Homework Help: Change in linear momentum

1. Sep 22, 2016

MPat

1. The problem statement, all variables and given/known data
A 15 000kg loader traveling east at 20km/h turns south and travels at 25km/h. Calculate the change in the loader's
a) kinetic energy
b) linear momentum

2. Relevant equations
ΔKE= 1/2mvf^2 –1/2mvi^2
Δp= mv2 –mv1
m= 15000
vi = 20km/h = 5.56m/s
vf= 25km/h = 6.94m/s
3. The attempt at a solution

Part A I think I got.

Part B is where I am confused.
I am using Δp= mv2 –mv1
I substitute the numbers in
Δp= 15000*6.94 -15000*5.56
The magnitude of Δp that I arrive at is equal to 20 700...is that correct?

Also, when I am trying to find the change in direction when subtracting vectors you simply add the opposite direction. so rather than using east, i add my vector pointing south and add the initial vector but pointing west?

I then use this to calculate the angle for the direction. Is this right?

I tried to add an image to show you my diagram...but can't seem to figure out how!

Thanks in advance!

2. Sep 22, 2016

Simon Bridge

hint: momentum is a vector, and so is velocity
... so $\Delta \vec p = \vec p_f - \vec p_i = m(\vec v_2-\vec v_1)$
... try subtracting the vectors by the heat-to-tail method.

But notice: you are not asked to find the magnitude of the change in momentum.

Last edited: Sep 22, 2016
3. Sep 23, 2016

MPat

Thank you!

I think I got it, it's essentially the hypotenuse formed by the triangle...my numbers weren't adding up because I was forgetting that I had to separate the vectors into components, N-S and E-W.

4. Sep 23, 2016

PeroK

Sometimes it's usful to look at vectors graphically, but you can also look at vectors as pairs (x, y) or triplets (x, y, z). In this case, you have:

Initial velocity $= (u_x, u_y)$

Final velocity $= (v_x, v_y)$

That might be an easier way to cacluate the change in velocity and momentum in this case.

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