# Conservation of momentum in two dimensions?

1. Sep 25, 2005

### twofish

Hi all,
I'm having a tough time with a below question and would like some hints on where I've gone wrong if my answer is incorrect.
I think the reason it's difficult is because it intuitively doesn't make sense, ...heading 90 degrees south after initially travelling due east, not to mention that it changes the hypotenuse of my vector addition triangle from the Pa to Pb.

Here is the question;

"Ball A, rolling west at 3.0 m/s, has a mass of 1.0 kg. Ball B has a mass of 2.0 kg and is stationary. After colliding with ball B, ball A moves south at 2.0 m/s. Calculate the momentum and velocity of ball B after the collision.

This is my answer and methodology so far..

pa + pb = pa + pb (vector sum) with pb = 0
pa = mava = (1.0 kg) * (3.0 m/s) = 3.0 kg*m/s (this is the end total that must be met to satisfy conservation of momentum law)
pa = ma+va = (1.0 kg) * (2.0 m/s) = 2.0kg*m/s

pa_x + pb_x = pa_x + pb_x
pa_y + pb_y = pa_y + pb_y

now I know the following.
pa_x = 3.0, pa_x = -3.0 (since ball a is now going due south and has no eastward momentum)
pb_x = 0, pb_x = ?

pa_y = 0, pa_y = 2.0 (since ball a is now going due south)
pb_y = 0, pb_y = -2.0 (since vector sum of vertical components must be 0)

Now I do this, I've changed signs to represent direction of travel for -x, +x, -y and +y ..they should all work out in the end?
pa + pb = pa + pb OR (pa_x + pa_y) + (pb_x + pb_y) = (pa_x + pa_y) + (pb_x + pb_y)
(-3.0 + 0) + (0 + 0) = (-3 + (-2)) + ( X + 2)
X = 0, or pb_x = 0 ..so I now know that Ball B has no horizontal momentum, in theory.

Great ..do I care? I have no idea where to go from here with this methodology...

Alternatively I don't have to do any of the above save for;
pa = ma+av = (1.0 kg) * (2.0 m/s) = 2.0kg*m/s.
Now I have two sides of a triangle, with pa = 3.0kg*ms and p'a = 2.0kg*m/s, p'b (Hypotenuse) uknown.
By Pythagoras theorem I get.. p'b = $$\sqrt{13}$$ or (3.6kg*m/s) which is the momentum of ball B.
Divide by 2.0 kg and I get 1.8m/s as the velocity.
Is this correct?
Thanks....

Last edited: Sep 25, 2005
2. Sep 25, 2005

### whozum

I'm not really sure how to interperet the information above, but the easiest way to do it (which is sort of the same as aove but simpler) is draw a vector diagram at the moment of collision and then find the total momentum vector. THen draw the vector diagram after the collision and draw the same total momentum vector again. You should have two 'known' vectors in your second diagram and one unknown vector. To find the unknown vector, subtract the 'known' balls vector from the net momentum vector.

3. Sep 26, 2005

### HallsofIvy

Staff Emeritus
No, you don't know that! Assuming that you are setting up a standard coordinate system with positive x to the east and positive y to the north, after the collision p'a_x= 0 because there is neither east nor west momentum. Further, pa_x= -3.0 since west is negative.
So -3.0+ 0= 0+ p'b_x. That's easy to solve.
Better is: p'a_y= -2.0 so 0+ 0= -2.0+ p'b_y and p'b_y= +2.0

FAR better to "put in the signs" as you calculate the momentum: mv and v has sign corresponding to its direction.

For this very simple case where B was not moving and A turned at a right angle, yes, that works. For more general collisions, it might not.

Last edited: Sep 26, 2005
4. Sep 26, 2005

### twofish

Okie, thanks much.
I will work on this a little more.