Elastic Collision with a Spring Constant

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SUMMARY

The discussion focuses on calculating the speed of a pinball after an elastic collision with a lever in a pinball machine. The lever, with a mass of 80g, is propelled by a spring with a spring constant of 1.4 N/cm, compressed by 2 cm. Using the equations of motion, specifically the conservation of momentum and kinetic energy, the initial velocity of the lever is calculated to be 0.37 m/s. The equal masses of the lever and pinball complicate the momentum equation, leading to confusion regarding the final velocities post-collision.

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Homework Statement



A pinball with mass 80g is struck by a lever with mass 80g in a pinball machine in an elastic collision. The lever was pulled back 2 cm by a spring with spring constant k=1.4N/cm. What is the speed of the ball just after collision?

Homework Equations


F=kx

1/2kx^2=1/2mv^2

M1V1 = M2V2



The Attempt at a Solution



1/2 (2.8)(4)=5.6 5.6=1/2 (80) v^2 <-- trying to find velocity of lever just before contact

5.6/40 = v^2 V=.37

now this is where i get even more stuck, the masses of the ball and the lever are equal and they cancel out in the equation M1V1 = M2V2, which would leave the velocities as being equal. However, the answer choices are as follows:

A) 1.1 m/s
B) 1.4 m/s
C) 11 m/s
D) 14 m/s

If someone can shed some light on how to solve this problem it would be greatly appreciated.
 
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First thing I would look at is units. g to kg

centimeters to meters, I would change both and see how it works out.
 
i tried that, and i am still stumped
 

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