# Homework Help: 2D Momentum Collision

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1. Feb 17, 2016

### aimen khattakk

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

One car of mass 1400 kg is travelling at 45km/h when it collides and becomes entangles with another car mass of mass 1300 kg travelling at 39km/h[E]. What is their velocity after collision

So MA = 1400 kg
VA = 45kh/h

MB = 1300 kg
VB = 39 kh/h [E]

and After Collision:
MAB = 2700kg

2. Relevant equations

P = mv
∑P = ∑P'

PA + PB = PAB'

3. The attempt at a solution

so I've always been told to solve momentum first so since car A is going South that means it only has a verticle momentum? so
that means PAY = m⋅v
so PAY = 63000kg⋅km/h

similarly for B

so only a horizontal momentum
so PAX = 50700kg⋅km/h [E]

so now im stuck on what to do, i tried to find the momentum of Pab in the x and y direction then find pab and solve for v but that doesnt work when i tried it. Pab in the x direction just turns out to be the same as pb in the x direction and pab in y direction turns out to be pa in y direction and then using pythaguras theorum to find the hyp and hence find Pab i get 37396 kg⋅km/h and then solving for V by diving the Pab by the total mass of both of the cars together gives me a velocity of 13.85km/h but the answer is 30.0km/h so i think i messed up after solving my momentums for before the collision but idk what i did wrong

Last edited: Feb 17, 2016
2. Feb 17, 2016

### Staff: Mentor

3. Feb 17, 2016

### Staff: Mentor

This number is not correct.

Edit: NascentOxygen beat me to it.

4. Feb 17, 2016

### aimen khattakk

okay so
PABY' = 63000kg⋅km/h and PABX = 50700 kg⋅km/h [E]

so

c2 = a2 + b2
so
c2 = 630002 + 507002
c2 = √630002 + 507002
so c = 37396.6576 kg⋅km/h so this is PAB '

now i need to find the velocity after collison so i took the combined mass and divided it with 37396.6576kg⋅km/h
and my v ends up being 13.85 km/h but the answer is 30 km SE

5. Feb 17, 2016

### aimen khattakk

i do not understand hot i
it is? i dont understand how because i did
507002 + 630002
and since 63000 is actually in the negative direction i basically did

507002 - 630002
and then square rooted it. do i not assign that negative direction? but its going south?

6. Feb 17, 2016

### Staff: Mentor

The negative sign isn't applicable here. You use Pythagoras to calculate magnitude.

7. Feb 17, 2016

### Staff: Mentor

What if A was going North instead? Do you expect a different result? Making a diagram of the situation would help.

8. Feb 17, 2016

### aimen khattakk

my calculator is fine, i've used it for all my courses and im in 2 other math courses so its not broken, but i took ur advice and tried it with another one and still getting the same answer

9. Feb 17, 2016

### aimen khattakk

oh wait, so since im looking for the overall momentum of Pab after the collision, i dont need to consider negative and positives? since im not going in a specific direction, rather im going in 2 directions not one?

10. Feb 17, 2016

### Staff: Mentor

NascentOxygen said that because
is not correct. But this is not what you calculated, as you stated in a subsequent post.

11. Feb 17, 2016

### aimen khattakk

and im basically looking for the sum of 2 vectors so it would be that the cars are going S of E not just S or just E so doing what i did earlier would have made the direction be just south right?

12. Feb 17, 2016

### Staff: Mentor

You already have the sum of the two vectors. What you are looking for is the magnitude of the resultant vector.

13. Feb 17, 2016

### aimen khattakk

that is the Pab not the velc
OH so since im looking for the magnitude i dont need to consider their directions so it would just be

c2 = 630002+ 507002
c2 = √630002 + 507002
so c = 80867.11 kg⋅km/h ??

so then velocity would be 80867.11 kg⋅km/h / 2700kg
so v = 29.9 km/h! thank you! that makes so much more sense!

14. Feb 17, 2016

### Staff: Mentor

You are adding two perpendicular quantities. The hypotenuse will be longer than either of the perpendicular components.