to the approximate value of g = 10 m/s^{2} acceleration due to gravity near Earth’s surface. Thus, assume our notional spaceship can expel enough high speed particles to increase velocity an additional 10 m/s for every second of spaceflight. The resultant force will simulate Earthlike gravity for ship and contents.
2nd, we further choose to expend one gram of particles in one second. I.e., set m_{fuel} to be a constant 1.0 gram for convenience.
3rd, our thot experiment uses an onboard particle accelerator to give these fuel particles extremely high speeds. Since current devices easily accelerate particles to 99% of light speed, our thought experiment will compare sustainable fuel exhaust velocity, v_{fuel}, to light speed, c. Furthermore, we now choose to name v_{Exh} to more clearly indicate exhaust velocity of particles exiting vessel. Finally, we choose to independently vary v_{Exh} by incrementally increasing from row to row in the following tables. Thus, v_{Exh} is our Independent Variable (IV).
4th, solve for the remaining term, M_{ship}, mass of ship, to determine how much ship can be propelled 10 m/s by a certain amount of fuel mass exiting vehicle at high speed in opposite direction of travel. Since M_{ship} depends on the value of v_{Exh}, M_{ship} is our Dependent Variable (DV).
M_{ship} = m_{fuel} * v_{Exh} / V_{ship}
(Note: Following tables will consider only Newtonian concepts and will disregard relativistic mass increase for now. Relativistic mass growth will be considered later
M_{ship} = m_{fuel} * v_{Exh} / V_{ship}  
M_{ship}  V_{ship}  m_{fuel}  _{} 
Ship's Mass  ship's vel  fuel mass  exhaust vel 
Grams  m/s  grams  m/s 
1,000,000  10.0  1.0  10,000,000 
2,000,000  10.0  1.0  20,000,000 
3,000,000  10.0  1.0  30,000,000 
DV  Const.  Const.  IV 
While above calculations are straight forward, two terms have a lot of digits which might prove inconvenient. We can streamline this process through a conversion contant.
TABLE2. Express v_{Exh} as d_{c }* c; decimal c times light speed
For convenience, use a mix different velocity units. In particular, we'd like to continue expressing ship's velocity increase ,V_{ship}, as 10 meters per second; but we'd like to change v_{Exh} from m/s to decimal light speed, d_{c} c, where d_{c} is decimal component of particle exhaust speed expressed in terms of light speed.
Recall light speed is approximately 300 million meters per second, c = 300,000,000 m/s.
Thus, Table1 proposes example value for v_{Exh} of .1 c, 30,000,000 m/s = 10% c. (Note: Disregard relativity effects for now; we'll discuss that later).
M_{ship} = m_{fuel} * K_{2} * v_{Exh}/ V_{ship}  
K_{1} = 300,000,000  
M_{ship}  V _{ship}  m _{fuel}  v_{Exh} 
Ship's Mass  ship's vel  fuel mass  exhaust vel 
Grams  m/s  grams  d_{c} c 
3,000,000  10.0  1.0  .1 c 
6,000,000  10.0  1.0  .2 c 
9,000,000  10.0  1.0  .3 c 
12,000,000  10.0  1.0  .4 c 
DV  Const.  Const.  IV 
Consistent values requires conversion constant. Note that V_{Exh} has equivalent values in last row of Table 1. and first row of Table 2. For consistent M_{ship}, determine conversion constant K_{2} such V_{Exh} can be expressed as decimal light speed.
TABLE3. Convert M_{ship} to metric Tonnes (mT)
By definition: 1 mT = 1,000,000 gms; thus, example value of M_{Ship}=3,000,000 gms = 3 mT.
K_{3}=  M_{ship} * V_{ship} m_{fuel }* v_{Exh}  =  3mT * 10m/s 1 gm * .1c  =  3 c 

Thus, k_{3} = 3/c, given following set of dimensions:


TABLE4. Convert to Rates
Changing to mixed units affects the terms: v_{Exh} and M_{ship}; thus, unaffected terms (V_{ship} and m_{fuel}) keep the same dimensions. However, those terms (V_{ship} and m_{fuel}) can be usefully converted to rates.
Start with the original shipfuel momentum equation:
M_{ship} * V_{ship}  =  m_{fuel} * v_{Exh} 

and divide both sides by one second:
M_{ship} * V_{ship} 1 second  =  m_{fuel} * v_{Exh} 1 second 

M_{ship} *  V_{ship} 1 sec  =  m_{fuel} 1 sec  * v_{Exh} 

M_{ship} * g = (m_{fuel}/1 sec) * v_{Exh}
Further, we can introduce new term, fuel flow per sec (ff_{sec}), and substitute for the expression: m_{fuel} /sec. Thus, far we've arbitrarily chosen the value of 1 gram per second for convenience, but we will eventually use this term for any value.
M_{ship} * g = ff_{sec} * v_{Exh}


k =  M_{ship} * g ff_{sec }* d_{c}c  =  3 mT * g 1 gm/sec * .1c  =  30 g c 

1. Microscale. Spread the amount of fuel ejected in a smooth continuous manner throughout the entire second.
2. Macroscale. Reliably repeat above item for each of the 86,400 seconds during each day of space flight.
Both above items are easily done now in various modes of transportation, but can they be done for particles accelerated to large portions of c, speed of light???
Summary
Give terms following values:
M_{ship} = (1.0 gm * 30,000,000 m/s) / (10 m/s) M_{ship}= 3,000,000 gm 

 1,000,000 gms = 1,000 kgms = 1.0 metric Tonnes = 1 mT M_{ship} = k_{3} * m_{fuel} * v_{Exh} / V_{ship} where k_{3} = 300 
Particle velocity, v_{Exh}, of .1 c contributes sufficient momentum so that one gram of exhaust particles can propel a 3 ton vessel to an additional 10 m/sec. Since momentum is directly proportional to velocity, increasing v_{Exh} another .1 c will propel another 3mT of ship's mass.  Recall: m_{fuel} * v_{Exh} = M_{ship} * V_{ship}.
Recall that we set V_{ship} equal to 10 m/s; thus, V_{ship}/ 1sec = 10m/s/s = g, acceleration due to gravity near Earth’s surface. Introduce new term, fuel flow per second: ff_{sec} = m_{fuel}/sec. Thus, we substitute to get: ff_{sec}* v_{Exh} = M_{ship} * g To continue expressing v_{Exh} as decimal c and M_{ship} in metric Tonnes, we'll introduce conversion constant, k = 300/c. Thus, k * v_{Exh} = 300 * d_{c}.

Further Study, Related Concepts: Impulse and Specific Impulse
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Specific Impulse
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