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:$h=\dfrac{kL^2}{8mg}$

Step-by-step 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

$mgh=\frac{1}{2}k(\frac{L}{2})^2$

$h=\dfrac{kL^2}{8mg}$

considering $h >> 0.5L$