Conservation of Momentum problem

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SUMMARY

The discussion focuses on a conservation of momentum problem involving a bullet and a block of wood. A 21.0g bullet is fired vertically at a speed of 310m/s into a 1.40kg block of wood at rest. The initial momentum equation m1v1 + m2v2 = m1v11 + m2v12 is applied, resulting in a final velocity of -4.65m/s for the block and bullet system. To find the maximum height the block will rise, kinetic energy principles are utilized, specifically the conversion of kinetic energy to gravitational potential energy.

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  • Ability to convert units (grams to kilograms)
  • Familiarity with basic algebra for solving equations
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Homework Statement



A gun is fired vertically into a 1.40kg block of wood at rest directly above it. If the bullet has a mass of 21.0g and a speed of 310m/s, how high will the block rise into the air after the bullet becomes imbedded into it?

Homework Equations


m1v1+m2v2= m1v11+ m2v12


The Attempt at a Solution


So I first converted 21.0 grams into .021kg because I know you use kg for this equation.
After substituing in a few things this is as far as I have gotten:

m1v1+m2v2= m1v11+ m2v12
(.021)(0)+(1.40)(0)=(.021)(310)+(1.40)(V)
V= -4.65m/s as the final velocity of the block

so from here I have a few questions first did I do this part correctly and from here how would I proceed in order to determine how high the block will go?
 
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Not sure where your - sign is from, but for positive y up, 4.6 m/s looks about right.
(Your equations look like there are typos.)

Now you know your mv for the combined block/bullet and that Kinetic energy = ½mv² , and that energy will eventually be consumed by gravity with mgh which is the height you are looking for.
 
Thank you ever so much! That makes so much sense!
 

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