Elastic Collision with a Spring Constant

AI Thread Summary
In an elastic collision involving a pinball and a lever, both with a mass of 80g, the lever is pulled back 2 cm by a spring with a spring constant of 1.4 N/cm. The initial calculations for the lever's velocity before impact result in 0.37 m/s. However, confusion arises because the equal masses seem to cancel out in the momentum equation, suggesting equal velocities post-collision. The problem solver is encouraged to convert units from grams to kilograms and centimeters to meters for accuracy. Further assistance is sought to clarify the correct approach to find the speed of the ball after the collision.
Drstevebrule
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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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