Calculate momentum and velocity of ball after collision

[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.

 

[cyby]

At the same time, without some of these assumptions, the problem becomes unsolvable. But you make a good point in saying that we know little about the collisions. I guess I was a bit too rash in assuming a point mass collision.. a lot of things would happen if you don't assume that...

 

[arildno]

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?

You must regard both balls as physical objects with non-zero radii.

Now, consider what happens at the CONTACT POINT :
The incoming ball has a non-zero component along THE VECTOR NORMAL at the contact point, i.e unless an impulse directly counteracts this collision velocity, the incoming ball would pass into the other ball.

A collision where the impulse is solely in the direction of the contact normal, is called a normal impact.
As long as the contact normal is not strictly horizontally directed, the normal impulse would generate a vertical momentum component in either ball.
(but the SUM of these remain zero).

EDIT:
An important conclusion which can be drawn from this simple example, is that UNLESS you know the specific geometry of the objects involved, you really have no a priori way to detemine the direction the objects may leave.
That is, the collision should essentially be regarded as a RANDOM collision.

 

[IGeekbot]

Ok, sorry to bring this back up, but I need to get someone to straighten me out.

My teacher is telling me that the velocity of Ball B in this problem would be 3.6m/s. He is figuring that ball B is moving southeast at around a 45 degree angle when it makes contact with ball a. This would explain Ball A moving the direction it did, but when the problem specifically states that Ball B is at rest when struck.

Could someone please explain this to me? Because I thought if the 2Kg Ball B was struck at the 1 KG ball A moving at 3.0 m/s, the Ball B would have to move at less than the 3.0 m/s. Not more than it.


And for the record, I use the Merrill Physics: Principles and Problems. It is the black version.

 

[Andrew Mason]

Ok, sorry to bring this back up, but I need to get someone to straighten me out.

My teacher is telling me that the velocity of Ball B in this problem would be 3.6m/s. He is figuring that ball B is moving southeast at around a 45 degree angle when it makes contact with ball a. This would explain Ball A moving the direction it did, but when the problem specifically states that Ball B is at rest when struck.

Could someone please explain this to me? Because I thought if the 2Kg Ball B was struck at the 1 KG ball A moving at 3.0 m/s, the Ball B would have to move at less than the 3.0 m/s. Not more than it.

The key to this problem is that for elastic (not head-on) collisions between equal masses, the two masses move at 90 degrees to each other (see footnote)(a very useful principle if you play pool for money). And for inelastic collisions, the angle has to be less than 90 degrees. The angle can never be more than 90. And if energy is lost, it can never be 90.

So it is apparent that the first problem provides impossible facts because momentum cannot be conserved.

For the second problem, to conserve momentum, the cue ball must move with $\vec v_{cue} = 2\hat i + 5\hat j$. But this represents a seriously inelastic collision. Either someone stuck their gum on the object ball, or the guy that devised the problem has never played pool.

AM

(Footnote: Since $\vec v_1 + \vec v_2 = \vec v_0$ (momentum conserved) and $v_0^2 = v_1^2 + v_2^2$ (energy conserved) the vector diagram defines a triangle in which v0 is the hypotenuse whose length is related to the sides by the pythagorean formula. This means the angle between v1 and v2 is a right angle).

AM