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Momemtum Revisited

  1. Sep 22, 2004 #1

    Arg I have another question on momentum again, :grumpy: . I have looked through the physics book some more after getting help from you guys, thanks!

    Q. If a collision occurs, and its intial momentum is x, then the final momentum of the two objects will also sum up to x. During the collision, each object will experience the same average force for the time of impact as its counterpart.

    since we know that [tex] F=ma[/tex], we can re-write ma as:

    [tex] m * (\frac {v_1} {t_1} - \frac {v_2} {t_2} )[/tex]

    and since the change in time is equal for both, we get the momentum equation,

    [tex] m_1_i v_1_i + m_2_i v_2_i = m_1_f v_1_f + m_2_f v_2_f [/tex]

    Now what if the inital momentum of a body was again x as stated previously, but the initial mass and velocity of the two objects that will collide are different. In both cases we have the same total value of momentum x, but different individual momenta values for each particle.

    In the first senario, the collision might happen for a time t.
    In the second senario, the collision might happen for a time t/2.
    In addition, the curve of the force vs time diagram will most likely be different.

    Yet despite all these differences, the text book claims that the value of the area underneath the curve will be exactly the same, provided that we started with the same intial momentum value x in both cases.

    What I am having trouble seeing is how this equation PROVES that the area will be the same. From what I see, it only tells us that the average force that each body exerts the other will be the same. I do see how this gives us the momentum equation for ONE PARTICULAR collision.

    What I do not see, is how this proves that the area underneath the force vs time curve will be exactly the same in a different type of collision that has the same starting value of x.

    In both collisions, the beginning momentum is x; however, where is the evidence to prove to us that the force vs time AREA will be the same in any different collision case?

    Is this something newton discovered from experimentation alone, or is there a mathematical method for showing that the two areas indeed will be the same in the two different collision cases. (I am looking for a proof where we do not assume things, like we assume we know the final velocities of the two objects and from there can infer that the momentum after is the same as when it started. I think that would be very inacurate, because if we did assume that, we could not perform that experiment in a practical sense. We could do it on ice where friction is low, but friction is still there, and if the equation happened to say that the final momentum was 99.9/100 the starting value, then even the smallest amount of friction in the ice would give us final velocity values that are wrong, plus I seriously doubt newton had a handy digital stop watch and tape measure to find the final velocity and distance that accurate in a collision, especially with two bodies going off in different directions!)

    Thanks for any insight you can provide.

  2. jcsd
  3. Sep 22, 2004 #2


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    Think of force as the change in momentum per unit time rather than mass times acceleration.
  4. Sep 22, 2004 #3
    Ok, but how does that explain the simliar force vs time area in different initial conditions for a given total inital momentum value x?
  5. Sep 22, 2004 #4

    Doc Al

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    Staff: Mentor

    Could it be that you are misinterpreting what your book is saying?

    I see no reason to think that the area under the Force vs time will be the same. What you can deduce is that since the two objects exert equal and opposite forces on each other (regardless of the details of the interaction) for equal times the change in momentum (the impulse) of each object will be equal and opposite. Thus there is no change in the total momentum of the two-object system.

    Here's an example to illustrate why the Force x time on each object will depend on the details of the interaction, not just the initial total momentum. Imagine two identical objects heading towards each other at a speed of 1 m/s. Total momentum = 0. Now imagine the same two objects heading towards each other at 1000 m/s. Total momentum = 0. Let's assume that the collisions are totally inelastic. The area under the F-t curve for the first collision will be 1 N-s (per object); for the second collision it will be 1000 N-s.
  6. Sep 22, 2004 #5


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    I'm not sure if the following helps.

    The impulse on a object A
    [itex]\Delta \vec p_\text{on A}=\int \vec F_\text{net on A} dt=\displaystyle\frac{\int \vec F_\text{net on A} dt}{\int dt} \int dt=\vec F_\text{avg net on A}\Delta t[/itex].

    Assume A has some initial momentum [itex]\vec p_\text{A, initial} [/itex].
    Consider two collisions in which:
    in one, A is brought to rest in 1 sec
    in the other, A is brought to rest in 10 sec.
    In each situation, A has received the same impulse.
    Thus, the area under the [itex]F_\text{net on A}[/itex]-vs-[itex]t[/itex] is the same in both situations.
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