budullewraagh Posted January 14, 2005 Share Posted January 14, 2005 hokay so tarzan is trying to save jane from an angry-angry hippo. he quickly climbs a tree and swoops down on a vine to pick her up and deliver her to the safety of a 9 meter high cliff. if tarzan has twice the mass of jane, how fast must he be going before impact to safely deliver her to the cliff? Link to comment Share on other sites More sharing options...
Mokele Posted January 14, 2005 Share Posted January 14, 2005 Use kinetic and potential energy to solve it. That should give you the answer pretty simply. Remember that the energy remains the same when tarzan grabs jane, so if his mass increases by picking her up, his speed will decrease. Link to comment Share on other sites More sharing options...
budullewraagh Posted January 14, 2005 Author Share Posted January 14, 2005 you know, i would do that except it is midnight and i have to do a physics lab first, then write an english paper, write maybe 5 pages of bio, then do a packet of questions for bio. all in all i'm thinking 5 hours of work in total. sad thing is that ive been working since i've come home but i came home late because i was busy representing my school at a competition for "smart" people. i could not have done anything last night because i had a physics test today and i was away for 5 hours after school due to a concert. ie, life is hell tonight. thank you for your help Link to comment Share on other sites More sharing options...
swansont Posted January 14, 2005 Share Posted January 14, 2005 Use kinetic and potential energy to solve it. That should give you the answer pretty simply. Remember that the energy remains the same when tarzan grabs jane, so if his mass increases by picking her up, his speed will decrease. The kinetic energy does not stay the same, because the collision is completely inelastic. You have to use conservation of momentum during the collision. Link to comment Share on other sites More sharing options...
Mokele Posted January 15, 2005 Share Posted January 15, 2005 Swansont: The Kinetic energy *does* stay the same in an inelastic condition. The object just "gains mass" via the collision. KE = 0.5*m*V^2 KE is the same before and after, but m increases, so v must decrease. If you're doing it right, you'll get the same answer as with conservation of momentum. budullewraagh: While I sympathize with your plight, that does not justify what amounts to copying your homework from a friend. If someone else wants to simply hand you the answer for you to copy down, that's fine, but I prefer to keep my academic integrity. Mokele Link to comment Share on other sites More sharing options...
ecoli Posted January 15, 2005 Share Posted January 15, 2005 Who ever said he was copying someone else's homework??? Link to comment Share on other sites More sharing options...
Mokele Posted January 15, 2005 Share Posted January 15, 2005 He's asking us to do the problem for him. In principle, this is the same as copying off someone else, just facilitated by the information age. Link to comment Share on other sites More sharing options...
ecoli Posted January 15, 2005 Share Posted January 15, 2005 Oh, gotcha...I didn't realize the question was HW related Link to comment Share on other sites More sharing options...
budullewraagh Posted January 15, 2005 Author Share Posted January 15, 2005 what a terrible existence. i can't even remember if i slept last night or not. i wasnt necessarily asking for work to copy per se, but i was asking for a few prods in the right direction Link to comment Share on other sites More sharing options...
fuhrerkeebs Posted January 15, 2005 Share Posted January 15, 2005 Find the velocity you need to get Jane and Tarzan safely to the cliff, and use the conservation of momentum. Just set (Mass of Tarzan + Mass of Jane)*(velocity needed to get safely to the cliff) = (Mass of Tarzan)*(velocity before impact), and solve for velocity before impact. Link to comment Share on other sites More sharing options...
swansont Posted January 15, 2005 Share Posted January 15, 2005 Swansont: The Kinetic energy *does* stay the same in an inelastic condition. The object just "gains mass" via the collision. KE = 0.5*m*V^2 KE is the same before and after' date=' but m increases, so v must decrease. If you're doing it right, you'll get the same answer as with conservation of momentum.[/quote'] Absolutely not. In an inelastic collision KE is converted into other forms. KE is conserved in elastic collision, by definition. KE is not conserved in an inelastic collision, by definition. Tarzan's mass is 2m, Jane's is m. Tarzan starts with a speed v1. His initial kinetic energy is mv12 2mv1 = 3mv2 from conservation of momentum so v2=2/3 v1 KE2 =1/2 * (3m) * (2/3 v1)2 = 2/3 mv12 1/3 of the original KE was lost. The collision made a sound, temeratures went up somewhat, there was work of deformation (haven't you ever seen someone break a bone in a collision? Ever wonder where that energy comes from?) Link to comment Share on other sites More sharing options...
chadn Posted January 15, 2005 Share Posted January 15, 2005 2mv1 = 3mv2 from conservation of momentum How the hell do you get v2=2/3 v1 from that? Shouldnt it be v2=(2mv1)/3m? Link to comment Share on other sites More sharing options...
fuhrerkeebs Posted January 15, 2005 Share Posted January 15, 2005 How the hell do you get v2=2/3 v1 from that? Shouldnt it be v2=(2mv1)/3m? The masses cancel out. Link to comment Share on other sites More sharing options...
Guest L33t Rappa S Posted January 15, 2005 Share Posted January 15, 2005 Oh dear guys it's time get a life. Link to comment Share on other sites More sharing options...
5614 Posted January 15, 2005 Share Posted January 15, 2005 3rd line of your sig applies to YOU Link to comment Share on other sites More sharing options...
budullewraagh Posted January 15, 2005 Author Share Posted January 15, 2005 funny how quickly he got banned Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now