Why is the momentum not conserved

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SUMMARY

The discussion centers on the conservation of momentum during a collision between two vehicles with specific masses and velocities. Vehicle 1 has a mass of 13158 kg and initial velocity of 0.0246 m/s, while Vehicle 2 has a mass of 1726 kg and an initial velocity of 4.374 m/s. After the collision, their velocities change to 0.49 m/s and 0.91 m/s, respectively. The participants emphasize the importance of calculating momentum as a vector quantity, requiring both magnitude and direction, to accurately assess momentum conservation.

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  • #31
Probie1 said:
x = -14195.1592
y = -2626.994

P=√(Px^2 + Py^2)
P = 14436.19209

I'm not sure about this other thing
θ=Tan-1(y/x)
θ= 10.48

I can't tell where your x and y values have come from, nor can I tell which particular momentum vector they're supposed to represent.

Let's start by clearing up the initial momentum. Forget the 'departure' values for a moment.
Write the initial momenta of the two vehicles in component form:

p1o = xxxx i + yyyy j

p2o = xxxx i + yyyy j

where i and j designate the vector x and y unit vectors.
 
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  • #32
At this point I am lost completely... I have no clue what I am suppose to do.
 
  • #33
Probie1 said:
At this point I am lost completely... I have no clue what I am suppose to do.

You want to start by making a clear distinction between initial values and final values. So far you've been mixing all the parts in various places in your calculations. This leads to confusion.

So start by taking your initial values for the two vehicles and writing the momentum vector components for each. You've already done the calculations of the values, but they haven't been organized in such a way that they are clear to work with. The format that I showed in my last post is a convenient way to display them. You will have two vectors written in the form xxxxx i + yyyyy j, where xxxxx is the x-component value and yyyyy is the y component value.

Then we can combine these two vectors to find the total initial momentum. After that, we'll do the same for the final values. The results can then be directly compared.
 
  • #34
Okay let's try again.

weight 13158kg
heading 0 deg
departure 179
impact velocity 0.0246 m/s
departure velocity 0.49 m/s

weight 1726 kg
heading 190
departure 245
impact velocity 4.374 m/s
departure velocity 0.91 m/s

*V1ix = in V1ox = out


V1ix = 13158 * .0246 = 323.6868
V2ix = 1726 * -4.30 = -7421.8
V1ox = 1318 * -0.489 = -6434.262
V2ox = 1726 * -0.384 = -662.784


V1iy = .0246 * sin 0 = 0
V2iy = 1726 * -0.759 = -1310.034
V1oy = 13258 * 0.008 = 105.264
V2oy = 1726 * -0.824 = -1422.224

P1i = xxxx i + yyyy j

P1i = 323.6868 = 323.6868 + 0 Veh1 in

P2i = -6111.766 = -7421.8 + -1310.034 Veh2 in

P1o = -6328.998 = -6434.262 + 105.264 Veh1 out

P2o = -2085.008 = -662.784 + -1422.224 Veh2 out
 
Last edited:
  • #35
Probie1 said:
Okay let's try again.

weight 13158kg
heading 0 deg
departure 179
impact velocity 0.0246 m/s
departure velocity 0.49 m/s

weight 1726 kg
heading 190
departure 245
impact velocity 4.374 m/s
departure velocity 0.91 m/s

*V1ix = in V1ox = out


V1ix = 13158 * .0246 = 323.6868
V2ix = 1726 * -4.30 = -7421.8
V1ox = 1318 * -0.489 = -6434.262
V2ox = 1726 * -0.384 = -662.784


V1iy = .0246 * sin 0 = 0
V2iy = 1726 * -0.759 = -1310.034
V1oy = 13258 * 0.008 = 105.264
V2oy = 1726 * -0.824 = -1422.224

P1i = xxxx i + yyyy j

P1i = 323.6868 = 323.6868 + 0 Veh1 in

P2i = -6111.766 = -7421.8 + -1310.034 Veh2 in

P1o = -6328.998 = -6434.262 + 105.264 Veh1 out

P2o = -2085.008 = -662.784 + -1422.224 Veh2 out

I'm not sure what the values in red are meant to represent, but you've got the components of the vectors displayed which is good. The values are a little imprecise, perhaps due to not keeping enough significant figures in intermediate results (such as values of sin or cos)? Also, you should keep include the "i: and "j" in the notation to make sure that it is clear which component is which. That way in some future problem when you're dealing with THREE components and not just two, you can keep them straight when, say, the "j" component happens to be zero...

I figure the vectors should look like:

p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

Note that I've kept three decimal places for these intermediate results. This will help prevent rounding errors from getting into the final results.

Now, in order to find the initial momentum vector you should add vectors p1i and p2i. That means summing their like components (add the x components, add the y components, and end up with Pi = xxxxxx i + yyyyyy j).

Do the same with the final momenta, adding p1o and p2o.

What do you get?
 
  • #36
I don't know why they came out in red, never mind them... I thought I was suppose to add them together that way.

p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1i = 7758.517 i + 1310.961 j


p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

p2o = -7110.273 i + -1310.978 j
 
Last edited:
  • #37
Probie1 said:
I don't know why they came out in red, never mind them... I thought I was suppose to add them together that way.
Actually, I added the red highlighting in the quote in order to draw attention to them :smile:

p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1i = 7758.517 i + 1310.961 j


p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

p2o = -7110.273 i + -1310.978 j

Oops. mind the signs in the first addition. The second one looks good.
 
  • #38
p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1i = -7111.143 i + 1310.961 j


p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

p2o = -7110.273 i + -1310.978 j
 
  • #39
Probie1 said:
p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1i = -7111.143 i + 1310.961 j


p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

p2o = -7110.273 i + -1310.978 j

There's still a sign issue with the j component of the initial momentum vector. When you add (0.000) to (-1310.961), what's the result?

When that's sorted, you will be in a position to directly compare the components of the initial momentum and final momentum vectors.
 
  • #40
p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1i = -7111.143 i + -1310.961 j

p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

p2o = -7110.273 i + -1310.978 j
 
  • #41
Probie1 said:
p1i = 323.687 i + 0.000 j
p2i = -7434.830 i - 1310.961 j

p1i = -7111.143 i + -1310.961 j

p1o = -6446.483 i + 112.523 j
p2o = -663.790 i - 1423.501 j

p2o = -7110.273 i + -1310.978 j

Much better.

So, now you can directly compare the components of the two momentum vectors, or if you wish, convert them to polar form (magnitude and angle) and compare them that way.
 
  • #42
That results are very close... probably be right on if I did not shorten my after decimal digits.
Thank you for all the patients you have shown me through this... you should get a medal for all you do.

Now if I can copy this stuff down properly I will never forget how to do it on my own.

You are the BEST... thank you so much!
 
  • #43
Probie1 said:
That results are very close... probably be right on if I did not shorten my after decimal digits.
Even using extended precision in the intermediate results, you can only expect the results to be as good as the initial data supplied in the problem. No doubt the angles and speeds given were rounded, too. So a couple of digits of accuracy is about all you can expect. In this case, it should be enough to see that momentum is being conserved, at least to the provided accuracy.
Thank you for all the patients you have shown me through this... you should get a medal for all you do.

Now if I can copy this stuff down properly I will never forget how to do it on my own.

You are the BEST... thank you so much!

It's my pleasure to have been of help. Good luck in your studies!
 
  • #44
Yes... you are correct about the velocities and angles being rounded, and I will never again say that (P)i doesn't = (P)o.

Thanks a Million.
 

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