Block and the bullet

1. Apr 9, 2015

Westin

1. The problem statement, all variables and given/known data
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

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

3. 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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2. Apr 9, 2015

SammyS

Staff Emeritus
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 ?

3. Apr 9, 2015

paisiello2

Maybe you could explain why you think it couldn't be the other two equations you excluded?

4. Apr 9, 2015

Westin

Because the other two equations are missing components to their equations

5. Apr 9, 2015

Westin

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

6. Apr 9, 2015

paisiello2

And what components do these two equations have that are missing that the other one isn't?

7. Apr 9, 2015

Westin

(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

8. Apr 9, 2015