
From the second law
F_{2} = m_{2}a_{2}
which removes forces from the picture. Also:
Divide left by left and right by right in the last equation, to get
which removes the time t. Going back to equation (1), divide both sides by a_{1}m_{2}. Then
Substitute from (2) to (3), or vice versa, and suddenly the accelerations, too, are gone. All that is left is
where everything is known except for the recoil velocity v_{2}. Its value is isolated by multiplying both sides by v_{1}
Our final equation (4) becomes more symmetric if all fractions are eliminated. Multiplying both sides by the product (m_{2}v_{1}) of the denominators gives
The quantity "mass times velocity" (or "the product of mass and velocity") is called momentum (plural: momenta) and is often denoted by the letter P. One way of interpreting the last equation (5) is to state that the momentum given to the cannon equals the momentum given to the shell.
Conservation of MomentumActually, the momentum P, like the velocity v, is a vector quantity. Suppose we regard velocities in one direction as positive and in the opposite direction as negative. Then

v_{1}= 1000 m/s
P_{1} = m_{1}v_{1} = 10 kg x 1000 m/s = 10,000 kgm/s v_{2}=  10 m/s P_{2} = m_{2}v_{2} = 1000 kg x ( 10 m/s) = 10,000 kg 
Before the gun was fired, neither mass had any velocity and therefore the total momentum P = P_{1} + P_{2} was zero. Afterwards, evidently, the total momentum remained zero. This is a general property (and yet another formulation of Newton's laws) and can be stated as In a system of objects subject to no forces from the outside, the vector sum of all momenta stays the same ("is conserved"). This also works when three or more objects are involved and each moves along a completely different direction. For instance, "the shells bursting in the air" of the US anthem had the same momentum as the collection of fragments and gases produced imediately after they burst, before air resistance had its say. This is also the principle by which a rocket operatesas it throws mass backwards in a fast jet of gas, it receives an amount of forward momentum equal to the backward momentum given to the jet.
EnergyWhen the cannon recoils, it receives as much momentum as the shell. How is the energy E shared? Since
and for the cannon
Very unequal sharing! The cannon, 100 times massive, receives 100 times less energy. Is that the rule?
Dividing
Substitute equation (1) in the numerator, then cancel (m_{2}v_{2}) above and below to get
By (4), inverted
hence
The lighter mass always receives the lion's share of the energy! 
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Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 9212004
Reformatted 24 March 2006