Question

The green puck's initial speed is 4.0m/s in the positive x-direction. The mass of the green puck is twice that of the blue puck. Predict the speed of both pucks after the collision

Answer 1

can anybody help me understand this. its an elastic collision

Answer 2

theEric

Answer 3

Hi!

http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

So, elastic collision just means that there is no loss or gain of kinetic energy. So the kinetic energy and momentum of the system is constant before, during, and after the collision.

Answer 4

ok. i've tried to work this a few ways. the best number i came up with for puck b is the around 7m/s. i understand that energy is going to be conserved. i just don't know where to go with the math or if the math i did is correct

Answer 5

what makes this different to what we've been studying is that one is sitting still till impact, then its asking VBf bascially.

Answer 6

here is how the full problem reads....Question 2.
The green puck's initial speed is 4.0m/s in the positive x-direction. The mass of the green puck is twice that of the blue puck. Predict the speed of both pucks after the collision. After your calculations, run the simulation to check your prediction.
Elasticity set at 1.0

Answer 7

And you'd have to use the @ symbol to notify me! :) Like, @a-higbee !

And no symbols before or after the `@a-higbee`.

I haven't done this in a while, if I ever did it! So I'm looking at the Wikipedia link I posted. It states that the center of mass has a constant velocity, I think. I'm trying to understand it, now.

But the picture on Wikipedia is helpful.
http://upload.wikimedia.org/wikipedia/commons/e/e5/Elastischer_sto%C3%9F3.gif
The common \(v\) here could be \(2\ [m/s]\) from a different frame of reference (one traveling \(2\ [m/s]\) to the right). We could use that, but I'm not sure what your course would want you to do.

Answer 8

ok. yea i'm not sure how to calculate a prediction. i see the examples on wikipedia and i understand the relative velocity. based on that, our relative velocity is 4.0. when you mention common v what are you referring to?

Answer 9

one our known velocities is zero, B is not moving initially.

Answer 10

i just tried to work it another way I just don't get it.

Answer 11

If you read this post, just know that I don't think this was the purpose of the assignment. It probably wants you to be able to arrive at the conclusion Wikipedia got to! I just want to explain what I min

If you were someone running \(2\ [m/s]\) towards the right, between the pucks, and considered yourself to be at rest, puck A would be coming at you at \(2\ [m/s]\) and puck B would be coming at you at \(2\ [m/s]\). So, if you are \(0\ [m/s]\), then puck A is \(2\ [m/s] \ \underrightarrow{\overrightarrow{right}}\) and puck B is \(2\ [m/s]\ \underleftarrow {\overleftarrow{left}}\)

Wikipedia says that puck A would have \(\frac{1}{3}(v)=\frac{2}{3}\ [m/s]\) and puck B would be \(\frac{5}{6}(v)=\frac{10}{6}\ [m/s]=\frac{5}{3}\ [m/s]\). That's in the frame where it seems like both masses have the same \(v\).

To go back to the other frame, we have to look at all the velocities how they were before, which is where you are running at \(2\ [m/s]\) right - everything gets the \(2\ [m/s]\) to the \(\underrightarrow{\overrightarrow{right}}\) so you see yourself as running \(2\ [m/s]\ \underrightarrow{\overrightarrow{right}}\). Then
Puck A = \(\left(\frac{2}{3}+2\right)\ [m/s]=\frac{8}{3}\ [m/s]\)
Puck B = \(\left(\frac{5}{3}+2\right)\ [m/s]=\frac{11}{3}\ [m/s]\)

I spent too much time on that, huh? I'll go back to Wikipedia to see what I can learn!

Answer 12

I see what your saying, i'm reading a ton of wiki right now also

Answer 13

i think i got it. i was doing almost everything correctly according to a yahoo similar answers, except i wasn't foiling. and other small algebra stuff involving a quadratic

Answer 14

http://answers.yahoo.com/question/index?qid=20110516144113AADIurW

Answer 15

thanks theEric

Answer 16

So, you have it now! I didn't know about that equation with the velocities!