A block of mass, m, is pushed up against a spring, compressing it a distance x, and is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case?

x₂ =6*x₁

Explanation:

In absence of friction, along a horizontal surface, all the elastic potential energy stored in the spring becomes kinetic energy, so we can write the following expression for the first case:

       

In the second case, we have a mass 4m, which reaches to a speed 3v due to the energy stored in the spring.We can write the same equation than for (1) replacing m by 4m and v by 3v, as follows:

       

Dividing (2) over (1) on both sides, and rearranging terms, we get:

       

Taking square roots on both sides, we get:x₂ = 6*x₁So, in this case, the spring was compressed six times the distance that was compressed in the first case.

The objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.