Vector subtraction with momentum

[PeterO]

So the vector subtraction we did is correct..
i think its correct since it doesn;t change much correct?

It was the correct subtraction

Many people - in order to ensure they do subtractions in the "correct" order do the following

Vector A - Vector B = Vector A + (-Vector B)

In other words you reverse the (appropriate) vector and do the standard head to tail resultant.
That avoids which particular tail to tail vector you are after.
since in this problem you wanted the ~19 - ~16, you would reverse the 16 vector and add.

That confirms that the direction you drew it was correct.

 

[TSny]

oh okayyy...

On my sheet with this question they give the equation

m2 = (v1`-v1 / v2 - v2`)m1

So does that mean the change in velocity should be with v2-v2`
not v2`-v2 ? that how we did it

The equation should read m₂ = |v₁'-v₁ |/ |v₂ - v₂'| m₁ where the numerator and denominator of the fraction are magnitudes of the velocity changes. So, it won't matter whether you write |v₂ - v₂'| or |v₂' - v₂|. They are equal.

 

[Lolagoeslala]

The equation should read m₂ = |v₁'-v₁ |/ |v₂ - v₂'| m₁ where the numerator and denominator of the fraction are magnitudes of the velocity changes. So, it won't matter whether you write |v₂ - v₂'| or |v₂' - v₂|. They are equal.

oh okay..
but see i found the experimental error...

so i did 27 - 32 / 32

and i get 15.6 % ...
but my teacher said u can have a maximum of 8% what is this? :(

 

[PeterO]

The equation should read m₂ = |v₁'-v₁ |/ |v₂ - v₂'| m₁ where the numerator and denominator of the fraction are magnitudes of the velocity changes. So, it won't matter whether you write |v₂ - v₂'| or |v₂' - v₂|. They are equal.

disagree. Absolute value not needed as one of the subtractions is written as final - initial while the other is written initial - final. This reinforces the idea that the gain in momentum of one object matches the loss in momentum of the other ie Δp₁ = -Δp₂, or if you prefer Δp₁ + Δp₂ = 0
It doesn't matter if you do include absolute value, but it is not necessary.

 

[TSny]

I don't think it's very meaningful to calculate a % error for an angle. For example, suppose you had an angle that was supposed to be 2 degrees but your experimental value was 3 degrees. That's pretty good - only one degree off. That's just as good as an experimental value of 101 degrees for an angle that should be 100 degrees. But the % error in the first case would be calculated as 100%*(3-2)/2 = 50 %. In the second case, 100%*(101-100)/100 = 1%.

Are you sure that you weren't suppose to find the % error in the mass m1?

 

[TSny]

It doesn't matter if you do include absolute value, but it is not necessary.

Hello, PeterO.

We are considering the equation m₂ = |v₁'-v₁ |/ |v₂ - v₂'| m₁

If you take away the magnitude symbols, then you would be dividing by a vector. Is that possible?

 

[Lolagoeslala]

I don't think it's very meaningful to calculate a % error for an angle. For example, suppose you had an angle that was supposed to be 2 degrees but your experimental value was 3 degrees. That's pretty good - only one degree off. That's just as good as an experimental value of 101 degrees for an angle that should be 100 degrees. But the % error in the first case would be calculated as 100%*(3-2)/2 = 50 %. In the second case, 100%*(101-100)/100 = 1%.

Are you sure that you weren't suppose to find the % error in the mass m1?

what do you find the Error in mass 1?

 

[TSny]

what do you find the Error in mass 1?

Ach. I meant to write m2. I think you said you needed to find m2. So, I was thinking maybe that's what you had to determine a % error for.

 

[Lolagoeslala]

Ach. I meant to write m2. I think you said you needed to find m2. So, I was thinking maybe that's what you had to determine a % error for.

so you mean to say... wait...

do you mean like this m2-m1/m1 x 100%?

 

[TSny]

so you mean to say... wait...

do you mean like this m2-m1/m1 x 100%?

No, % error in m2 would be to compare your experimental value of m2 to the known value of m2 (assuming you have a known value). So, % error = (m2_experiment - m2_known)/m2_known * 100. If you don't know the actual value of m2, then you can't get a percent error for it.