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(18a) Momentum

A cannon weighing 1 ton (metric ton, 1000 kilograms) fires a 10 kg shell at 1000 meters/second. With what velocity does the cannon recoil?

Recoil

At first sight something is missing here. How can Newton's laws be applied to a problem where neither force nor acceleration are given? But hold on!

Let us start the solution process by collecting all the given information. Denoting with subscript 1 quantities related to the shell and with subscript 2 those related to the cannon, we have

Shell: m₁ = 10 kg v₁ = 1000 m/s
Cannon: m₂ = 1000 kg v₂ = ?

When a problem involves unspecified quantities, the best strategy usually is to give each of them a name and hope for the best: either their actual values can be found, or they will "drop out," showing they are not essential to the solution.

Such "miracles" may happen if all the information needed for the answer is already given. You will need some experience, though, to judge whether or not all vital pieces of information are already on hand.

What missing quantities are needed by the equations? For starters, the forces (F₁, F₂) on the shell and cannon, the magnitudes of their accelerations (a₁, a₂) which we assume to be constant (we know they are in directions opposite to each other), and the length of time t during which they act. We now collect equations:

From the second law

₁

₂

The third law says the forces are equal in magnitude, so

m₁a₁ = m₂a₂ (1)

which removes forces from the picture. Also:

v₁ = a₁t v₂ = a₂t

Divide left by left and right by right in the last equation, to get

v₂/v₁ = a₂/a₁ (2)

which removes the time t. Going back to equation (1), divide both sides by a₁m₂. Then

m₁/m₂ = a₂/a₁ (3)

Substitute from (2) to (3), or vice versa, and suddenly the accelerations, too, are gone. All that is left is

m₁/m₂ = v₂/v₁ (4)

where everything is known except for the recoil velocity v₂. Its value is isolated by multiplying both sides by v₁

v₂ = v₁ (m₁/m₂) = 1000 (10/1000) = 10 m/sec

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

m₁v₁ = m₂v₂ (5)

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

Actually, 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

v₁= 1000 m/s P₁ = m₁v₁ = 10 kg x 1000 m/s = 10,000 kg-m/s

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.