Calculate momentum and velocity of ball after collision

[IGeekbot]

HOw do I do these two problems. I know it has something to do with angles the ball trike each other, but I am so Lost.

Problem 1.
Ball A, rolling west at 3.0m/s, has a mass of 2.0 kg. Ball B is 2.0 kg,and is stationary. After colliding with ball b, ball a moves south at 2 m/s. calculate the momentum and velocity of ball B after the collision.


Problem 2.
A cue ball, moving with 7.0 Ns of momentum strikes the nine ball at rest. The nine ball moves off with 2.0Ms in the original direction of the cue ball and 2.0Ns perpendicual to that direction. What is the momentum of the cue ball after the collision.


I have no idea what to do with these.

 

[cyby]

Are you familiar with 2d collisions and the principles of conservation of momentum?

 

[IGeekbot]

yes, i have my book here in front of me with equations, but I am not sure which one to use.

 

[cyby]

Well, consider using conservation of momentum for two dimensions (meaning that you break them up into x and y components)...

Show us what you are trying to do in equations, and we'll work from there :D

 

[IGeekbot]

thats just it, i don't know what to do. Our teachers idea of introducing a new concept is go through it in five minutes to move on to a lab. He justrecites the book so it is no help. I know we have to use the sin and cos of one of the veloctiies, but i am not sure which one. When i get osm calculations going here, ill show ya

 

[cyby]

In your particular case, you don't have to. Are you familiar with unit vector notation to express vector quantities?

 

[cyby]

Let's just look at #1 for now.

By conservation of momentum, you know that the momentum in each directional component has to be equal before and after the collision. Right?

Now, what is the original momentum:

1. Of ball A in the x (east-west) component?
2. Of ball A in the y (north-south) component?

For simplicity purposes, let's just east as +x, and north as +y.

 

[IGeekbot]

ive probably seen them, but that is not ringing a bell. could you please explain what it is?

 

[cyby]

The unit vector notation basically breaks down a vector quantity into its directional components (i, j, k) or (x, y, z), depending on what notation you like to use. This way, you can express a quantity say, 3 units to the east and 4 units to the north, as 3i + 4j, for instance. This quantity is the same as saying 5 units (~51 degrees north of east)... familiar?

 

[IGeekbot]

The ball would have the original momentum of 3kg x m/s moving west. I got that, but this is where I get lost, because this is all new to me.

 

[cyby]

Ok. Why don't we start even further back. Can you type out the conservation of momentum equations at least? :-)

 

[cyby]

Oh boy. Let's see here.

http://chortle.ccsu.edu/VectorLessons/vch08/vch08_6.html

Here is a quick briefing on unit vectors. Your textbook should have a few things on unit vectors, and I imagine that you must have worked on some of that stuff during kinematics too...

 

[cyby]

you mean 3.0m/s * 2.0kg = 6.0 kg*m/s...

The ball would have the original momentum of 3kg x m/s moving west. I got that, but this is where I get lost, because this is all new to me.

 

[IGeekbot]

To calculate momentum, it is P=mv.
In a closed system you have Pa+Pb=P'a+P'b

Then we have the ones for elastic and inelastic equations, but i don't think those are needed here.

 

[IGeekbot]

you mean 3.0m/s * 2.0kg = 6.0 kg*m/s...

yeah, wrong key.

 

[IGeekbot]

and no, we have never done vectors like that. I think i will read on that more.

 

[cyby]

Now. Note that momentum is a vector quantity (because velocity is).

So, you can further break it down to:
Pa_x + Pb_x = P'a_x + P'b_x
Pa_y + Pb_y = P'a_y + P'b_y

Now, in your case, you know that
Pa_x = 6, and Pb_x = 0, and Pb_y = 0

You also know...
Pa'_y = 4

With units in kg*m/s

Do you see where I was getting my numbers?

To calculate momentum, it is P=mv.
In a closed system you have Pa+Pb=P'a+P'b

Then we have the ones for elastic and inelastic equations, but i don't think those are needed here.

 

[cyby]

Check the beginning of your book - almost all books have a section on that stuff at the very beginning - I suppose you are on the high school/intro college level?

and no, we have never done vectors like that. I think i will read on that more.

 

[IGeekbot]

so I need to use the horizontal and vertical numbers? Then, when you get those, they should equal the initial momentum of the ball, which is 6 kg*m/s?

 

[IGeekbot]

Check the beginning of your book - almost all books have a section on that stuff at the very beginning - I suppose you are on the high school/intro college level?

Yeah, its an intro class, and the teacher expects you to pick it up the frist time, and if you ask for help, he still won't really explain it. He just repeats what he says. And I foudn that section, and we have just used it a little bit.

 

[cyby]

Yes, you need to break them down into the horizontal and vertical components. Whenever you work with more than one dimension, do that.

When we get those, the *individual* components should conserve.

And a slight correction on my notation earlier.

P_ax should be -6 and P'_ay should be -4, because I specified +x and +y to be east and north, respectively.

so I need to use the horizontal and vertical numbers? Then, when you get those, they should equal the initial momentum of the ball, which is 6 kg*m/s?

 

[cyby]

You actually also know that P'_ax is 0, because after the collision, the ball only moves south at 2m/s, so it's not traveling in the east-west direction at all. As a result, after the collision, ball a has a horizontal momentum of 0.

Yes, you need to break them down into the horizontal and vertical components. Whenever you work with more than one dimension, do that.

When we get those, the *individual* components should conserve.

And a slight correction on my notation earlier.

