What are the speeds and compression in a collision with a spring?

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AI Thread Summary
The discussion focuses on a physics problem involving a 100 g block attached to a spring and a 20 g ball colliding with it. For a perfectly elastic collision, participants emphasize using conservation of momentum and kinetic energy to determine the ball's speed post-collision. The maximum compression of the spring requires consideration of spring potential energy, which is independent of the block's mass. In the case of a perfectly inelastic collision, similar principles apply, but the final velocities will differ. The conversation highlights the importance of distinguishing between kinetic energy and potential energy in solving the problem.
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Homework Statement


A 100 g block on a frictionless table is firmly attached to one end of a spring with k = 20 N/m. The other end of the spring is anchored to the wall. A 20 g ball is thrown horiontally toward the block with a speed of 5.0 m/s.
a. If the collision is perfectly elastic, what is the ball's speed immediately after the collision?
b. What is the maximum compression of the spring?
c. Repeat parts a and b for the case of a perfectly inelastic collision.

Homework Equations


Us=.5k(delta s)2
K = .5mv2

The Attempt at a Solution


a. I'm thinking i have to use momentum of the ball.. but because of the spring, i guess i have to use energy equations..
Ki+Ui=Kf+Uf
but for some reason for the Ui, there is no place for me to put in the mass of the block since that is potential energy..do i need a mass?

i've found a thread that's the same..nm
https://www.physicsforums.com/showt...ed+direction+of+each+ball+after+the+collision
 
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loststudent0 said:
a. I'm thinking i have to use momentum of the ball.. but because of the spring, i guess i have to use energy equations..
Ki+Ui=Kf+Uf
To find the speed of the block after the collision, you'll need to apply both conservation of momentum and kinetic energy. For part (b) you'll need to consider spring potential energy.
but for some reason for the Ui, there is no place for me to put in the mass of the block since that is potential energy..do i need a mass?
Presumably the table is horizontal, so there are no changes in gravitational PE. The only PE you need to worry about is spring PE, which does not depend on mass.
 
Thank you for you guidance Doc Al..i think i understand now
 

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