Which Equation Best Describes the Block's Height After a Bullet Impact?

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Homework Help Overview

The discussion revolves around determining the appropriate equation to calculate the height a block reaches after being impacted by a bullet. The equations presented involve gravitational potential energy, kinetic energy, and spring potential energy, indicating a focus on energy conservation principles in a mechanics context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore which of the provided equations is suitable for calculating the block's height post-impact. There are attempts to justify the exclusion of certain equations based on perceived missing components. Questions arise regarding the definitions of variables and the correct interpretation of terms in the equations.

Discussion Status

The discussion is ongoing, with participants questioning the validity of different equations and seeking clarification on variable definitions. Some guidance has been offered regarding the interpretation of the equations, but no consensus has been reached on the best equation to use.

Contextual Notes

Some participants note that certain variables in the equations require definitions for clarity. There is also mention of potential confusion regarding the notation used for the spring constant and other variables.

Westin
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Homework Statement


Which equation could adequately be used to determine how high the block goes after being hit by the bullet (a height h)? (see figure) (m+M)gh+ksh=1/2(m+M)V^2
(m+M)gh+1/2ksh^2=1/2(m+M)V^2
(m+M)v+ksh=(m+M)V

Homework Equations


KE=1/2mv^2
Ugrav=Mgh
PE of spring = 1/2ks(s^2 final - s^2 initial)

The Attempt at a Solution



Based off of the equations, I believe the answer should be the second equation. I don't understand what else it could be..

I have only one attempt
 

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Westin said:

Homework Statement


Which equation could adequately be used to determine how high the block goes after being hit by the bullet (a height h)? (see figure)(m+M)gh+ksh=1/2(m+M)V^2
(m+M)gh+1/2ksh^2=1/2(m+M)V^2
(m+M)v+ksh=(m+M)V

Homework Equations


KE=1/2mv^2
Ugrav=Mgh
PE of spring = 1/2ks(s^2 final - s^2 initial)

The Attempt at a Solution



Based off of the equations, I believe the answer should be the second equation. I don't understand what else it could be..

I have only one attempt
Some of the variables in those equations need to be definrd for us.

I see L1, m, v, and h in the figure.

What are k, s, M, V, an s without k ? Is that ks rather than k⋅s ?
 
Maybe you could explain why you think it couldn't be the other two equations you excluded?
 
paisiello2 said:
Maybe you could explain why you think it couldn't be the other two equations you excluded?

Because the other two equations are missing components to their equations
 
SammyS said:
Some of the variables in those equations need to be definrd for us.

I see L1, m, v, and h in the figure.

What are k, s, M, V, an s without k ? Is that ks rather than k⋅s ?

1/2ks(s^2 final - s^2 initial)

ks = spring constant (k_s_)
(s^2 final - s^2 initial) = final stretch - initial stretch
v= velocity
m=mass1
M=mass2
 
Westin said:
Because the other two equations are missing components to their equations
And what components do these two equations have that are missing that the other one isn't?
 
paisiello2 said:
And what components do these two equations have that are missing that the other one isn't?
(m+M)gh+ksh=1/2(m+M)V^2 the spring constant should be 1/2ksh^2 not ksh
(m+M)gh+1/2ksh^2=1/2(m+M)V^2
(m+M)v+ksh=(m+M)V Ugrav should be (m+M)gh instead of velocity multiplying the masses (m+M)v, also (m+M)V is missing V^2
 
I think you have your answer then.
 

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