Question

A 4.70 kg chunk of ice is sliding at 15.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 4.70 kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down.
Part A
How fast are they moving when they reach the valley floor again?

Answer 1

inertia before collision:
P = 70*15 + 4.7 * 0

Inertia after collision also =
P = (74.7)v
compute v

kinetic energy after collision = Ke = (1/2)(74.4)v^2, do not bother to compute because
at max distance up slope 74.4 g h = that Ke
and when it comes back down
Ke is that Ke again
and
v = -v

Answer 2

By the conservation of momentum
4.7*15+4.7*0=(4.7+4.7)*V
V=7.5
As there is no friction,
So no energy is lost there.
So the speed is the same, 7.5 m/s

P.S. due to inelastic collision, kinetic energy is not conserved here.