physics help!?

[wolfpacker]

Two blocks of masses m1 = 2.00 kg and m2 = 4.55 kg are each released from rest at a height of y = 4.10 m on a frictionless track, as shown in Figure P6.56, and undergo an elastic head-on collision.
(a) Determine the velocity of each block just before the collision. Let the positive direction point to the right.
v1im/s
v2im/s

(b) Determine the velocity of each block immediately after the collision.
v1fm/s
v2fm/s

(c) Determine the maximum heights to which m1 and m2 rise after collision.
y1fm
y2fm

http://img354.imageshack.us/my.php?image=p656altfk4.gif

[david]

If you want people to answer your questions, post them 1 at a time.

[odu83]

a) Using energy
v=sqrt(2*g*h)
v1i=v2i=sqrt(2*9.81*4.10)
8.97 m/s


b) Since the collision is elastic
momentum and energy are conserved
2*8.97-4.55*8.97=2*v1f+4.55*v2f

.5*2*8.97^2+.5*4.55*8.97^2=
.5*2*v1f^2+.5*4.55*v2f^2

do a bunch of messy algebra
2*8.97-4.55*8.97=2*v1f+4.55*v2f
v1f=-2.275*v2f-11.44
v1f^2=5.1756*v2f^2+52.04*v2f+130.8

.5*2*8.97^2+.5*4.55*8.97^2=
.5*2*v1f^2+.5*4.55*v2f^2

263.51=v1f^2+2.275*v2f^2
substitute
0=7.45*v2f^2+52.04*v2f-132.71

v2f=1.9857 m/s
v1f=-15.95 m/s

c)
Using energy
h=.5*v^2/g

Block 1
h=.5*15.95^2/9.81
h=13 m

Block 2
h=.5*1.9857^2/9.81
h=0.20 m

_______________-