Conservation of Momentum problem

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The problem involves a bullet being fired into a block of wood, requiring the application of conservation of momentum to find the height the block will rise after the bullet embeds itself. The bullet's mass is converted from grams to kilograms for calculations, and the initial momentum equation is set up correctly. The final velocity of the block after the bullet embeds is calculated as -4.65 m/s, with a discussion on the sign indicating direction. To determine the height the block will rise, the kinetic energy of the combined system can be equated to gravitational potential energy. The conversation emphasizes understanding the transition from momentum to energy equations to solve for height.
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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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