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Confused on impulse

  1. Apr 3, 2014 #1
    I am very confused on these questions...help please :)
    stuff in () is what I think the answer is but am not sure.
    A 100 kg cart and a 200 kg cart move toward each other. The 100 kg cart initially moves at +5 m/s while the 200 kg cart moves at -10 m/s.

    1. What is impulse in terms of momentum? (Impulse occurs as force is applied to an object over a period of time. Thus, produces a change in momentum.)

    2. How does impulse change if the carts are in contact for a shorter period of time? (if the cart is in contact for shorter period of time, then the impulse would decrease or be shorter.)

    3. What is the impulse of the 100kg cart if it moves -3 m/s after collision?

    4. What is the of the 200 kg cart after the collision?

    5. what would a graph look like to exemplify this scenario?
  2. jcsd
  3. Apr 3, 2014 #2

    Simon Bridge

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    I usually get impulses when I'm confused: best to resist them but then.... where's the fun in that?!

    1. concise statement of the definition of impulse.
    this answer sounds like you don't know what impulse is.
    you are almost there: impulse = change in momentum.

    2. you think there is less change in momentum in a short contact?

    3+ can you calculate momentum?
  4. Apr 3, 2014 #3
    Your right, Simon Bridge, I don't know what it is. I missed my first semester of physics and my school is less than helpful with catching me up. I was just trying to find somewhere that someone could explain these things to me. But thanks for your help, I guess.
  5. Apr 3, 2014 #4


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    I disagree with Simon, your wording is better than his.
    what impulse IS: "Impulse occurs as force is applied to an object [strike]over[/strike]a period of time.
    careful with using "over", as it might suggest division ... try thru a duration of time, which suggests multiplication
    what impulse DOES: "Thus, produces a change in momentum.

    Is it the time, or the Force, or their product, that produces the momentum change?

    I suspect they want you to calculate the numerical change in momentum that these carts will undergo during their collision, even in part (1).

    Do you know what momentums to expect, after the collision?
  6. Apr 3, 2014 #5

    Simon Bridge

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    ... but I can always back up my assertions :)
    That is when impulse happens, that is not what impulse is.

    You can also define impulse as the product of the average force and the time over which it is applied, or the area under the force-time graph. This can be shown to be the same as the change in momentum, thus impulse is change in momentum" in the same way that force is the rate of change in momentum.
    ref: http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html

    ... but emma was very close, and you are correct: mentioning a multiplication between force and time would have gained the mark.
    It is as accurate to say that a change in momentum produces an impulse.

    The derivation is from Newton's laws where the force is defined as the rate of change of momentum ... not as something that produces a rate of change of momentum.

    Remember, "force" (like impulse) has a special meaning in physics.
    Your intuition is that you apply a force and then, right after that, things move ... thus the force causes the change in the motion. Here you'd think of an impulse as a quick shove.

    In physics, the force is simultaneous with the acceleration, thus force is ma.
    It is also possible for a change in momentum to result in a force on another body, whose momentum subsequently changes - which spoils the intuitive cause and effect but is fine in physics ... with small issues about propagation of energy as a wave and Newton's cradle etc.

    Which way it is taught at this level is a matter of taste.
    As far as the person doing the marking is concerned, any of the three would be acceptable.
    Last edited: Apr 3, 2014
  7. Apr 3, 2014 #6

    Simon Bridge

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    How it's taught is mostly a matter of taste: I go by

    "impulse is change in momentum, which is calculated by finding the area under the F-t graph for the interaction."

    F-t graphs for simple head-on collisions tend to look like inverted parabolas. F(t)=-At(t-T) where T is the "contact time" for the collision and A is a constant of proportionality. [A]=[F]/[T]2.

    ... it's OK, I have to work out where to start from before I can answer your questions helpfully.

    This link should help you catch up:
    ... so I'll stick to what that link does not cover.
    Hopefully you know about forces and Newton's laws?

    (2) This question is about how to calculate the impulse.
    You need the contact time and the average force to know the momentum, so knowing only the time is shorter is not enough information. You need to explain this.

    I have a feeling that the question refers to something in the course that you missed so everyone elses answers will be shorter. You will have to cover the bases to get the mark.

    (3) change in momentum - you know how to do that right?

    (4) what was the impulse applied to the 200kg cart?

    (5) what sort of graph do they have in mind? a v-t graph or a F-t graph? Something else?
    ... the proper context is in the lesson you missed.

    I suspect they want a v-t graph for both carts on the same axes.
    At this level is is common to show the change in velocity as a uniform acceleration during the contact time.
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