How Does Momentum Change in a Two-Dimensional Collision?

[Donnie_b]

A 5.00 kg Ball A moving at 12.0 m/s collides with a stationary 4.00 kg Ball B. After
the collision, Ball A moves off at an angle of 38.0º right of its original direction. Ball B 52.0º left of Ball A's original direction. After the collision, what is the momentum
of Ball A?

P1 + P2 = P1' + P2'
60 + 0 = P1' + P2'

After this I have no idea where to go. We would need a velocity however there isn't one given.

I'm assuming a vector diagram would be useful, but I am not too sure what that would look like.

 

[hage567]

You must break up the momentum into the x and y components, and sum up in each direction. Draw a diagram to keep track of everything.

 

[Donnie_b]

You must break up the momentum into the x and y components, and sum up in each direction. Draw a diagram to keep track of everything.

I don't have the hypotenuse to be able to do that. The only thing I could think of would be x =(60-y) and y=(60-x)

 

[Donnie_b]

Ok so I also asked this question on Yahoo answers, and some person gave me this:

60=P’[sin(38) + cos(38)]
60= P’[1.25144222]
P'=60 / 1.25144222
P'=47.9446826 =
P'= 48 kgm/s

Now I have never seen anything alike this taught to me. Does this have any truth to it?

I was also given this which seems more... legit

PX(x) = 5kg*12 m/s = 60
PX(y) = 5kg*0 m/s = 0
PY(x) = 4kg * 0 m/s = 0
PY(y) = 4kg * 0 m/s = 0

PX'(x) = cos(38)*PX'
PX'(y) = sin(38)*PX'
PY'(x) = cos(52)*PY'
PY'(y) = sin(52)*PY'

Then use conservation of momentum in each direction
60 = PX(x) + PY(x) = PX'(x) + PY'(x)
0 = PX(y) + PY(y) = PX'(y) + PY'(y)

from this we see
PX'(y) = - PY'(y)
and
PX'(x) = 60 - PY'(x)

then plug in the sine formulas for each component.
you now have two equations with two variables.
solve one for PY' and plug into the other to solve for PX'

 

[asd1249jf]

I don't have the hypotenuse to be able to do that. The only thing I could think of would be x =(60-y) and y=(60-x)

You don't need the hypotenuse to break into components. You have the angle given after the impact, use trigonometry.

[EDIT]

The first response is completely off. The second one seems right to me as well.