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What are momentum and impulse?

Learn what momentum and impulse are, as well as how they are related to force.

What is momentum?

Momentum is a word that we hear used colloquially in everyday life. We are often told that sports teams and political candidates have "a lot of momentum". In this context, the speaker usually means to imply that the team or candidate has had a lot of recent success and that it would be difficult for an opponent to change their trajectory. This is also the essence of the meaning in physics, though in physics we need to be much more precise.
Momentum is a measurement of mass in motion: how much mass is in how much motion. It is usually given the symbol p.
By definition, start box, p, equals, m, dot, v, end box, point
Where m is the mass and v is the velocity. The standard units for momentum are k, g, dot, m, slash, s, and momentum is always a vector quantity. This simple relationship means that doubling either the mass or velocity of an object will simply double the momentum.
The useful thing about momentum is its relationship to force. You might recall from the kinematic equations that change in velocity delta, v can also be written as a, dot, delta, t.
We can then see that any change in momentum following an acceleration can be written as
Δp=mΔv=maΔt=FΔt\begin{aligned} \Delta \mathbf{p} &= m\cdot \Delta v\\ &= m\cdot \mathbf{a}\cdot \Delta t \\ &= \mathbf{F}\cdot \Delta t\end{aligned}

What is impulse?

Impulse is a term that quantifies the overall effect of a force acting over time. It is conventionally given the symbol start text, J, end text and expressed in Newton-seconds.
For a constant force, J, equals, F, dot, delta, t.
As we saw earlier, this is exactly equivalent to a change in momentum delta, p. This equivalence is known as the impulse-momentum theorem. Because of the impulse-momentum theorem, we can make a direct connection between how a force acts on an object over time and the motion of the object.
One of the reasons why impulse is important and useful is that in the real world, forces are often not constant. Forces due to things like people and engines tend to build up from zero over time and may vary depending on many factors. Working out the overall effect of all these forces directly would be quite difficult.
When we calculate impulse, we are multiplying force by time. This is equivalent to finding the area under a force-time curve. This is useful because the area can just as easily be found for a complicated shape—variable force—as for a simple rectangle—constant force. It is only the overall net impulse that matters for understanding the motion of an object following an impulse.
The concept of impulse that is both external and internal to a system is also fundamental to understanding conservation of momentum.

Momentum in space

Most people are familiar with seeing astronauts working in orbit. They appear to effortlessly push around freely floating objects. Because astronauts and the objects they are working with are both in free-fall, they do not have to contend with the force of gravity. However, heavy moving objects still possess the same momentum that they do on earth, and it can be just as difficult to change this momentum.
Suppose that an emergency occurs on a space station and an astronaut needs to manually move a free-floating 4,000 kg space capsule away from a docking area. On earth, the astronaut knows she can hold a 50 kg weight above herself for 3 seconds. How quickly could she get the capsule moving?
We first calculate the total impulse that the astronaut can apply. Note that the astronaut is pushing vertically in both cases so we don't need to keep track of the direction of the force.
J=(mg)Δt=50 kg9.81 m/s23 s=1471.5 Ns\begin{aligned} J &= (mg)\cdot \Delta t\\ &= 50~\mathrm{kg} \cdot 9.81~\mathrm{m/s^2} \cdot 3~\mathrm{s} = 1471.5 ~\mathrm{Ns} \end{aligned}
And, by the impulse-momentum theorem, we can find the velocity of the spacecraft:
start fraction, 1471, point, 5, space, N, s, divided by, 4000, space, k, g, end fraction, equals, 0, point, 37, space, m, slash, s

What is specific impulse?

Specific impulse—I, start subscript, S, P, end subscript—is a specification commonly given to engines which produce a thrust force. Jet engines and rocket engines are two common examples. In this context, specific impulse is a measure of the efficiency of using fuel to produce thrust and is one of the most important specifications of such an engine.
When the prefix specific is used in physics, it means "relative to" a particular quantity. Specific gravity and specific heat are two examples of where you may have seen this prefix used. Specific impulse is impulse measured relative to the weight of fuel—on earth—used to produce the impulse.
I, start subscript, S, P, end subscript, equals, start fraction, F, dot, delta, t, divided by, m, start subscript, f, u, e, l, end subscript, g, end fraction
Because we are dividing an impulse by a force—the force on the fuel due to earth's gravity—the force units cancel out, and the units for specific impulse are simply seconds.
A rocket might have a specific impulse of 300 s. This means that it could use fuel weighing 1 N to produce 1 N of thrust for 300 s. In practice, the rocket might have some minimum thrust, say 100 N. In this case it could use fuel weighing 1 N to produce the 100 N thrust for 3 s.

Impulse of an aircraft

A Boeing 747 aircraft has four engines, each of which can produce a thrust force of up to 250 kN. It takes around 30 s for the aircraft to get up to take-off speed. The thrust produced by the engines during take off is approximated by the force-time curve shown below.
CF6 Engine thrust during take off of Boeing 747 [1]
CF6 Engine thrust during take off of Boeing 747 [1]
Exercise 1a: What is the total impulse produced by the aircraft in getting up to take-off speed?
Exercise 1b: The specific impulse of the jet engines is known to be around 6000 s. How many kilograms of fuel were burned in getting the aircraft up to take-off speed?

