Ablock of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. at the bottom, it strikes a block of mass m = 6.80 kg which is at rest on a horizontal surface, (assume an elastic collision, and ignore friction.)
(a) determine the speeds of the two blocks after the collision.
lighter block
m/s
heavier block
m/s

(b) determine how far back up the incline the smaller mass will go.
m

(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:

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come closer and become far apart and become steam/gas