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

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The discussion centers on identifying the correct equation to calculate the height a block reaches after being struck by a bullet. The second equation, (m+M)gh + 1/2ksh^2 = 1/2(m+M)V^2, is favored for its inclusion of gravitational potential energy and spring potential energy. Participants highlight that the other equations lack essential components, such as the correct representation of the spring constant and gravitational energy. Clarifications are requested regarding the definitions of variables like k, s, M, and V, emphasizing the need for precise terminology. Ultimately, the consensus suggests that the second equation is the most comprehensive for solving the problem.
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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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