P_ax should be -6 and P'_ay should be -4, because I specified +x and +y to be east and north, respectively.

 

[Nylex]

Yeah, its an intro class, and the teacher expects you to pick it up the frist time, and if you ask for help, he still won't really explain it. He just repeats what he says. And I foudn that section, and we have just used it a little bit.

Doesn't sound like much of a teacher then. Welcome to PF, both of you .

 

[cyby]

Thanks bunches! Yeah, I remember having teachers like that. I guess one of the things I quickly learned was how to teach myself. It was the most important thing I learned over the years...

Doesn't sound like much of a teacher then. Welcome to PF, both of you .

 

[cyby]

Yeah, its an intro class, and the teacher expects you to pick it up the frist time, and if you ask for help, he still won't really explain it. He just repeats what he says. And I foudn that section, and we have just used it a little bit.

I really suggest you know your unit vectors really well, as I bet this stuff will help you in your future physics and mathematics courses.

 

[IGeekbot]

ok, I think I am getting this. Ball B, it will have a velocity of -1 m/s, since it is moving east, and it has a mass of 2kg.


Because the original velocity is 6kg*m/s east when the ball makes contact, the two balls afterward will have to have a 6kg*m/s momentum. Since the angle is 90degrees, the hypotenuse of that triangle will be the 6kg*m/s. if you take the original momentum then, 6kg*m/s, and subtract the momentum of ball A after the collision, you will have the total momentum of Ball B. That is part of the answer, and if you divide that answer by the 2 kg, it will give you a velocity which will be 1.


I know that is garbled a bit, but am i on the right track?


And nylex, thanks for the welcome.

 

[cyby]

You know, hold on a minute for this one. What I *really* want to know, is how a ball that strikes a stationary ball with no vertical momentum develops vertical momentum. Ball A only had horizontal momentum, and Ball B orginally had no momentum. It should therefore even be impossible for Ball A or B to develop vertical momentum, unless there is some external force. Anyone want to take a stab at this, or am I just suffering from dementia?

 

[Nylex]

You know, hold on a minute for this one. What I *really* want to know, is how a ball that strikes a stationary ball with no vertical momentum develops vertical momentum. Ball A only had horizontal momentum, and Ball B orginally had no momentum. It should therefore even be impossible for Ball A or B to develop vertical momentum, unless there is some external force. Anyone want to take a stab at this, or am I just suffering from dementia?

Good point! I was kinda missing the point when you asked if IGeekbot knew about momentum in 2D, cos it just seemed like a 1D collisions problem to me.

Oh and IGeekbot, if it's easier, you can write the units of momentum as Ns.

 

[cyby]

In the mean time

Let's look at #2

Problem 2.
A cue ball, moving with 7.0 Ns of momentum strikes the nine ball at rest. The nine ball moves off with 2.0Ms in the original direction of the cue ball and 2.0Ns perpendicual to that direction. What is the momentum of the cue ball after the collision.

I suppose you also mean by 2.0N*s, right? (Nine ball moves off with 2.0 ...)

With that assumption..We can use those previous equations earlier.

Pa_x + Pb_x = P'a_x + P'b_x
Pa_y + Pb_y = P'a_y + P'b_y

Let a = cue ball and b = nine ball

I will just skew the coordinate system such that +x = direction of the cue ball, and +y = direction perpendicular to the cue ball. So:

Pa_x = 7, Pa_y = 0
Pb_x = 0, Pb_y = 0 (Because they're at rest)

What are P'a_x, P'b_x, P'a_y, and P'b_y? That is what you have to figure out. Show us what you're doing :-)

 

[cyby]

Good point! I was kinda missing the point when you asked if IGeekbot knew about momentum in 2D, cos it just seemed like a 1D collisions problem to me.

Oh and IGeekbot, if it's easier, you can write the units of momentum as Ns.

It seemed like a 1D collision problem to me, but the end condition is impossible given the starting conditions to start with. The second problem is far more plausible, of course. Nevertheless, it may be a good idea to check the first problem again and to see if it was typed in incorrectly.. cause I really think it may be impossible to do.

And yea, Nylex, I am kinda distracted at the moment, I should have caught that problem far earlier.

 

[IGeekbot]

I went back and checked. I tpyed what the book says. It seemed odd to my friend and I today as well. We wondered how a ball A could roll along completely horizontally, hit Ball B, have ball B continue on horizontally, and have Ball A deflect to a completely 270 degree direction.

 

[cyby]

Better check with your teacher then, otherwise a lot of physics will have to be rewritten :P

I went back and checked. I tpyed what the book says. It seemed odd to my friend and I today as well. We wondered how a ball A could roll along completely horizontally, hit Ball B, have ball B continue on horizontally, and have Ball A deflect to a completely 270 degree direction.

 

[IGeekbot]

Welll, get out your pens, because i took a picture of the problem, and other than having a wrong number in my typing, which I fixed in my work, I typed it word for word. Apparently, this book has the power to defy physics. I will return later with the answers to question 2, but I need to go do some other work now. Thank you cyby for the help.

 

[cyby]

Well, then the book is wrong. I trust physics over your book any day.

What book do you use, by the way? I am interested now :-)

 

[Skomatth]

Well as I understand it, we don't have a lot of information about the details about the collision. The balls could have struck each other on the edge instead of through the center of mass. They could also have an electric charge or be irregularly shaped as far as we are concerned. In my book, collisions are sometimes denoted as "head-on" to specify that the momentum stays in one dimension. I don't see any laws of physics being broken here, it just may be messing with your intuition because you assume the collision to be head-on.