A 75g ball is dropped onto a spring that is fixed to the ground. The spring is 35cm long at equilibrium. (continued below)

Question

 If the ball is dropped from 1.5m above the spring, and the spring constant is 350 N/m, how much will the spring compress when the ball comes to a stop at maximum compression? (use the ground as your reference point)

Answer 1

TL;DR

m=mass of the ball, h is the height above the spring at rest length, L is the rest length of the spring, k is the spring constant, x is the maximal compression.

T₁=T₂

mg(h+L)=1/2 kx²+mg(L-x)

mgh+mgL=1/2 kx²+mgL-mgx

1/2 kx²-mgx-mgh=0

x=(mg/k)*[1+/-sqrt(1+2kh/(mg))]

x=0.081543m

Detailed answer:

The best approach to this problem will be conservation of energy. We can be sure that this approach is valid since (if we neglect air resistance) our only force are gravity and a spring, both of which are conservative forces.

So, we need to employ our techniques for solving energy conservation problems. That is, we need to identify two states in the problem where can deduce the form of the total energy, set the total energies of these two states equal to each other, and then solve for our unknown quantity. The two states for this problem are 1.) The moment the ball is released and allowed to fall, and 2.) The moment the spring is fully compressed. We'll tackle these in order.

First up is the moment the ball is released. At the exact moment it is dropped, the ball is momentarily at rest, as it has not yet at the chance to accumulate any velocity. Hence, the kinetic energy is 0. There are no springs acting on the ball at its release point, so the spring potential is 0 at this state as well. The only energy the ball has at its release, therefore, is gravitational potential energy. We're given the mass (m=0.075kg), and we always know g (g=9.81m/s²), so we just need the height. The height is measured relative to our origin, which the problem instructs us to use as the ground. The initial height of the ball then is its height above the spring (h=1.5m) plus the length of the spring (L=0.35m). Taken together, we have

T₁=mg(h+L)=mgh+mgL

We could put in numbers here to get a value for the total energy, but it's often preferable to wait until the end of a problem for that, if for no other reason than if you make a mistake, it's easier to correct later.

Now we need the total energy at the final state. The final state is the moment the spring is as compressed as its going to be, which is the moment it has stopped the ball and is about to push it back up. Since the ball is momentarily stopped, it once again does not have any kinetic energy. This time, however, there is a spring acting on it (with spring constant k=350 N/m), compressed by some unknown amount which I shall label x. Written out, the spring potential energy then is

1/2 kx²

But be careful, that's not the only energy at play! Because the origin is on the ground, and the ball is not on the ground, we still have a gravitational potential to deal with. This time, the height of the ball relative to the ground is the length of the compressed spring, which is the same as the uncompressed length MINUS the amount the spring has been compressed, L-x. Therefore the gravitational energy is

mg(L-x)=mgL-mgx

And so the total energy in state 2 is the sum of these

T₂=1/2 kx²+mgL-mgx

Because energy is conserved, the total energy in state 1 and state 2 are equal, and so we can write

T₁=T₂

mgh+mgL=1/2 kx²+mgL-mgx

Which we can now solve for x. Notice that there's an mgL on both sides which just cancels anyway; if we'd put in the numbers from the start, we'd have had to compute the same quantity twice and subtract it, adding the risk for an error for no reason!

mgh+mgL=1/2 kx²+mgL-mgx

mgh=1/2 kx²-mgx

1/2 kx²-mgx-mgh=0

This is a quadratic equation, and unfortunately the best approach here is probably going to be the quadratic formula. In the end you get

x=[mg+/-sqrt[(mg)²-4(1/2 k)(-mgh)] ] / [2*1/2 k]

x=[mg+/-sqrt[m²g²+2kmgh] ] / k

x=[mg+/-mg*sqrt[1+2kh/(mg)]] / k

x=(mg/k)* [1+/-sqrt[1+2kh/(mg)] ]

There are two solutions there, depending on whether you use the + or the - in +/-. However, if you take the - solution, then the x you get will be negative. Physically, this solution would suggest that the dropped mass hit the spring and caused it to extend, rather than compress. Obviously that wouldn't be a physical result, so we'll elect to keep only the + solution:

x=(mg/k)* [1+sqrt[1+2kh/(mg)]]=0.081543, or a bit more than 8 cm.