Solving for Maximum Spring Length With a Dropped Clay Ball

[maki]

A massless pan hangs from a spring that is suspended from the ceiling. When empty, the pan is 50cm below the ceiling. If a 100g clay ball is placed gently on the pan, the pan hangs 60cm below the ceiling. Suppose the clay ball is dropped from the ceiling onto an empty pan. What is the pan's distance from the ceiling when the spring reaches its maximum length?

I figured I'd start by finding the final velocity of the ball using the gravitational potential energy.
Ugi=mgy
Ugi=(0.100kg)(9.8m/s)(0.5m)
Ugi=0.49 J
Ugf=0 J
Vf=sqrt[(-2 * (Ugf - Ugi)) / m]
Vf=sqrt[(-2 * -0.49) / 0.5)]
Vf=sqrt(1.96)
Vf=1.4 m/s

Next, I found the spring constant (k) using the stationary weight of the ball in the pan.
Fsp=mg
Fsp=(0.1)(9.8)=.98 N
Fsp=k*(Delta S, or change in spring length)
.98 N=k*0.1
k=9.8

Now I'm stuck, can't seem to find the right formula to use to get any farther, I should have just the one more step I think. If someone could post a formula or explain where I need to go next I would appreciate it. I have a lab in 15 minutes, but I'll be back after that to work on this some more.

Thanks (=

 

[Päällikkö]

Potential energy of a spring:
$$W = \frac{1}{2}kx^2$$

 

[quicknote]

Thank you for sharing your work and thought process so far. To solve for the maximum length of the spring, we need to use the concept of conservation of energy. The energy of the system (spring, pan, and ball) remains constant throughout the motion, so we can equate the initial energy to the final energy.

Let's start by finding the initial energy of the system when the ball is dropped onto the pan. At the point of impact, the ball has a kinetic energy of 1/2mv^2, where m is the mass of the ball and v is its final velocity (which you have already calculated). The pan, being massless, does not contribute to the system's energy. Therefore, the initial energy of the system is just the kinetic energy of the ball, which is 1/2(0.1kg)(1.4m/s)^2 = 0.098 J.

Now, let's find the final energy of the system when the spring reaches its maximum length. At this point, the ball has zero velocity, so its kinetic energy is zero. The spring, however, has potential energy due to its stretched length. The potential energy of a spring is given by 1/2kx^2, where k is the spring constant and x is the displacement from its equilibrium position. In this case, the displacement of the spring is the maximum length it reaches, and we'll call it x_max. Therefore, the final energy of the system is 1/2kx_max^2.

Since the initial and final energies are equal, we can set them equal to each other and solve for x_max:

0.098 J = 1/2(9.8 N/m)(x_max)^2

x_max = sqrt(0.098 J / 4.9 N/m) = 0.2 m

Therefore, the maximum length of the spring is 0.2 m, which means the pan will be 0.2 m + 0.5 m = 0.7 m from the ceiling when the spring reaches its maximum length.

I hope this helps and good luck with your lab!