Block move down the curved hill and hit attached block spring

AI Thread Summary
A 100-gram block moving at 2 m/s collides elastically with a stationary 400-gram block attached to a spring with a spring constant of 500 N/m. To solve for the maximum spring compression and the height the 100-gram block reaches post-collision, conservation of energy principles and momentum equations should be applied. Users are encouraged to provide a diagram and show their work for more effective assistance. The discussion emphasizes the importance of understanding the equations of motion and the relationship between force and spring compression. Engaging with the community and sharing progress will facilitate quicker solutions.
cecico
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A 100grams block is moving at 2m/s down the curved hill. The block slides along the smooth surface and collides elastically with the 400gram block. The 400gram block is initially at rest and is attached to an ideal spring with spring constant of 500N/m
a)Find the maximum distance the spring compresses
b)Find the maximum height the 100gram block reaches after the collision.

I have no idea with this problem...someone help me...
 
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Conservation of energy.
Give that a go - let us know what happens.
 
In order to receive help with this problem I suggest that you

a) Provide a diagram
b) Show the work you have tried thus far. Forum rules.
 
dont know for sure but collisions usually go by the equation

m1u1 + m2u2 = m1v1 +m2v2

m1 being mass 1 = 0.1kg
m2 being 0.4 kg

u1 being =2 m/s
u2 = 0 m/s as its stationary

then work out the momentum of the second block as this would be equal to the force on the spring which would follow the equation

F = kx

where k is 500 N/m and thus u have the distance x

and then its pray someone with more brains than me gives u an answer otherwise ul b stuck on it all night long...lol...sorry mate
 
how to post diagram?
 
cecico said:
how to post diagram?

Scanner, photograph, MS Paint. Free account at Photobucket.com will host images.
I know it's a little work... but it will get your question answered much quicker.
 
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