Collisions?

Question

Consider a frictionless track ABC as shown in Figure P9.26. A block of mass m1 = 5 kg is released from A. It makes a head-on elastic collision at B with a block of mass m2 = 10 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.


Here is the diagram http://www.webassign.net/sb5/p9-26.gif


I know that KE and Momentum are conserved....and this is what i have so far...

m1v1=m2v2f + m1v1f

mgh = 1/2mv1^2 + 1/2mv2^2

I don't know where to go from here, if this is even right

Answer 1

You're on the right track
One small clarification:


m1gh = 1/2m1v1^2 + 1/2m2v2^2

m1gh=5*9.8*5
=245

You can also calculate the velocity of m1 at the moment of collision:
m1gh=1/2m1v1^2
=sqrt(9.8*5*2)
=9.9
so
5*9.9=10*v2f+5*v1f
9.9=2*v2f+v1f
v1f=9.9-2*v2f
Substitute into the quadratic:
245*2=5*(9.9-2v2f)^2+10*v2f^2
245*2/5=
9.9^2-4*9.9V2f+4V2f^2+2v2f^2
6v2f^2-39.6V2f+.01=0
Solve
for v2f
0, 6.6
using 6.6
v1f=9.9-2*6.6
v1f=-3.3
again,
using mgh=1/2mv^2
h=.5*4.4*3.3/9.8
h=.74m


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