# 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!

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.