Attributions

  1. Data from (a) W.A. Fasching 9/1979 NASA-CR-159564 CF6 Jet Engine Performance Improvement Program (b) Project for the Sustainable Development of Heathrow, Ch 3 – Emission Sources. 7/2006.

Want to join the conversation?

  • old spice man green style avatar for user spaceboytimi
    the problem in the paragraph momentum in space , i dont think the acceleration should be 9.8 ms^-2 cuz gravitational acceleration decreases with increasing altitude ! 9.8 ms^-2 is only applicable for objects on the earth under 100m altitude ! am i right ?
    (12 votes)
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    • leaf green style avatar for user Richard Graham
      The first part of the question requires us to find out how much impulse the astronaut is capable of applying which depends on how strong she is. This is found based on her ability to lift weights on earth as given in the problem.. Knowing this we can find the effect of the same impulse on the spacecraft.
      (63 votes)
  • blobby green style avatar for user nj.nicole14
    I as well am not really understanding what impulse is exactly.
    (14 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      so....
      if you want to change the momentum of an object, you have to apply a force. OK??


      But if you apply that force for a long time then the momentum will change more.

      and the effect of the force AND the time the force is applied, is the impulse of the force.

      so the change in momentum = size of force x time for which the force was applied

      ok??
      (43 votes)
  • hopper jumping style avatar for user Sam D
    So from what I understand, Impulse is change in momentum but it is a bit different. For instance Acceleration is rate of change of velocity v/t ,velocity is rate of change of position x/t but impulse is not F/t it's F*t , furthermore p= f*t ==> f=p/t This indicates that force is rate of change of momentum. Can someone help me here? (P.S. I haven't done Calculus yet but I think that integral is a function for area under the curve and derivative would be for slope {rate of change} So what's the derivative of momentum, since its integral is impulse )
    (4 votes)
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    • blobby green style avatar for user Surya Bhushan Tripathi
      Yes! That's entirely correct.
      Impulse is a certain amount of force you apply for a certain amount of time to cause a change in momentum. That is why it is F*t. For example, when you hit a ball with a cricket bat, you apply a force for a time(a very short period in this case) to cause a change (or transfer) of momentum in the ball.
      Also, you saying that F=p/t is correct. When we focus on the mathematics a little bit, it is quiet obvious. F=p/t => F=m*v/t => F=m*a (here the change in the quantities is considered).
      (6 votes)
  • blobby green style avatar for user wafteacher77
    Hi, what are some of the most important everyday uses of momentum? I mean how can I benefit from my simple physics knowledge about momentum in my daily life?
    (3 votes)
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    • duskpin ultimate style avatar for user Cindy Phan Nguyen
      I think airbag is one of of it . Force and time is inversely proportional in momentum calculation so if time is increased, force is decreased. The airbag is designed so that it can increase the time required to stop our body momentum in a collision, reducing force impact, minimise injury to our body.
      (11 votes)
  • mr pants teal style avatar for user ritesh
    what is the difference between force and thrust?
    (4 votes)
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  • starky seedling style avatar for user Arundhati
    I still don't understand how to calculate the change in momentum? Can someone help?
    (2 votes)
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    • male robot donald style avatar for user Aaron Ghosh
      Hi Arundhati,
      Δp = F_net * Δt is the equation to calculate the change in momentum. F_net is the net external force,
      Δp is change in momentum, and Δt is the time over which a net force acts. Change in momentum is proportional to the net external force and the time over which a net force acts.
      Hope this helped!
      Aaron
      (1 vote)
  • blobby green style avatar for user Jyoti Uppal
    Hello. In the "Impulse of an aircraft", it says that the aircraft "can produce a thrust force of up to 250 kN" and that it "takes around 30 s for the aircraft to get up to take-off speed". I understand how to take the area under the curve, but we know the force and we know the time. Why can't I just multiply the thrust force of 250 kN and 30 s to get impulse? What is the difference between the impulse from the multiplication I just described and the area under the curve? Thank you.
    (2 votes)
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  • duskpin seedling style avatar for user Nicolette Foster
    A question on the force-time curve (in the Impulse of an Aircraft paragraph): the instantaneous slope of the curve is the momentum, right? Or is it something else
    (2 votes)
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  • blobby green style avatar for user digestingquasar
    If impulse is the change in momentum, how would one take its derivative? Would it be J/dt = F, since J approaches zero and is thus the same as dp in dp/dt = F?
    If that's the case, then it would also hold for work and kinetic energy, right, since W = ΔKE. That is, W/dt = P because dKE/dt = P.
    (2 votes)
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    • male robot hal style avatar for user Satwik Pasani
      When we take a derivative of anything (let us call it X) with respect to time, we always do dX/dt. What this derivative means depends on the X we begin with. dp/dt is the rate of change of momentum. dJ/dt is the rate of change of the impulse J, or, the rate of change of the change in momentum (which does not make much intuitive sense). If J is the momentum change that happens in a very tiny amount of time (dt), then J/dt will be dp/dt =F, but J/dt is not the derivative of J.
      (1 vote)
  • duskpin ultimate style avatar for user Aquila Mandelbrot
    A question on the force-time curve (in the Impulse of an Aircraft paragraph): the instantaneous slope of the curve is the momentum, right? Or is it something else?
    (1 vote)
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