Question

"If a ball of mass M is dropped from a height h onto a spring with spring constant k (whose equilibrium positions is at height 0), compresses the spring an additional distance (L/2), and then rebounds, what height will the ball reach? Express your answer symbolically."

So I was thinking something along the lines of h= M*k*(L/2) but I'm not sure :/ Any guidance?

Answer 1

Answer:
`h=(kL^2)/(8mg)`

Explanation:

Given

Mass of ball is M is dropped from a height h

spring constant is k

Here we can use conservation of energy

Mass has a Potential energy of E=mgh

and when it falls on the spring then it compresses it a length of 0.5 L

Now potential energy is converted into Elastic potential energy and then spring unleashes this energy to the mass and again provide kinetic energy to mass to move upward .

In this way energy is conserved. So, mass must move to an initial height h

`mgh=(1)/(2)k((L)/(2))^2`

`h=(kL^2)/(8mg)`

considering
`h >> 0.5L`