# (18b) Momentum

From the second law

F1 = m1a
F2 = m2a2
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.

## Energy

When the cannon recoils, it receives as much momentum as the shell. How is the energy E shared? Since

E = mv2/2
we have, for the shell

E1 = 10 kg (1000m/s)2/2 = 5,000,000 joule

and for the cannon

E2 = 1000 kg (10 m/s)2/2 = 50,000 joule

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

E1 = m1v12/2

E2 = m2v22/2

Dividing

E1/E2 = m1v12/m2v22

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

E1/E2 = v1/v2

By (4), inverted

v1/v2 = m2/m1

hence

E1/E2 = m2/m1                 (6)

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

Next Stop: #18c Work

Author and Curator:   Dr. David P. Stern
Mail to Dr.Stern:   stargaze("at" symbol)phy6.org .

Last updated: 9-21-2004
Reformatted 24 March 2006