Answer:
(a) The speed of the lighter block is
.
The speed of the heavier block is
.
(b) The smaller block goes up to 0.69 m.
Explanation:
We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.
In order to solve this question, I will assume that the smaller mass is initially at rest.
Part A:
We will use conservation of energy.

This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.
Part B:
Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.

During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.

The following relation will be used when combining the two equations:

The following equation is useful for combining the two equations:

Therefore from the first equation,

Part C:
We will again use the conservation of energy to find the highest point that the mass can go:
