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 guidiance?

Answer 1

h ’= k (h +L/2)²/ (2 M g )

Step-by-step explanation:

For this exercise, one of the best methods to solve it is with energy conservation.

Starting point. Lower, spring with maximum compression

Em₀ = Ke = ½ k x²

Final point. Higher after bounce

= U = M g h ’

Em₀ = Em_{f}

½ k x² = M g h’

the compressed distance is

x = h+ L / 2

where h is the distance that compresses the spring by the height where it comes from and L/2 the additional compression

h ’= ½ k x² / M g

we calculate

h ’= k (h +L/2)²/ (2 M g )

h' = (k/2Mg) h2 (1 + L/2h)2