Vector subtraction with momentum

[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.