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Hi there! I have a question about an experiment that was conducted. It is related to momentum.

2 carts were put on a track on opposite sides. They were then propelled towards one another at approximately the same speed.Please note that on each end of each cart, there was placed a negative magnet, so as soon as they came close to each other, repulsion commenced.

Two sensors were set on a frequency of 20 Hz to record their motion. After collecting the data, the data for the difference in momentum was graphed through the equation m2v2 - m1v1. The group did this to calculate the approximate average frictional force of the system by producing a linear fit. The slope of this line proved to be the average friction of the system.

This was an extremely simple experiment and did not prove at all difficult.

However, there is one question I have been asking myself that is driving me crazy. I can't seem to find the answer (Please note that our instructor did not ask us for this).

Why are there spikes in the graph of Δp (change in momentum) very near the moment of collision (actually there is no "collision" because of the magnetic repulsion). Is it because the frequency was set only to 20Hz and thus since the instance of collision was so tiny, the sensors couldnt pick up data correctly? Is it because for the brief moment that the carts cease to move friction with the rail is static instead of kinetic and thus requires more force?

Some useful information: mass of cart1: 255.9g mass of cart2: 251.1g

Please note the equation: Δp= Δt*Favg and p = mv

An answer would be greatly appreciated as this question has been driving me crazy.

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Is the vertical scale of the graph position, or what? (There is no hope that you will get help without being clear about such things.) If it is, why do you think that it is a good idea to fit a single line to the before data before and after the interaction? –  dmckee Nov 19 '12 at 0:16
    
Please read the question very carefully before accusing me of being unclear. I mentioned twice what the y-axis represents: "the data for the difference in momentum was graphed through the equation m2v2 - m1v1" and "Why are there spikes in the graph of Δp (change in momentum)" implying it's clearly a Δp-t graph. Also please look at the title which clearly states IMPULSE. you missed it three times. –  Outlier Nov 19 '12 at 0:31
    
If you are serious about that then you seem to have an error in your analysis up to this point: that quantity should change signs during the interaction (which is why I was confused). Were you careful about the sign of the velocities? –  dmckee Nov 19 '12 at 0:42
    
yes. please note that this is the graph of change in momentum. this is not the graph of velocities, in which case, yes the signs of the velocities would change. –  Outlier Nov 19 '12 at 1:27
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1 Answer

up vote 1 down vote accepted

Let's consider this a bit. Label the carts with 1,2, and let the initial velocities be $v_{1,2}$ and the final velocities be $v'_{1,2}$.

So, the impulse delivered to cart 1 is $$ I_1 = \Delta p_1 = m_1 \delta v_1 = m_1(v'_1 - v_1) $$ and likewise $$ I_2 = \Delta p_2 = m_2 \delta v_2 = m_1(v'_2 - v_2) $$ but by Newton's law of reaction $I_2 = -I_1$

Now if you consider the quantities $\Delta P = m_1 v_1 - m_2 v_2$ and $\Delta P' = m_1 v'_1 - m_2 v'_2$ (where I have chosen the capital p simple to distinguish the difference of momenta from the individual momenta of the carts).

We construct $$\Delta P' - \Delta P = m_1 (v'_1 - v_1) - m_2 (v'_2 - v_2) = I_1 - I_2 = 2I_1 .$$

You would expect the graph that you use to show a discontinuity at the time of the interaction equal to twice the impulse.

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I understand now! Thanks so much! So are the spikes in the graphs supposed to represent the discontinuity of which you speak? –  Outlier Nov 19 '12 at 1:52
    
Excellent answer. But can I ask why you have subtracted the momenta of the carts? –  Outlier Nov 19 '12 at 6:58
    
Er...because you said you were plotting the difference in momenta and that got me started thinking along those lines. –  dmckee Nov 19 '12 at 15:03
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