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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,

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
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Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
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David
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emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
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Maurice
answer
Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
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Source:  OpenStax, Unit 6 - momentum. OpenStax CNX. Jan 22, 2016 Download for free at https://legacy.cnx.org/content/col11961/1.1
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