Putty dropped on a frame suspended from a spring.

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A 0.150 kg frame suspended from a spring stretches it by 0.050 m, and a 0.200 kg lump of putty is dropped onto the frame from 30.0 cm, resulting in a perfectly inelastic collision. The discussion focuses on finding the maximum downward distance the frame moves after the collision, involving conservation of momentum and energy equations. Users emphasize that while momentum is conserved during the collision, mechanical energy is not, complicating the calculations. The energy equation leads to a quadratic form, causing confusion when solving for the distance, with some attempts yielding imaginary or negative results. The conversation highlights the challenges of integrating momentum and energy concepts in this scenario.
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Homework Statement



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

The collision between the two is perfectly inelastic.

Find the maximum distance the frame moves downward from its initial position.

Homework Equations



\Delta{K} + \Delta{U} + \Delta{U_o} = 0

The Attempt at a Solution



I plugged in the energies into the above equations, and it didn't bring me closer to an answer. I'm not sure how I can relate momentum to energy in this, with a completely inelastic collision.

Any help is appreciated. Thank you.
 
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hello Jax988! welcome to PF!

remember that momentum is conserved. so find the common velocity of putty ad frame.

then with this velocity as 'v', and m = mass of putty + frame, find kinetic energy (initial)

then equate it with 1/2.k.x^2 + mgx
k = spring constant.

x is the required distance.
 
jax988 said:
I plugged in the energies into the above equations, and it didn't bring me closer to an answer. I'm not sure how I can relate momentum to energy in this, with a completely inelastic collision.

In inelastic collision mechanical energy is not conserved. So you cannot directly use energy conservation. But there is no harm in using it when collision is completed.
 
I guess my problem is the quadratic as well. If I carry out the energy equation after the collision, I have the distance x included. However, it takes the form of a quadratic equation.

When using the equation given.

\frac{1}{2}Mv_0^2 - Mgd + \frac{1}{2}kd^2 - \frac{1}{2}kx_f^2 = 0

Where M is the combined mass of the frame and putty, x_f is the initial stretch of the spring due to the weight of the frame, d is the max distance due to collision the spring stretches, and v_0 is the initial velocity.

So now d is the variable of a quadratic equation, and when I try to solve using the quadratic formula, I get imaginary numbers.

Edit: Attempted again with freshly made calculations. The results after plugging the numbers into the equation and attempting to solve with the quadratic formula resulted in real numbers, albeit negative.
 
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