A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0 ∘
incline which is 3.60 m high. At the bottom, it strikes a block of mass M=7.00M=7.00 kg which is at rest on a horizontal surface, (Assuming a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored, determine the speeds of the two blocks (v_mv
m

and v_Mv
M

) after the collision.

(moving toward the incline.)

(moving away from the incline.)

(Assumption: .)

Explanation:

If , the potential energy of the block of would be when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this block right before the collision would also be approximately .

Calculate the velocity of that based on its kinetic energy:

.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

.

.

Sum of the momentum of each block right before the collision: approximately .

Sum of the momentum of each block right after the collision: .

For momentum to conserve in this collision, and should ensure that .

Kinetic energy of the two blocks right before the collision: approximately and . Sum of these two values: approximately .

Sum of the energy of each block right after the collision:

.

Similarly, for kinetic energy to conserve in this collision, and should ensure that .

Combine to obtain two equations about and (given that whereas .)

.

Solve for and (ignore the root where .)

.

The collision flipped the sign of the velocity of the block. In other words, this block is moving backwards towards the incline after the collision.

this could be newton#'s law of cooling. the cooling curve is exponential