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  • Describe the principle of conservation of momentum.
  • Derive an expression for the conservation of momentum.
  • Explain conservation of momentum with examples.
  • Explain the principle of conservation of momentum as it relates to atomic and subatomic particles.

Momentum is an important quantity because it is conserved. Yet it was not conserved in the examples in Impulse and Linear Momentum and Force , where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum conserved?

The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is constant, even if momentum changes for components of the system. If a football player runs into the goalpost in the end zone, there will be a force on him that causes him to bounce backward. However, the Earth also recoils —conserving momentum—because of the force applied to it through the goalpost. Because Earth is many orders of magnitude more massive than the player, its recoil is immeasurably small and can be neglected in any practical sense, but it is real nevertheless.

Consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—for example, one car bumping into another, as shown in [link] . Both cars are coasting in the same direction when the lead car (labeled m 2 ) size 12{m rSub { size 8{2} } \) } {} is bumped by the trailing car (labeled m 1 ) . size 12{m rSub { size 8{1} } \) "." } {} The only unbalanced force on each car is the force of the collision. (Assume that the effects due to friction are negligible.) Car 1 slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total momentum of the two-car system remains constant.

A brown car with velocity V 1 and mass m 1 moves toward the right behind a tan car of velocity V 2 and mass m 2. The system of interest has a total momentum equal to the sum of individual momentums p 1 and p 2. The net force between them is zero before they collide with one another. The brown car after colliding with the tan car has velocity V 1prime and momentum p 1 prime and the light brown car moves with velocity V 2 prime and momentum p 2 prime. Both move in the same direction as before collision. This system of interest has a total momentum equal to the sum p 1 prime and p 2 prime.
A car of mass m 1 size 12{m rSub { size 8{1} } } {} moving with a velocity of v 1 size 12{v rSub { size 8{1} } } {} bumps into another car of mass m 2 size 12{m rSub { size 8{2} } } {} and velocity v 2 size 12{v rSub { size 8{2} } } {} that it is following. As a result, the first car slows down to a velocity of v′ 1 size 12{ { {v}} sup { ' } rSub { size 8{1} } } {} and the second speeds up to a velocity of v′ 2 size 12{ { {v}} sup { ' } rSub { size 8{2} } } {} . The momentum of each car is changed, but the total momentum p tot size 12{p rSub { size 8{"tot"} } } {} of the two cars is the same before and after the collision (if you assume friction is negligible).

Using the definition of impulse, the change in momentum of car 1 is given by

Δ p 1 = F 1 Δ t , size 12{Δp rSub { size 8{1} } =F rSub { size 8{1} } Δt} {}

where F 1 size 12{F"" lSub { size 8{1} } } {} is the force on car 1 due to car 2, and Δ t size 12{Δt} {} is the time the force acts (the duration of the collision). Intuitively, it seems obvious that the collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. This assumption must be modified for objects travelling near the speed of light, without affecting the result that momentum is conserved.

Similarly, the change in momentum of car 2 is

Δ p 2 = F 2 Δ t, size 12{Δp rSub { size 8{1} } =F rSub { size 8{1} } Δt} {}

where F 2 is the force on car 2 due to car 1, and we assume the duration of the collision Δ t size 12{?t} {} is the same for both cars. We know from Newton’s third law that F 2 = F 1 size 12{F rSub { size 8{2} } = - F rSub { size 8{1} } } {} , and so

Δ p 2 = F 1 Δ t = Δ p 1 . size 12{Δp rSub { size 8{2} } = - F rSub { size 8{1} } Δt= - Δp rSub { size 8{1} } } {}

Thus, the changes in momentum are equal and opposite, and

Δ p 1 + Δ p 2 = 0 . size 12{Δp rSub { size 8{1} } +Δp rSub { size 8{2} } =0} {}

Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is,

Practice Key Terms 3

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Source:  OpenStax, Physics subject knowledge enhancement course (ske). OpenStax CNX. Jan 09, 2015 Download for free at http://legacy.cnx.org/content/col11505/1.10
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