Clay falling onto a pan attached to a vertical spring

[ultimateman]

Homework Statement

"A spring is hung from the ceiling. A pan of mass 100.0 g is attached to the end, which causes it to stretch 5.00 cm. Find the maximum distance the pan moves downward when a lump of clay of mass 120.0 g is dropped from a height of 40.0 cm onto the pan."

The correct answer, according to my teacher is 0.233 meters.

Homework Equations

Conservation of momentum, perfectly inelastic collision.

m1v1i + m2v2i = (m1 + m2)vf

Conservation of mechanical energy.

MEi = MEf

PEg = mgh
PEe = 1/2 kx^2
KE = 1/2 mv^2

Work kinetic energy theorem.

Fnet d cosθ = 1/2 m (vf^2 - vi^2)

The Attempt at a Solution

I tried a perfectly inelastic conservation of momentum + work-KE theorem approach.

m(ball)v0 + 0 = M(combined)v(final)

Fnet d cos(θ) = 1/2 M(vf^2 - vi^2)

where Fnet was the average spring force minus gravity, F = [k(0.05) + k(x)] / 2 - M(combined)g

d was (x+0.05)

and vf = 0, vi = root(2gh)

UPDATE:

OK so I tried conservation of mechanical energy after the collision to no avail. I did:

1/2 (.22kg) (1.53 m/s)^2 + (.22 kg)(9.81) x = 1/2 (19.62 N/m) (x+0.05)^2

Solving for x I got x = 0.225 m.

I still can't get x = 0.233m. I'm sure it's not a rounding error of some sort. What am I missing?

 

[Doc Al]

I have tried using a conservation of energy approach. This equation looked like:

PEg(ball) + PEg(pan) + PEe(spring) = PEe(spring, (x+0.05))

Mechanical energy is not conserved during the collision, so that won't work.

I also tried a perfectly inelastic conservation of momentum + work-KE theorem approach.

m(ball)v0 + 0 = M(combined)v(final)

So far so good.

Fnet d cos(θ) = 1/2 M(vf^2 - vi^2)

where Fnet was the average spring force, F = [k(0.05) + k(x)] / 2

d was x

and vf = 0, vi = root(2gh)

You forgot about gravity.

Hint: After the collision, mechanical energy is conserved.

 

[ultimateman]

Actually I forgot to include gravity in my attempted solution post, but I did include it in my actual calculations already and I still got the wrong answer. : / But Fnet was the average spring force and gravity.

Thanks for pointing out that mechanical energy is not conserved during the collision. Obviously I should have seen that with it being perfectly inelastic.

Trying conservation of mechanical energy after the collision.

EDIT: OK so I tried conservation of mechanical energy after the collision to no avail. I did:

1/2 (.22kg) (1.53 m/s)^2 + (.22 kg)(9.81) x = 1/2 (19.62 N/m) (x+0.05)^2

Solving for x I got x = 0.225 m.

I still can't get x = 0.233m. I'm sure it's not a rounding error of some sort. What am I missing?

 

[Doc Al]

EDIT: OK so I tried conservation of mechanical energy after the collision to no avail. I did:

1/2 (.22kg) (1.53 m/s)^2 + (.22 kg)(9.81) x = 1/2 (19.62 N/m) (x+0.05)^2

Solving for x I got x = 0.225 m.

I still can't get x = 0.233m. I'm sure it's not a rounding error of some sort. What am I missing?

The left side of your equation is incomplete. You forgot the spring potential energy.

 

[ultimateman]

The left side of your equation is incomplete. You forgot the spring potential energy.

Derp...Ty!