max veloicty after collision

[FistLength]

---

Members do not see this ad.

The question involves a bob swung from height h and allowed to collide with bob m2 which is at rest.

EK physics 407
As the mass of m1 becomes much greater than m2, what is the maximum velocity that can be reached by m2? (assume completely elastic collision)

A: 0 m/s
B: v
C: 2v
D: The maximum velocity is infinite



Answer is C, 2v

Can someone please explain how?

 

[1TB4RKSB4CK]

http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html
Can't really explain but I hope this link can. Once you see the concept and the equation you should be able to get it.

 

[FistLength]

OK, I searched every nook and cranny and finally found the answer...it is pretty complicated to go through but it makes sense
http://books.google.com/books?id=IZ...llision if m2 is much greater than m1&f=false

 

[sps27]

OK, I searched every nook and cranny and finally found the answer...it is pretty complicated to go through but it makes sense
http://books.google.com/books?id=IZIWAAAAQBAJ&pg=PA258&lpg=PA258&dq=collision if m2 is much greater than m1&source=bl&ots=fxForoL-tR&sig=1LzOGbltoSsco_m51VOvL9qn8Fk&hl=en&sa=X&ei=6DajUoeeFIznsASJ3ILYCg&ved=0CE8Q6AEwBg#v=onepage&q=collision if m2 is much greater than m1&f=false

That was a good question. I had been thinking about it. The eqns are a bit tedious to go through but it does work out. I doubt EK would have taken the pains to go through all of that to explain.....

 

[patrickmbyrne]

They didn't really show the derivation of equations 9.21 and 9.22... the just kinda throw them out there

 

[sps27]

They didn't really show the derivation of equations 9.21 and 9.22... the just kinda throw them out there

V1(i) + V1(f) = V2(i) + V2(f) (shown in the derivation)
V1(f) = V2(i) + V2(f) - V1(i). Now apply this to 9:16 eqn and solve for V2(f). Likewise V2(f) = V1(i) + V1(f) - V2(i) and apply that to eqn 9:16. Solve for V1(f). You will get those 9:21 and 9:22 eqns.