Recent content by Stahs

Well I've handed it in now and I'd like to thank you OlderDan for all your help. It was greatly appreciated.

Wow, this is starting to make sense now. So going by your equation: -(100kg)v1' = (20kg)(v1' - 10m/s) then v1' = 1.667 m/s Which is the speed of Jean relative to the ice after he throws the box. And v2' would be: 0 = m1v1' + m2v2' 0 = (100)(1.667) + (20)v2' v2' = -8.335 m/s Which...

But i still have no idea how to get his velocity relative to the ice. I've tried to use: Pt = Pt' 0 = Pt' 0 = m1v1' + m2v2' -m1v1' = m2v2' -(100)v1' = (20)(8) v1' = -1.6 m/s But it still doesn't get me the right answer. I guess I'm just not...

Ok so... Jean's velocity relative to the box is still 2 m/s right? Which means that his velocity relative to the ice is less than 2 m/s and that the box's velocity relative to the ice is 8 m/s?

Hmm, I don't really know how to do it then. I can't think of any way of getting Jean's speed. Can I maybe get a hint?

Ohhh I see, so wouldn't that make the carton's velocity 8m/s since Jean moves at 2m/s? That still doesn't give the right answer.

First of all I would like to explain that this is the last question I need to answer to complete an assignment that consisted of 20 University of Waterloo SIN Contest questions. This question is from the 1999 SIN contest. I would ask my Physics teacher for help but it is now the weekend and the...