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 a₁m₂. 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 v₂. Its value is isolated by multiplying both sides by v₁

Our final equation (4) becomes more symmetric if all fractions are eliminated. Multiplying both sides by the product (m₂v₁) 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 Momentum

v₂= – 10 m/s     P₂ = m₂v₂ = 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 = P₁ + P₂ 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 operates--as 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.

Energy

and for the cannon

Very unequal sharing! The cannon, 100 times massive, receives 100 times less energy. Is that the rule?
    Yes, it is. We have, using symbols

Dividing

Substitute equation (1) in the numerator, then cancel (m₂v₂) above and below to get

By (4), inverted

hence

The lighter mass always receives the lion's share of the energy!