# Calculate the change in the loader’s kinetic energy

1. Apr 3, 2014

### jgray

Linear Momentum question - help please

1. The problem statement, all variables and given/known data

A 15,000 kg loader traveling east at 20 km/h turns south and travels at 25 km/h. Calculate the change in the loader’s
kinetic energy.
linear momentum.

2. Relevant equations

KE2-KE1 = change in KE
p=mv

3. The attempt at a solution

a.
20km/hr = 5.56m/s
25km/hr=6.94 m/s

KE2-KE1
1/2mv2^2 - 1/2mv1^2
1/2 (15000)(6.94)^2 - 1/2(15000)(5.56)^2
361227-231852
129375J

b. p(east) = mv1
(15000)(5.56)
83400

p(south) = mv2
(15000)(6.94)
104100

using pythagorean theorem for the two sides I know I got 133388 kg m/s for the magnitude of the momentum..
Am I doing this right??? I also don't know if I can just use p(east) and p(south) to find the angle of the momentum - does 83400/104100 = 0.801152737 then inv cos - 36.8 degrees work? Doesn't sound right to me… Or can I just show the direction using N/E/S/W

Last edited by a moderator: May 12, 2015
2. Apr 3, 2014

### Simon Bridge

You have noticed that momentum is a vector - so the change in momentum involves subtracting vectors.
Drawing the vectors head-to-tail, shows you a right-angle triangle: so the magnitude of the change is, indeed, going to be the length of the hypotenuse by pythagoras. Well done.

To get the direction, remember that subtracting a vector is the same as adding the negative of the vector ... do it head-to-tail. Specify angles with respect to one of the compass directions. You could finesse it though by specifying the change as "so much" East and "so much" South (i.e. by components of the vector).

3. May 12, 2015

### Brendan Webb

Hi, I think I have got the right answer for the question but I was really hoping someone could verify it for me. My work is shown on the attachment. Final answer = 5.0 * 10^4 kg*m/s 51 degrees south of east direction. ( I forgot to add the 51 degrees on the attachment final answer) Thanks a bunch :)

4. May 12, 2015

### BvU

Hello BW.

For part a:. It's not meaningful to work in seven digits after first rounding off to three (for v). What you do at the end (bring back to 2 digits) is fine. because the first digit is a 1, you might consider giving one digit more (matter of taste/habit).

For part b. you might draw a wind rose to begin with. Most folks would have East pointing to the right.

Now go back to what Simon says: just like with the kinetic energy scalars, you want the difference. What you draw is the sum of the two momenta.
Instead, p1 + change in p = p2 , so $\ \$ change in p = p2 $-$ p1 .

So draw p1 starting at the origin. Draw p2 starting at the origin. The vector "change in p" is from the tip of p1 to the tip of p2 ! If you want to see that starting at the origin, you have to add up p2 and $-$p1. Post the drawing: to get your head around this the right way is an important learning step.

Another tip: the m and the 1/3.6 are common factors. Much easier to work with symbols and do all the calculation work with the 20 and 25 , then apply these factors at the end.

5. May 12, 2015

### Brendan Webb

Hi BvU

Thanks for the reply. I am still a little confused about the change in P though. I re-did my vector diagram. Both vectors start at the origin. But from here I am confused. Is the change in vectors the hypotenuse, from what I understand by Simon this is true but you state "p1 + change in p = p2 , so change in p = p2 − p1" which makes more sense to me because I don't know how the change in momentum would be larger than p1 or p2. If I do it this way I get a change in momentum of 2.1 * 10^4 kg *m/s in the 51 degrees south of east direction. I'm just not sure what the hypotenuse has to with at all anymore though. your help is greatly appreciated

6. May 12, 2015

### haruspex

To see this, say the two vectors you drew are from point O. To represent -p1 you need your p1 pointing the other way. To then add this to p2, i.e. p2+(-p1), you need to start the -p1 vector at arrow end of the p2 vector. So you will have p2 pointing 'down' from O to P, say, then -p1 pointing from P to the right, reaching point Q.
The resultant, p2-p1 is then the vector from O to Q. This looks exactly like the hypotenuse you have drawn, but shifted over to start at O.

7. May 13, 2015

### Brendan Webb

Hi haruspex

Thanks for the reply. Ok so the vector looks like this. I'm not sure why though -p1 is facing east though because I thought its initial vector was in the east direction. Don't we turn around a vector when subtracting? sorry to keep asking but I feel as though I am missing something.

8. May 13, 2015

### haruspex

Sorry, I based my post on your diagram and did not check back with the original question. So flip left/right in you diagram in post #5 and in my reply in post #6.

9. May 13, 2015

### BvU

In a picture what I described in #4

And I persist in using "to the right" as East (Couldn't find a windrose agreeing with your picture )

What happened to Haru is a nice demonstration of the importance of a clarifying sketch !

As you see:

| the change | is bigger than either |p1| or |p2| .
(but it doesn't have to be: if p2 = p1 then |thechange| is 0
and if p2 = -p1 then |thechange| is 2|p1|

--

10. May 13, 2015

### Brendan Webb

Thank you both for your time and patience! I finally understand! :)

Cheers