calculate height of block after collision?

Question

A block of mass 4.764 kg is relased on the track at a height of 4.46 m above the level surface. It slides down the track and makes a head on elastic collision with a block of mass 14.292 kg, initially at rest. The acceleration of gravity is 9.8 m/s^2. Calculate the height to which the block with mass 4.764 kg rises after rebounding from the collision.

Answer 1

I will assume that there is no friction in the system

First, in any collision momentum is conserved, so the center of mass velocity before the collision is equal to the center of mass after
or
m1v1i+m2v2i=m1v1f+m2v2f
as stated the v2i=0
also keep in mind that the velocities are vectors with magnitude and direction.

Since the collision is elastic, energy is also conserved so
.5m1v1i^2+.5m2v2i^2=
.5*m1*v1f^2+.5*m2*v2f^2

again, v1i=0

to calculate v1i, use conservation of energy for the vertical drop
m1g*h=.5*m1*v1i^2
m1*g*h=4.764*9.8*4.46
m1*g*h=208.225 J

v1i=sqrt(2*g*h)
v1i=sqrt(2*9.8*4.46)
v1i=9.35 m/s

plug in and solve the quadratic

then use conservation of energy to calculate
m*g*h2=.5*m1*v1f^2

j