What is the maximum spring compression in a bullet and block collision?

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Homework Help Overview

The problem involves a collision between a projectile and a block, focusing on the maximum compression of a spring attached to the block. The subject area includes concepts from momentum conservation and energy transfer in mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply momentum conservation to find the final velocity of the block after the collision. Some participants suggest using energy conservation to relate kinetic energy to spring potential energy, while others question the setup of the equations and whether all relevant factors have been considered.

Discussion Status

The discussion is ongoing, with participants providing guidance on the relationships between momentum and energy. There is an exploration of the equations set up by the original poster, and some participants express uncertainty about the correctness of these equations and the inclusion of all necessary factors.

Contextual Notes

There is mention of the bullet's mass needing to be included in the calculations, indicating potential missing information in the original setup. The problem is constrained by the requirement to consider energy and momentum principles without providing a complete solution.

huskydc
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A projectile of mass 20 g has an initial horizontal velocity of 100 m/s when it hits and stops in a wood block of mass 0.403 kg. The block is sitting on a horizontal frictionless surface and is attached to a massless spring, initially relaxed, with spring constant 148 N/m. What is the maximum compression of the spring?

since momentum is conserved, i used the equation
mv(initial) = mv (final), and found the final velocity of the block, as a result of the mass hitting it, is 4.963,

but I'm not sure where to go from here? ..or am i heading the right direction ?
 
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You have the velocity of the block. Now you can use energy to solve for the compression of the spring. (KE = .5mv^2; Es=.5kx^2)
You're on the right track. Just try to think of the relationships between the various quantities (often there is more than one, i.e. velocity is related to momentum, but it is also related to energy)
 
ok, I've set up the equations as follow:

.5kx^2 = KE (in this case, also KE final)
thus,

.5kx^2 = .5m(4.963^2)

and solve for x... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?
 
huskydc said:
ok, I've set up the equations as follow:

.5kx^2 = KE (in this case, also KE final)
thus,

.5kx^2 = .5m(4.963^2)

and solve for x... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?
Make sure you include the mass of the bullet since it embeds in the block.
 

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