Conservation of Energy: Kinetic to Elastic Potential

[awelex]

Homework Statement

A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.050m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0m.

mass of frame: mf
mass of putty: mp
height: h

Homework Equations

The Attempt at a Solution

I tried the following, which was wrong:

0.150 * 9.8 = k * 0.050 --> k = 29.4

velocity of putty when it touches frame:
0.200*g*0.3=(1/2)*0.200*v^2 --> v^2 = 2hg

(1/2)(mp+mf)*v^2 = (1/2)*k*x^2
--> x = 0.2m

So the final answer is either 0.2m or 0.25m (depending on whether one adds the initial 0.05m). Both answers are wrong, though. What am I doing wrong?

Thanks.

 

[vela]

Homework Statement

A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.050m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0m.

mass of frame: mf
mass of putty: mp
height: h

Homework Equations

The Attempt at a Solution

I tried the following, which was wrong:

0.150 * 9.8 = k * 0.050 --> k = 29.4

velocity of putty when it touches frame:
0.200*g*0.3=(1/2)*0.200*v^2 --> v^2 = 2hg

You're fine up to here. I take it that h=30 cm, not 30 m like you wrote above.

(1/2)(mp+mf)*v^2 = (1/2)*k*x^2
--> x = 0.2m

So the final answer is either 0.2m or 0.25m (depending on whether one adds the initial 0.05m). Both answers are wrong, though. What am I doing wrong?

Thanks.

You're assuming that both the frame and the putty are initially moving with the speed the putty had just before hitting the frame. So right before the collision, the putty has a certain amount of kinetic energy, and immediately after the collision, it still has that amount of kinetic energy because it still has speed v. But now the frame is moving too, so there's even more kinetic energy. Does that make sense?

You need to find the speed of the putty-frame combination immediately after the collision. If you use that speed in your calculations instead, you should get the right answer.